18.305 lecture notes: power tools for de nite integrals...

18.305 Lecture Notes:

Power Tools for Definite Integrals, Part 1:

Asymptotic Approximations

Homer Reid

November 12, 2015

Contents

1 Motivation 31.1 Asymptotic approximations . . . . . . . . . . . . . . . . . . . . . 51.2 Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Elementary methods 72.1 Series expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Integrands with rapidly decaying exponential factors: Watson’sLemma and Laplace’s method 103.1 Integrands containing e−xt . . . . . . . . . . . . . . . . . . . . . . 103.2 Generalization: Integrands containing exf(t) . . . . . . . . . . . . 113.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Integrands with rapidly oscillating sinusoidal factors: Elemen-tary methods, Wick rotation, and stationary phase 164.1 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . 164.2 Change of variables . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Wick rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.4 Method of stationary phase . . . . . . . . . . . . . . . . . . . . . 224.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5 The method of steepest descent 245.1 Identifying steepest-descent contours . . . . . . . . . . . . . . . . 255.2 Applications to Fourier integrals: Saddle points . . . . . . . . . . 28

6 When functions defined by definite integrals aren’t functions 32

1

18.305 Lecture Notes 2

7 Auxiliary variables: Adding complexity in the service of findingsimplicity 337.1 Feynman parameters . . . . . . . . . . . . . . . . . . . . . . . . . 337.2 Schwinger parameters . . . . . . . . . . . . . . . . . . . . . . . . 347.3 Hubbard-Stratonovich Transformation . . . . . . . . . . . . . . . 35

A Integral representations of special functions 36A.1 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36A.2 Modified Bessel functions . . . . . . . . . . . . . . . . . . . . . . 36

B Proof that∫C2eixt

p

dt→ 0 as R→∞ for p > 1 37

C Delta sequences 38

D The curious case of the sinc integrals 39


1 Motivation

As engineers, scientists, and applied mathematicians, none of us actually needmuch motivation for learning tricks for evaluating complicated integrals: weencounter functions defined by nontrivial definite-integral expressions in ourwork all day every day, and we hardly need somebody to explain to us how niceit would be to have at-the-ready intuition for the behavior of these functions.

Nonetheless, for the sake of completeness, let’s tick off just a few of the manyscenarios in which simple tools for estimating integrals can make the differencebetween rapid research progress and confused stumbling about in the dark.

Transform integrals. The Fourier transform of a function f(x) is

f(ω) =1

2πe−ikxf(x) dx.

The Fourier transforms of simple functions may be computed using high-school techniques; for example, the Fourier transform of f(x) = e−α|x| is

f(ω) = απ(α2+k2) . But for more complicated functions we must resort to

approximate (or numerical) techniques. For example, consider the follow-ing C∞ bump function:

B(x) =

0, x < −1

e− α

1−x2 , −1 < x < 1

0, x > −1.

What1 is the Fourier transform of B(x)?

Special functions. Many functions that arise frequently in mathematics, sci-ence, and engineering are defined by definite integral expressions; examplesinclude the Gamma function and the exponential integral function:

Γ(z) ≡∫ ∞

0

tz−1e−t dt

E1(z) ≡∫ ∞x

e−t

tdt

The task of evaluating special-function integrals is not an abstract aca-demic one; many numerical techniques for PDEs and other problems relyheavily on accurate evaluation of special-function integrals of this type.

1This question is interesting because the possible behavior of B(k) as |k| → ∞ is squeezedinto a narrow range of possibilities by two bounds coming from the Paley-Wiener theorem ofFourier analysis. First, because B(x) is C∞ (the function and its derivatives of all orders exist

everywhere), the magnitude of its Fourier transform |B(k)| must decay more rapidly than anypower of k for large |k|. However, because B(x) is not analytic (it has no Taylor expansionaround x = ±1), its Fourier transform must decay less rapidly than e−α|k| for any constant

α. To identify the particular function B(k) that somehow manages to wend its way into thisnarrow strait of possible behaviors for large k is a task for the methods of these notes.


Definite-integral expressions for solutions to ODEs. In other cases, func-tions that are defined as the solutions to certain ODEs may be expressedas definite integrals; examples include the following expressions for theBessel and Airy functions:

Jν(z) =1

inπ

∫ π

0

eix cos θ cos(nθ) dθ

Ai(z) =1

2πi

∫C

et3

x −zt dt

The methods discussed in these notes thus offer a complement to themethods we considered earlier in the course for identifying the asymptoticbehavior of ODE solutions.

Definite-integral expressions for solutions to PDEs. The general solutionof Laplace’s equation ∇2φ in 3D cylindrical coordinates (r, ϕ, z) reads, forproblems with cylindrical symmetry,

φ(r, ϕ, z) =

∫ ∞0

A(k)J0(kρ)e±kz dk

where A(k) is a function determined by boundary conditions. As just oneexample, the potential in the vicinity of an infinitely thin disc of radius Rlying in the z = 0 plane and centered at the origin is2

φ(ρ, z) = V0 ·2

π

∫ ∞0

e−k|z|J0(kρ)sin kR

kdk (1)

Once we have determined this function, we have an expression for thepotential everywhere in space—but how do we evaluate this expression?

In passing we note that an alternative but equivalent way to write thegeneral Laplace solution is

φ(r, ϕ, z) =

∫ ∞0

A(k)I0(kρ)e±ikz dk

where I0 is now a modified Bessel function.

Definite-integral expressions for Green’s functions. In scalar quantum fieldtheory in four-dimensional Minkowski spacetime, the vacuum expectationvalue of the correlation function for the scalar field between points withspacetime coordinates x1 = (x1, t1) and x2 = (x2, t2) reads

⟨0∣∣Φ(x1)Φ(x2)

∣∣0⟩ =

∫dp

(2π)3

e−i[p0(t1−t2)−p·(x1−x2)]

2p0, p0 ≡

√p2 +m2

2I find (1) to be an amazing expression. For z = 0 it somehow manages to achieve theconstant value φ ≡ V0 for all ρ ≤ R, but then for ρ > R suddenly ceases to be constant andbecomes an ordinarily wiggly varying function! How does it do that?!


where m is the mass of the scalar field. When the spatial distance betweenthe points |x2−x1| is large compared to their temporal displacement t2−t1,the points are causally disconnected (no signal traveling at or below thespeed of light can connect them), so we expect the fields to be uncorrelated.Is that expectation born out by the above expression?

1.1 Asymptotic approximations

In many cases of interest it is impossible to evaluate the integral in question ex-actly, and we must content ourselves with approximate solutions. In these noteswe will be interested in a certain type of approximation known as an asymptoticapproximation. Let Iexact(x) be a function defined by an integral expressioninvolving a parameter x. Then Iasymp(x) is an asymptotic approximation toIexact as x→ x0 if the condition

limx→x0

Iasymp(x)

Iexact(x)= 1 (2)

is satisfied. Basically, an asymptotic approximation picks off the dominant termin the exact solution as x→ x0, ignoring subleading terms. For example, the Γfunction defined above has the following asymptotic approximations near x = 0and near x =∞ : Γasymp(x) =

1

x, x→ 0

Γasymp(x) = x→∞.

Of course, there is no need to stop at just the leading term; the differencebetween Iexact and Iasymp(x) defines the subleading term in the exact function,and we can look for an asymptotic approximation to that:

limx→x0

Isubleading(x)

Iexact(x)− Iasymp(x)= 1 (3)

and so on.

1.2 Methods

We will discuss a succession of methods, of increasing sophistication and power,for obtaining asymptotic approximations to definite integrals.

(a) In many cases we can pick off the leading terms in asymptotic approxima-tions by applying methods of high-school calculus, including integrationby parts and series expansions.

(b) In cases involving rapidly decaying exponential factors whose overall decayrate is proportional to a parameter x, we can obtain asymptotic approxi-mations in the limit x→∞ by retaining only contributions to the integralfrom regions in the immediate vicinity of certain points (namely, the pointsat which the argument of the decaying exponential attains its maximumvalue, i.e. minimal attenuation).


(c) In cases involving rapidly oscillating sinusoidal factors whose overall oscilla-tion frequency is proportional to a parameter x, we can obtain asymptoticapproximations in the limit x → ∞ by playing games similar to thoseof item (b), but with some novel new twists that make the method lessstraightforward (and less powerful) than that case.

(d) The methods of cases (b) and (c) are special cases of a comprehensive andpowerful method called the method of steepest descent.


2 Elementary methods

In some cases, methods of high-school calculus suffice to determine useful asymp-totic approximations to functions defined by definite integrals. Here are someexamples. (Many of the examples in these notes come from the text of Benderand Orszag referenced on the course website.)

2.1 Series expansions

Consider the function

I(x) =

∫ x

0

e−t4

dt (4)

If we are looking for an asymptotic approximation in the limit x → 0, we cansimply go like this:

=

∫ x

0

(1− t4 +

1

2t8 + · · ·

)dt

= x− x5

5+x9

18+ · · · (5)

So that was easy.Another example of a series expansion is the behavior of the Γ function near

x = 0:

Γ(x) =

∫ ∞0

tx−1e−t dt

Integrate by parts:

=1

x

∣∣txe−t∣∣∞0︸︷︷︸

0

+1

x

∫ ∞0

txe−t dt︸︷︷︸Γ(x+1)

The boundary term vanishes, while the integral in the second term is just Γ(x+1), so we have derived the functional relationship xΓ(x) = Γ(x+ 1).

To evaluate the integral here in the limit of small x, we go like this:

tx = ex log t ≈ 1 + x log t+1

2x2[log t]2 + · · ·

so the integral reads

Γ(x) =1

x

∫ ∞0

txe−t dt

=1

x

[∫ ∞0

e−t dt

]︸︷︷︸

1

−[∫ ∞

0

(log t)e−t dt

]︸︷︷︸

γ

+x

2

[∫ ∞0

(log t)2e−t dt

]︸︷︷︸

γ2+π2

6

+ · · ·


where γ ≈ 0.577 is Euler’s constant. Thus we have found an asymptotic ap-proximation to Γ(x) near x = 0:

Γ(x) ≈ 1

x− γ +

x

2

(γ2 +

π2

6

)+ · · ·

2.2 Integration by parts

Now suppose we want an asymptotic approximation to the function (4) functionas x → ∞. The series (5) converges for all x, but it is not useful for large x.Instead, we can go like this:

I(x) =

∫ x

0

e−t4

dt

=

∫ ∞0

e−t4

dt︸︷︷︸Γ( 5

4 )

−∫ ∞x

e−t4

dt (6)

To estimate the second term on the LHS, we might consider integrating by parts.Recall that the integration by parts formula reads, in schematic form,∫

(fg′) = |fg| −∫

(f ′g)

To apply this to the integral (6), we need two functions f(t) and g′(t) whose

product is e−t4

. it is tempting to take f = e−t4

and g′ = 1. This yields∫ ∞x

e−t4

dt =∣∣∣te−t4∣∣∣∞

x︸︷︷︸xe−x4

+4

∫ ∞0

t3e−t4

dt

Integrating by parts again in the second term yields

= xe−x4

+ x4e−x4

+ 4

∫ ∞0

t7e−x4

dt

and so on. We could keep playing this game all day, and it would never be usefulfor determining the large-x behavior of our function, because each successiveterm that is picked off by the integration by parts is larger than the last forlarge x! We’re trying to find a series of terms in which each successive term issmaller than the previous one. So that wasn’t a very happy outcome.

However, we have better luck if we instead go back and reverse the choice off and g′ functions in the integration-by-parts of (6)—that is, if we make e−t

4

look like the g′ factor instead of the f factor. Of course, there is no function g(t)

whose derivative is g′(t) = e−t4

[if there were, we could just evaluate the integral(6) directly], but that is actually no deterrent—we just choose g(t) to be any

function whose derivative involves e−t4

times other factors, and then we choose


f(t) to be the inverse of those other factors so that the product fg′ = e−t4

. Achoice that works is

g(t) = e−t4

, g′ = −4t3e−t4

, f(t) = − 1

4t3,

and with respect to this choice the integration by parts reads∫ ∞x

e−t4

dt = −∣∣∣∣ 1

4t3e−t

4

∣∣∣∣∞x︸︷︷︸

14x3

e−x4

+3

4

∫ ∞x

t−4e−t4

dt

Integrating by parts in the second term yields

=1

4x3e−x

4

− 3

16x7e−x

4

+21

16

∫ ∞x

t−8e−t4

dt

Success! Applying integration by parts in the correct direction has picked off thefirst two terms in the asymptotic approximation to our function, and we couldkeep playing this game all day to obtain an increasingly accurate approximation.

Going back to our original problem (6), we have found∫ x

0

et4

dtx→∞−−−−−→ Γ

(5

4

)− 1

4x3e−x

4

+3

16x7e−x

4

+O(x−11)

Summary

To summarize the results of this section, we have found that

• Elementary series-expansion methods suffice to determine the small-x be-havior of Γ(x) and

∫ x0e−t

4

dt.

• Elementary integration by parts suffices to determine the large-x behaviorof∫ x

0e−t

4

dt.

However, as it turns out, neither of the elementary methods we have consid-ered suffices to determine the large-x behavior of Γ(x); for that we must turnto more sophisticated methods.


3 Integrands with rapidly decaying exponentialfactors: Watson’s Lemma and Laplace’s method

3.1 Integrands containing e−xt

There exists a very simple method for determining the asymptotic behavior offunctions containing rapidly decaying exponential factors. To illustrate the idea,we start by evaluating a high-school integral: for any nonnegative integer p,∫ ∞

0

tpe−xt dt =p!

xp+1(p ∈ N)

Actually, by replacing p! with Γ(p + 1) we obtain a result that is valid for anyvalue of p, including non-integer powers, but with the restriction that p > −1(for p = −1 the integral diverges):∫ ∞

0

tpe−xt dt =Γ(p+ 1)

xp+1, p ∈ R, p > −1 (7)

Now suppose g(t) is a function whose Taylor polynomial around t = 0 has aninfinite radius of convergence and consider the function

I(x) =

∫ ∞0

g(t)e−xt dt (8)

Insert the Taylor series of g:

I(x) =

∫ ∞0

[g0 + g1t+ g2t

2 + · · ·]e−xt dt

Integrate each term using (7):

=g0

x+g1

x2+

2g2

x3+ · · ·+ Γ(n+ 1)gn

xn+1+ · · · (9)

where gn ≡ dngdxn

∣∣∣x=0

. So that identifies (9) as the asymptotic expansion of (8)near x→∞.

What happens if we replace the upper limit of the integral in (8) with somefinite number b? Answer: Nothing! The reason is that changing the upperlimit from ∞ to b changes the value of the integral by a quantity that decaysexponentially with x∫ b

0

tpe−xt dt =

∫ ∞0

tpe−xt dt+ terms decaying like e−xb

Since we are looking for an asymptotic approximation in the limit x → ∞, wecan simply neglect the error terms here. In fact, we can take the upper limit bas small as we like and we will still get the same asymptotic approximation.


Among other things, this means that we can relax the condition that theTaylor polynomial of g(t) at t = 0 has an infinite radius of convergence; theabove method will work as long as g(t) may be represented near t = 0 by a sumof powers of t, i.e. g ∼

∑p∈P gpt

p for some set P of powers, as long as the most

negative power pmin satisfies pmin > −1. Thus the full statement of this methodis

if I(x) =

∫ b

0

g(t)e−xt dt with g(t) =∑p∈P

gptp (10)

then Iasymp(x)→∑p∈P

Γ(p+ 1)gpxp+1

as x→∞ (independent of b)

(11)

This set of ideas for evaluating integrals involving the factor e−xt is known asWatson’s Lemma.

3.2 Generalization: Integrands containing exf(t)

It is easy to generalize the method outlined above to cases involving more com-plicated arguments of the exponential. Thus, consider the function of x definedby

I(x) =

∫ tb

ta

exf(t)g(t) dt (12)

where f(t), g(t) are some functions of t. (In the case we considered above,f(t) = −t and [ta, tb] = [0, b].)

The take-home message of the previous section is that, in the x→∞ limit,the integral will dominated by the contribution coming from the portion of theinterval [ta, tb] at which the function f(t) attains its maximum value. For thecase f(t) = −t and the interval [0, b] considered above, f(t) attains its maximumvalue at the left endpoint of the interval, whereupon we were able to obtain theasymptotic expansion of the integral by considering only the behavior of g(t)near t = 0. For a more general function and a more general interval, there arethree possibilities:

1. f(t) attains its maximum value at the left endpoint of the interval ta.

2. f(t) attains its maximum value at the right endpoint of the interval tb.

3. f(t) attains its maximum value at some point t0 lying in the interior ofthe interval.

Case 1 The procedure to follow when f attains its maximum value at ta is toTaylor-expand f around t = ta:

f(t) = fa + f ′a(t− ta) + · · ·


where f ′a < 0 is negative. In what follows I will use the shorthand notationα ≡ −f ′a, so that α > 0. The integral (12) becomes, upon changing variables tou = α(t− ta) and putting b ≡ α(tb − ta),

I(x) = exfa∫ ub

0

e−xug(ta +

u

α

) [g0 + αug1 + α2u2g2 + · · ·

]α−1du

Replace the upper limit by ∞, incurring an error that is suppressed by theexponential factor e−xb, and replace g by its series expansion near ta, i.e. g(t) =∑p gp(t− ta)p:

= exfa∫ ∞

0

e−xu

[∑p

(uα

)pgp

]α−1du

= exfa∑ Γ(p+ 1)gp

(αx)p+1(13)

Equation (13) defines the asymptotic approximation to (12) as x→∞ for caseswhere f(t) attains is maximum at the left endpoint of the interval.

Case 2 The procedure to follow when f attains its maximum value at the rightendpoint is similar to the previous case with only the obvious replacements.

Case 3 When f attains its maximum value at an interior point t0, the Taylorexpansion of f around t0 will look like

f(t) = f0 +1

2f ′′0 (t− t0)2 + · · · (14)

where the linear term is missing (because we are at an extremum) and wheref ′′b < 0 is negative (because the extremum is a maximum). In what follows Iwill use the shorthand notation β ≡ −f ′′0 , so that β > 0. The integral (12)becomes, upon changing variables to u = (t− t0),

I(x) = exf0∫ ub

−uae−

12βxu

2

g (t0 + u) du

with ua,b ≡ (ta,b − t0). Replace the upper and lower limits by ±∞ and replaceg by its series expansion near t0:

= exf0∫ ∞−∞

e−12βxu

2

[g0 + g1u+

1

2g2u

2 + · · ·]du (15)


with gp ≡ dp

dtp g∣∣t=t0

. These are just ordinary Gaussian integrals, which we

evaluate using standard methods3 to find

I(x)x→∞−−−−−→ exf0

[g0

√2π

βx+

1

2g2

√2π

(βx)3+O

(x−5/2

)](16)

(Note that the linear term in the series expansion of g around t0 makes nocontribution.) Equation (16) defines the asymptotic approximation to (12) asx→∞ for cases where f(t) attains is maximum in the interior of the interval.

Some notes

• In our discussion of the case in which f attains its maximum at an endpointof the interval, we assumed that f ′ 6= 0 at that point. If in fact we havef ′ = 0 there then the right formula to use is (15), but with one or the otherof the integration limits replaced by 0. In this case, the final result (15)is modified in two ways: (a) It acquires an overall factor of 1

2 . (b) Thelinear term in the series expansion of g(t) now does contribute, becausewe don’t get the cancellation from the positive and negative lobes.

• If f(t) attains the same maximum value multiple times within the intervalof integration, then you have to include contributions from each one ofthese maxima. An example of such a situation is the integral

I(x) =

∫ π/2

−π/2

1

(t+ 5)2e−x cos t dt

which receives contributions from regions near ±π2 . These contributionsare not equal, but they are of the same asymptotic order of magnitudewith respect to x and thus must both be retained in an asymptotic ap-proximation.

• On the other hand, if f has one or more local maxima within the intervalof lesser value than the global maximum—say, a point at which f attainsa local maximum value that is a quantity ∆ less than the global maximumof f over the interval—then the contribution of the local maximum will besuppressed by a factor of e−x∆ relative to the contribution of the globalmaximum and thus irrelevant for an asymptotic approximation as x→∞.

• The expressions for the subleading term in equation (16) may not be theend of the story: if there is a third-derivative term in (14) then we needto account for that as well. This will be discussed in the examples below.

3We have∫ ∞−∞

e−σt2dt =

√π

σ,

∫ ∞−∞

te−σt2dt = 0,

∫ ∞−∞

t2e−σt2dt =

√π

4σ3


3.3 Examples

Modified Bessel function The modified Bessel function Iν(x) has the inte-gral expression (Appendix A)

Iν(x) =1

π

∫ π

0

ex cos t cos(νt) dt

This is of the form (12) with the function f(t) = cos t, which attains its maxi-mum value at the left endpoint of the interval and has the expansion

f(t) = 1− 1

2t2 +O(t4)

near that point. Following the discussion of Section 3.2, in the limit x→∞ weexpect the integral to be dominated by the contribution from t near 0, so weapproximate

Iν(x)x→∞−−−−−→ ex

π

∫ ∞0

e−12xt

2

[1− 1

2(νt)2 +

1

24(νt)4 + · · ·

]dt

=ex√2πx

[1− ν2

x+ · · ·

]Note that, compared to our work in Section 3.2, the integrals acquire a factor of12 because the Gaussian integrals are evaluated over the interval [0,∞] insteadof [−∞,∞].

Gamma function We noted previously that elementary methods fail to de-termine the large-x behavior of the Γ function:

Γ(x) =

∫ ∞0

tx−1e−t dt.

To make this look like an integral of the form (12), I go like this:

=

∫ ∞0

e(x−1) log t−t dt

This is of the form (12) with g(t) = 1 and f(t) = (x− 1) log t− t.Proceeding as per the discussion above, I now look for maxima of f(t) over

the interval [0,∞]:

df

dt

∣∣∣∣t=t0

= 0 =⇒ t0 = x− 1

The difficulty with this result is that the location of the minimum depends onthe parameter x, which makes things unwieldy. To scale out this dependence


and obtain a variable with respect to which the minimum occurs at a constantx-independent place, I define s = t

x , whereupon the integral becomes

Γ(x) = xx∫ ∞

0

e−x(s−log s) ds

s

which is of the form (12) with f = −(s − log s), g = 1s . The maximum of f is

f0 = −1 at the point s0 = 1, where we find f ′′0 = −1. The series expansion of garound this point is g(s) = g0 + g1(s− 1) + g2(s− 2)2 + · · · with {g0, g1, g2} ={1,−1, 1.} Then equation (16) reads

Γ(x)x→∞−−−−−→ xxe−x

√2π

x+O

(x−3/2

)


4 Integrands with rapidly oscillating sinusoidalfactors: Elementary methods, Wick rotation,and stationary phase

In this section we consider asymptotic approximations to functions of the form

I(x) =

∫ b

a

eixf(t)g(t) dt (17)

in the limit x → ∞. The basic intuition is that, in this limit, the exponentialfactor oscillates rapidly over intervals of the t axis over which g(t) is nearlyconstant, ensuring large cancellations of the integrand which make the integralvanish as x→∞.

For the simple case f(t) = t, which arises in evaluating Fourier transforms,this intuition is quantified by a result known as the Riemann-Lebesgue Lemma,which says that if g(t) is absolutely integrable over the interval [a, b] then thefunction (17) with f(t) = t vanishes as x→∞:

if

∫ b

a

|g(t)| dt <∞ then limx→∞

∫ b

a

eixtg(t) dt = 0.

However, this lemma doesn’t tell us how rapidly the function I(x) tends toinfinity as x increases. In this section we will consider a number of tricks forquantifying the rate of this vanishing.

4.1 Integration by parts

In many cases it is possible to pick off the leading terms in the asymptoticbehavior of (17) by the simple method of integration by parts. The idea is toarrange things in such a way that each successive integration by parts bringsdown a factor of x in the denominator, so that the resulting sequence of termsgives the leading, subleading, etc. terms in the x→∞ limit.

For example, consider

I(x) =

∫ 1

0

e−ixt

1 + tdt. (18)

Integrate by parts:

I(x) = − 1

ix

∣∣∣∣e−ixt1 + t

∣∣∣∣10

− 1

ix

∫ 1

0

e−ixt

(1 + t)2dt

=1

ix

(1− 1

2e−ix

)− 1

ix

∫ 1

0

e−ixt

(1 + t)2dt


Integrate by parts again:

=1

ix

(1− 1

2e−ix

)+

1

(ix)2

∣∣∣∣ e−ixt

(1 + t)2

∣∣∣∣+2

(ix)2

∫e−ikx

(1 + t)3dt

=1

ix

(1− 1

2e−ix

)− 1

(ix)2

(1− e−ix

4

)+

2

(ix)2

∫e−ikx

(1 + t)3dt

and so on. I can keep playing this game all day to pick off as many terms as Ilike in the x→∞ asymptotic expansion of (18).

More generally, if the oscillatory factor is of the form eixf(t), then integrationby parts brings down factors of xf ′(t) into the denominator. In this case wecan continue to play the above game unless the function f ′(t) vanishes at somepoint in the interval in question. In cases of this sort we must resort to morecomplicated tricks, as discussed below.

4.2 Change of variables

Another essentially elementary technique that is often useful is to identify achange of variables that exposes some structure we can exploit to obtain anapproximate integral evaluation in some asymptotic limit.

Here’s an example: Consider the function

I(x) =

∫ ∞0

f(t)sinxt

xtdt (19)

for large values of x. The integrand here may be thought of as the functionsinc(xt) ≡ sin xt

xt with a locally-varying amplitude (or “envelope”) given by f(t)[see plots in Figure 1(a)]. The idea for approximating this integral in the large-xlimit is the following: In the limit of large x—where “large” is defined relativeto the characteristic lengthscale over which the function f(t) varies—the sincfunction lives out almost its entire useful life and decays to zero before thefunction f(t) has had a chance to start changing much from its value at t = 0. Toexpose this behavior more explicitly, we change variables to u = xt, whereuponthe integral becomes

I(x) =1

x

∫ ∞0

f(ux

) sinu

udu. (20)

The two factors in the integrand of (20) are plotted vs. u in Figure 1(b) forvarious values of x. In terms of the variable u, the curve of the sinc factor isindependent of x, while the effect of larger and larger x is to render the factorf(ux

)more and more slowly varying over the effective life of the sinc factor. In

the limit x → ∞, this factor remains fixed at the value f(0) over the entire u


-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

y

x

(a)

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

y

x

(b)

Figure 1: (a): Integrand of (19) plotted as a function of the variable t. (b):The two factors in the integrand of (20) plotted as a function of the new variableu = xt. The idea is that, for large x, the sinc function lives almost its entirelife before the function f

(ux

)changes appreciably from its value at u = 0, so we

can approximate f(ux

)= 0 and pull this function out of the integral. [In this

figure we are assuming f(0) = 1.]


axis, suggesting the approximation

I(x)x→∞−−−−−→ 1

x

[f(0) +

(ux

)f ′(0) +

1

2

(ux

)2

f ′′(0) + · · ·]

sinu

udu (21)

=f(0)

x

∫ ∞0

sinu

udu︸︷︷︸

π2

+ subleading (22)

(I will evaluate the subleading terms later in these notes.)To summarize, I have found

limx→∞

∫ ∞0

f(t)

(sinxt

t

)dt =

π

2f(0). (23)

One way to write this is to say

limx→∞

2

π

sinxt

t= δ(t). (24)

Higher-order terms

Here’s a generalization of the above method that makes it easier to pick offhigher-order terms in the asymptotic approximation of (19). Consider the func-tion defined by (19) but now with an exponentially-decaying factor thrown intothe integrand:

I(x) =

∫ ∞0

f(t)e−ytsinxt

xtdt (25)

As above, change variables to u = xt and expand the quantity f(ux

)in a Taylor

series around u = 0:

=1

x

∫ ∞0

[f(0) +

(ux

)f ′(0) +

1

2

(ux

)f ′′(0) + · · ·

]e−

yxu

sinu

udu (26)

=∞∑p=0

f (p)(0)

p!xp+1Ip(x, y) (27)

where I defined

Ip(x, y) ≡∫up−1e−

yxu sinu du.

These integrals may be evaluated in closed form; the first few read

I0(x, y) = atan

(x

y

)I1(x, y) =

x2

x2 + y2

I2(x, y) =2x3y

(x2 + y2)2


In the limit y → 0 one finds

limy→0

Ip(x, u) =

π

2, p = 0

1, p = 1

0, p = 2

−2, p = 3

and thus one finds

limx→∞

∫ ∞0

f(t)sinxt

xtdt =

π

2xf(0) +

1

x2f ′(0)− 2

x4f ′′′(0) + · · ·

We can think of the higher-order terms here as finite-x corrections to equation(24):

2

π

sinxt

t= δ(t)− 2

πxδ′(t) +

4

πx3δ′′′(t) +O

(1

x5

)where the nth derivative of a δ function is a gadget that, when integrated againsta function f(t), picks out nth derivatives of f at t = 0 (with a plus or minussign that you can determine by formally integrating by parts).

Integrals involving Bessel functions

The Bessel function J0(r) is qualitatively similar to the sinc function—it oscil-lates and decays for large r—so a game similar to the game we played abovealso works to evaluate integrals of the form∫ ∞

0

f(t)e−ytJ0(xt) dt.

The large-x expansion of this integral is given by (27), except that now the Ipintegrals are defined by

Ip(x, y) ≡∫upe−

yxuJ(u) du

These integrals may be evaluated in closed form; the first few read

I0(x, y) =x√

x2 + y2

I1(x, y) =x2y

(x2 + y2)3/2

I2(x, y) =2x3y2 − x5

(x2 + y2)5/2


Figure 2: Contour in complex z plane used to evaluate the integral in equation(28).

4.3 Wick rotation

Before moving on we pause to cover one piece of computational technology thatwe will use in the following section. This is the evaluation of integrals of theform

Ip(x) ≡∫ ∞

0

eixtp

dt. (28)

If the argument of the exponential here were −xtp, then I could just evaluate

the integral directly to obtain4 1x1/pΓ

(p+1p

). This suggests interpreting (28) as

a contour integral over one segment of a closed contour in the complex t plane,as shown in Figure (2):

Ip(x) ≡∫C1

eixtp

dt

4 ∫ ∞0

e−xtpdt

u=xtp−−−−→1

x1/p·

1

p

∫ ∞0

u1p−1e−u du︸︷︷︸

Γ(

1p

)=

1

x1/pΓ

(p+ 1

p

).


where C1, C2, C3 are the contours parameterized by

C1 : t = u, u ∈ [0, R]

C2 : t = Reiu, u ∈ [0, θ]

C3 : t = ζu, u ∈ [R, 0].

Here R is a large real number which I will eventually take to ∞; also, ζ = eiθ isa complex number of unit magnitude which I choose in such a way as to ensurethat the argument of the exponential in the integrand is a negative real numberon C3:

ixtp = ix(ζu)p = −xup =⇒ ζp = i =⇒ ζ = eiπ2p .

In other words: Whatever the given value of p in (28), I choose the contour C3

to consist of a straight line running through the origin and making an angle ofπ2p radians with the Re t axis.

Because the integrand of (28) is analytic (no poles), its integral over anyclosed contour vanishes, and thus∫

C1+C2+C3

eixtp

dt = 0 (29)

For any p > 1, the integral over C2 vanishes in the limit R → ∞. (Thiscalculation is done explicitly in Appendix B.) Thus the integral over C1 equalsminus the integral over C3, or

Ip(x) =

∫C1

eixtp

dt = −∫C3

eixtp

dt = ζ︸︷︷︸=e

πi2p

∫ ∞0

e−xup

du = eπi2pΓ

(p+ 1

p

)1

x1/p.

(30)

4.4 Method of stationary phase

Armed with the result (30), we can now compute the leading terms in theasymptotic approximations of oscillatory integrals. Consider an integral of theform

I(x) =

∫ tb

ta

eixf(t)g(t) dt. (31)

In the limit x → ∞ we expect the rapid oscillations of the sinusoidal factor in(31) to ensure vanishingly small contributions to the integral from most regionsof the t axis, with the exception being regions near stationary points of f [pointst0 such that f ′(t0) = 0]; in these regions the sinusoidal factor accumulates phaseat a relatively slow rate compare to everywhere else, whereupon we expect thecontributions of these regions to dominate the integral in the large x limit.

We will here consider the case in which f has a stationary point at oneof the endpoints of the integration interval, say ta. (More general cases in


which the stationary point lies in the interior of the interval may be handledsimply by splitting the integral into two portions, each having an endpoint atthe stationary point.) In the vicinity of ta we write

f(t) ≈ fa −β

2(t− ta)2 + · · ·

where f0 = f(ta) and β = |f ′′(ta)| are the value and magnitude of the secondderivative of f at ta. Making this approximation in the integral yields

I(x)x→∞−−−−→ eixfa

∫ tb

ta

e−i12xβ(t−ta)2

[g(ta) +O(t− ta)

]dt (32)

Use (21) with p = 2:

= g(ta)ei(fax+π4 )√

π

2βx. (33)

Equation (33) gives the leading term in the asymptotic approximation of (17)as x→∞ for the particular case in which f has a single stationary point at anendpoint of the interval.

4.5 Examples

Bessel functions The Bessel function of order 0 has the integral expression

J0(x) =1

π

∫ π

0

eix cos t dt.

Equation (33) applies here with fa = 1, β = 12 , g(t) = 1, yielding the contribu-

tion [equation (33)]

eix+π4

√1

πx

However, this contribution from the left endpoint of the integration interval isonly half the story; in this case the right endpoint of the integration interval(t = π) is also a stationary point, and adding its contribution yields the fullasymptotic expression for J0(x):

J0(x)x→∞−−−−→

√2

πxsin(x+

π

4

)


5 The method of steepest descent

The method of stationary phase suffices to pick off the leading x→∞ behaviorof oscillatory integrals of the form (31), but it is an unwieldy method of obtain-ing subleading corrections. This is one way of motivating the more powerfultechnique known as the method of steepest descent. This is a generalization ofthe technique we used above to evaluate

∫∞0eixt

p

dt. As we did in that case,the general idea is to interpret an integral like (31) as a contour integral over acontour C consisting of a (possibly infinite) line segment lying on the real axisin the complex t plane. We (a) make a clever choice for how to extend the

open contour C into a closed contour C + C (where C may consist of multiple

segments, as it did in Section 4.3), (b) evaluate the contour integral over C,

then (c) use Cauchy’s integral formula to relate the contour integral over C tothe contour integral over C, which is our ultimate goal. (If the integrand isanalytic then this relation will look like

∫C

+∫C

= 0, while if the integrand haspoles inside the closed contour then we will get residues on the RHS.)

How should we choose the contour C? In Section 4.3 we chose C in such away as to ensure that the argument of the exponential was purely real-valuedon C, whereupon the oscillatory exponential factor in the original integral wastransmogrified into a decaying exponential factor. More generally, we can expectsimilar success using any contour on which the argument of the exponential hasconstant (not necessarily zero) imaginary part on C, for then we can simply pullthat portion out of the integral.

To make this more precise, let’s introduce notation for the real and imaginaryparts of the function of t that enters the argument of the exponential:

I(x) =

∫ b

a

ex[U(t)+iV (t)] dt (34)

=

∫C

exF (t) dt (35)

where C is the contour in the complex t plane running from a to b on the realaxis. Now suppose we extend C into a closed contour C + C and choose C insuch a way that the imaginary part of F (t) is constant along C, i.e. V (t) ≡ V0

for all t ∈ C. Then I can write simply∫C

exF (t) dt = eixV0

∫C

exU(t) dt.

Now the integrand involves only an exponential with a real-valued argument, soin the limit x → ∞ I can apply Laplace’s method and pick off arbitrary manyterms in the asymptotic approximation by zeroing in on the neighborhood ofwherever U(t) obtain its maximum.

There is another way to interpret this game of identifying contours alongwhich the imaginary part of F (t) is constant. Suppose F (t) is analytic. Thismeans that it satisfies the Cauchy-Riemann equations; putting t = α+ iβ these


read∂U

∂α=∂V

∂β,

∂U

∂β= −∂V

∂α

which I can express as the vanishing of the dot product of two vectors:∂U

∂α

∂U

∂β

T

︸︷︷︸∇U

·

∂V

∂α

∂V

∂β

︸︷︷︸∇V

The two vectors in question here are just the two-dimensional gradients of thescalar functions U and V , considered as functions of two real variables α, β; thusan analytic function satisfies the property that the gradient of its real part isorthogonal to the gradient of its imaginary part. In particular, suppose I am ata point t = α+ iβ and I identify a direction dt = dα+ idβ in which I can movewhile keeping V (t) constant, i.e. ∇V (t)dt=0. That means that dt points in the(positive or negative) direction of the gradient of U(t), and is thus the uniquedirection in which I can move to maximize the change in U .

Thus identifying contours on which the imaginary part of the exponent isconstant is equivalent to identifying contours on which the real part of theexponent is maximally rapidly varying, i.e. maximally steeply ascending anddescending at t varies. This is the origin of the term method of steepest descent.

5.1 Identifying steepest-descent contours

As an example of the practice of identifying steepest-descent contours, let’srevisit the integral we evaluated in Section 4.3, but now over a finite interval.Thus consider

I(x) =

∫ b

0

eixt2

dt (36)

=

∫C1

eixt2

dt (37)

where C1 is the segment [0 : b] of the real axis. The argument of the exponentialhere is of the form (35) with F (t) = it2 = −2αβ + i(α2 − β2). The contours onwhich the imaginary part of F is constant are those for which

α2 − β2 = constant. (38)

The contour C3 in Section (4.3) satisfies this constant-phase condition (it is thecontour α = β). We can use this contour as part of our solution to the problemof how to complete C1 in (37) into a closed contour; this contour joins with C1 atthe point t = 0. To complete the story, we need another contour satisfying theconstant-phase condition (38) that joins with C1 at its other endpoint, which is


the point t = α + iβ = b, i.e. (α, β) = (b, 0). This point lies on the constant-phase contour C4 defined by the equation

α2 − β2 = b2

and a parameterization of this contour is

t =√b2 + s2 + is, dt =

(s√

b2 + s2+ i

)ds, 0 ≤ s ≤ ∞.

Figure 3: Closed contour for the evaluation of∫ 1

0eixt

2

dt.

Note that C4 meets C3 at s =∞, so C1 + C4 + C3 is a closed contour over


which the full integral of our analytic function vanishes:∫C1+C3+C4

eixt2

dt = 0 =⇒ I(x) =

∫C1

eixt2

dt =

∫C3

eixt2

dt︸︷︷︸eiπ4√

π4x

−∫C4

eixt2

dt

(The integral over C3 was evaluated in Section 4.3.) On the contour C4 we have

ixt2 = ixb2 − 2s√b2 + s2

(note the constant imaginary part, as ordered) and the integral over this contourthen reads ∫

C4

eixt2

dt = eixb2

∫ ∞0

(i+

s√b2 + s2

)e−2xs

√b2+s2 ds︸︷︷︸

I(x)

The integral here is a Laplace integral for which we can easily obtain all termsin the asymptotic expansion by retaining contributions from the region of inte-gration near the maximum of the argument of the exponential at s = 0:

I(x) =

∫ ∞0

exf(s)g(s) ds

with

f(s) = −2xsb

√1 +

s2

b2

= −2xbs

[1 +

s2

2b2+O(s4)

]g(s) = i+

s√b2 + s2

= i+s

b− s3

2b3+O(s5)

where we have used the series expansions of f and g about the maximum of fat s = 0. The asymptotic expansion of I(x) is then

I(x) =

∫ ∞0

e−2xbs

[1−

(xs3

b

)+O(s5)

] [i+

s

b+O(s3)

]ds =

i

2xb+

1

4b3x2+O(x−4)

Putting it all together, I find

I(x) =

∫ b

0

eixt2

dtx→∞−−−−−→ e

iπ4

√π

4x︸︷︷︸T0

− ieixb2

2xb︸︷︷︸T1

− eixb2

4b3x2︸︷︷︸T2

+ · · · (39)

Figure 4 compares the exact value of the modulus of I(x) (as computed usingnumerical integration) and the asymptotic approximations obtained by keeping


the first 1, 2, and 3 term on the RHS of (39). The first term is what we wouldhave obtained using the method of stationary phase; it correctly captures thedecay rate, but misses the wiggles that arise from interference between the termsin the full asymptotic approximation.

0.1

1

0.1 1 10 100

y

x

Figure 4: The function I(x) defined by (39) as evaluated by numerical quadra-ture (solid yellow line) and as approximated by the first 1, 2, and 3 terms in theasymptotic approximation.

5.2 Applications to Fourier integrals: Saddle points

One useful application of the method of steepest descent is to the asymptoticevaluation of Fourier transforms. As a motivating problem, consider the half-Lorentian pulse function

P (x) =

{0, x ≤ 0

e−κx, x > 0

whose Fourier transform is

P (k) =1

2π

∫ ∞−∞

e−ikxP (x) dx =1

2πi(k + κ).

Note that |P (k)| decays algebraically with k (like |k|−1 for large k) as requiredby the Paley-Wiener theorem for the discontinuous function P (x).


Now suppose we smooth out the discontinuity at x = 0 by defining a moresophisticated C∞ pulse function that ramps up smootly for x > 0 in a way thatensures infinite differentiability:

P(x) =

{0, x ≤ 0

e−γx e−κx, x > 0.

The Fourier transform of P(x) is

P(k) =1

2π

∫ ∞−∞

e−ikxP(x) dx

=1

2π

∫ ∞0

e−ikx−1x−x dx

or

P(k) = =1

2π

∫C

eF (x) dx (40)

where C is the positive real axis and

F (x) = U(x) + iV (x) = −ikx− 1

x− x. (41)

The saddle-point method

I will evaluate (40) in the k → ∞ limit using a version of the steepest-descentmethod known as the saddle-point method. The rationale goes like this:

• For reasons described above, I want to extend the contour C to a closedcontour C + C with the property that F (x) has constant imaginary part

on C, i.e. dVdx = 0 on C.

• Having done this, I will evaluate the integral over C by retaining onlycontributions from regions of C near points x0 that maximize U(t)—that

is, points5 x0 ∈ C at which dUdx = 0.

• But this identifies x0 as a point at which the derivatives of both the realand imaginary parts of F vanish—that is, the derivative of the full functionF (x) vanishes at x0, i.e.

F ′(x0) = 0.

These points are called saddle points.6

5I say “points” here in case there are multiple local maxima at which U attains the samemaximum value. In many cases there will only be one point x0 to consider.

6More generally, thinking in terms of multivariable calculus, saddle points are points atwhich the gradient of a multivariable function vanishes and the second derivatives with respectto the multiple variables do not all have the same sign. In ordinary calculus over R2 there areother types of points at which the gradient vanishes, namely maxima (all second derivatives


• In the vicinity of a saddle point t0, I can then write

F (x) = F (x0) +1

2F ′′(x0)(x− x2

0) + · · ·

and evaluate the integral using Laplace’s method.

Saddle-point evaluation of Fourier transform

To apply the saddle-point method to the evaluation of (40), I first look for saddlepoints of F (x):

F ′(x0) = −ik +1

x20

− 1 = 0 =⇒ x0 = ±√

1

1 + ik

k→∞−−−−→ ±e−iπ4 k−1/2.

I will take the + sign here to get a point in the lower right quadrant, x0 =√2k2 (1− i). At this value of x0 I have (keeping only leading terms in the k →∞

limit)

F (x0) = −2e−iπ4

√k

= −√

2k + i√

2k

F ′′(x) = −2e−3iπ8 k−3/2

= −2(C1 + iS1)k−3/2

where C1 = cos 3π8 , S1 = − sin 3π

8 . I then approximate the integral (40) by itscontribution near x = x0:

B(k) =e−√

2kei√

2k

π

∫ ∞−∞

e−C1k−3/2s2 e−iS1k

−3/2s2︸︷︷︸1+O(s2)

ds

It is tricky to keep track of all the various factors here, but the key point is thatwe have a Gaussian integral with variance k−3/2, which will integrate to yieldfactor of k−3/4 :

∼ e−√

2kk−3/4.

The following plot shows the exact (numerical) value of |P(k)| together with thefirst and second factors in the asymptotic approximation obtained by saddle-point integration.

negative) or minima (all second derivatives positive). However, in calculus over C an analyticfunction cannot have maxima or minima in the interior of a domain, so interior points atwhich F ′(x0) = 0 must be saddle points.Why can’t analytic functions have maxima or minima in the interior of a domain? For the samereason that no point in the interior of a drum or trampoline can be higher (or lower) than thehighest (or lowest) point on its boundary—the real and imaginary parts of analytic functions,like the surface heights of drums and trampolines, satisfy Laplace’s equation ∇2U = ∇2V = 0,and Laplace solutions always attain their extrema at boundary points.


1e-05

0.0001

0.001

0.01

0.1

1

10

0.1 1 10

y

x

Figure 5: Comparison of exact (numerical) and asymptotic approximations to

|P(k)|


6 When functions defined by definite integralsaren’t functions

Consider the following two-variable functions defined by definite integral expres-sions:

I1(x, y) =

∫ ∞0

tJν(xt)Jν(yt) dt (42a)

(where Jν is the Bessel function of order ν)

I2(x, y) =

∫ ∞−∞

Ai(t− x)Ai(t− y) dt (42b)

where Ai(t) is the Airy function. What would you do if you an integral of thissort arose in your research? Upon discovering that You might try to obtainexpressions for these functions using the analytical tricks we have discussed forevaluating integrals, or you might try to evaluate these functions numerically—but either way, you’ll be sorry! The difficulty is that the functions defined by(42) are actually not integrals at all, but rather distributions; more specifically,we have

I1(x, y) =δ(x− y)

xI2(x, y) = δ(x− y).

You should think of (42) as analogues of the familiar integral

I3(x, y) =1

2π

∫ ∞−∞

ei(x−y)t dt (42c)

which arises in Fourier analysis. Equations (42) are really completeness rela-tions expressing the completeness of the set of eigenfunctions of an eigenvalueproblem, i.e. they should be interpreted as expressions of the form∑

n

un(x)u∗n(y) = δ(x− y) (43)

where the un are the eigenfunctions of a Sturm-Liouville ODE boundary-valueproblem defined over an interval of width L, considered in the limit L→∞ sothat the discrete sum in (43) goes over to an integral.


7 Auxiliary variables: Adding complexity in theservice of finding simplicity

We are all familiar with examples of definite integrals in which the integranddepends in a simple way on one or more parameters, but the final integraldepends in a complicated way on these parameters.

Turning this observation on its head, a function F (x1, · · ·xn) = F (x) thatdepend in complicated ways on its input variables x can sometimes be usefullyexpressed as an integral over one or more new variables, i.e.

F (x) =

∫f(x,α)dα

where the integrand f depends in a simple way on the x arguments. The αvariables here are called auxiliary variables.

If we are trying to evaluate an integral whose integrand involves F, it issometimes convenient to express this integrand as an integral over auxiliaryvariables, then interchange the order of integration and perform the integralover the original variables, leaving behind an integral over only the auxiliaryvariables—in other words, to go like this:∫

F (x) dx =

∫ ∫f(x,α)dα dx =

∫ ∫f(x,α)dx︸︷︷︸f(α)

dα =

∫f(α)dα

where f is some new function of the α. There are a few different variants ofthis procedure.

7.1 Feynman parameters

Feynman parameters are a type of auxiliary variable that are useful when thedenominator of an integrand involves a product of complicated factors. Theidea is to rewrite the product as the average of a certain weighted sum of thefactors, where the averaging is over the weights. (The weights are the auxiliaryvariables.)

The simplest version is based on the following identity:

1

Q1Q2=

∫ 1

0

dαdα

[αA+ (1− α)B]2(44)

which we could write in the alternative form

=

∫ 1

0

dα1

∫ 1

0

dα2δ(α1 + α2 − 1)dα

[α1Q1 + α2Q2]2(45)

More generally, (45) may be extended to express denominators involving a prod-uct of N factors, each raised to arbitrary powers, as an N -dimensional integral


over auxiliary variables:

1

Qp11 Qp22 · · ·Q

pNN

=

∫∆

J(p;α)dα

[α1Q1 + α2Q2 + · · ·+ αNQN ]P(46)

where

P =

N∑n=1

pn, J =

[Γ(p1 + p2 + · · ·+ pN )

Γ(p1)Γ(p2) · · ·Γ(pN )

]xp1−1

1 xp2−12 · · ·xpN−1

N ,

and where ∆ denotes the unit N -dimensional simplex; this is the unit trianglefor N = 2, the unit tetrahedron for N = 3, etc, i.e.∫

∆

dα =

∫ 1

0

dα1

∫ 1

0

dα2 · · ·∫ 1

0

dαN δ

(1−

N∑n=1

αn

)

7.2 Schwinger parameters

Schwinger parameters are based on identities like

1

Q=

∫ ∞0

e−Qαdα (47a)

1√Q

=2√π

∫ ∞0

e−Qα2

dα (47b)

Here’s an an example of a four-dimensional integral that reduces to a one-dimensional integral by clever application of Schwinger parameters:

Problem: Find the electrostatic potential at one vertex of a four-dimensional box of dimensions L1 × L2 × L3 × L4 throughoutwhich exists a constant unit-strength charge density.

φ =

∫ L1

0

dx1

∫ L2

0

dx2

∫ L3

0

dx3

∫ L4

0

dx41√

x21 + x2

2 + x23 + x2

4

(48)

Use equation (47b):

=2√π

∫ L1

0

dx1

∫ L2

0

dx2

∫ L3

0

dx3

∫ L4

0

dx4

∫ ∞0

e−α2(x2

1+x22+x2

3+x24)dα

The introduction of the Schwinger parameter decouples factors involving eachxi variable, so the xi integrals may now done separately:

=2√π

∫ ∞0

[∫ L1

0

e−α2x2

1 dx1

]︸︷︷︸

√π2 erf(αL1)

[∫ L2

0

e−α2x2

2 dx2

]︸︷︷︸

√π2 erf(αL2)

[∫ L3

0

e−α2x2

3 dx3

]︸︷︷︸

√π2 erf(αL3)

[∫ L4

0

e−α2x2

4 dx4

]︸︷︷︸

√π2 erf(αL4)

dα


My final expression for the potential reads

φ =π3/2

8

∫ ∞0

erf(αL1) erf(αL2) erf(αL3) erf(αL4)

α4dα (49)

where erf is the usual error function:

erf(x) ≡ 2√π

∫ x

0

e−t2

dt. (50)

Although this procedure has not yielded a closed-form expression, the integral(49) is significantly easier to handle than the original integral (48), for tworeasons:

• It is a one-dimensional integral instead of a four-dimensional integral.

• It is a nonsingular integral, in contrast to (49), which has a singularity atthe origin. [To see this, note that erf(x) ∼ x as x → 0, so the integrandof (49) tends smoothly to a finite constant as α→ 0.]

These improvements make the calculation vastly more tractable for numericaltreatments.

7.3 Hubbard-Stratonovich Transformation


A Integral representations of special functions

A.1 Bessel functions

The usual Bessel functions {Jν , Yν} are defined as the two linearly independentsolutions of the ODE[

d2

dx2+

1

x

d

dx+

(1− ν2

x2

)]{Jν(x)Yν(x)

}= 0

where the two functions are distinguished by the condition that

Jν(x) = regular at x = 0.

An integral expression for Jν is

Jν(x) =1

inπ

∫ π

0

eix cos θ cos(nθ) dθ

Alternative expressions for J and Y are

Jν(x) =2

π

∫ ∞0

sin(x cosh t− νπ

2

)cosh(νt) dt

Yν(x) = − 2

π

∫ ∞0

cos(x cosh t− νπ

2

)cosh(νt) dt

A.2 Modified Bessel functions

The modified Bessel functions {Iν ,Kν} are defined as the two linearly indepen-dent solutions of the ODE[

d2

dx2+

1

x

d

dx−(

1 +ν2

x2

)]{Iν(x)Kν(x)

}= 0

where the two functions are distinguished by the condition

Iν(x) = finite at x = 0.

The functions I and K are just the usual Bessel functions evaluated at imaginaryarguments: {

IνKν

}(x) =

{JνYν

}(ix)

Integral expressions are

Iν(x) =1

π

∫ π

0

ez cos θ cos(νθ) dθ

Kν(x) =

∫ π

0

e−z cosh t cos(νt) dt


B Proof that∫C2

eixtp

dt→ 0 as R→∞ for p > 1

The integral over C2 reads∫C2

eixtp

dt = iR

∫ π2p

0

eixRpeipu+iu du (51)

(note the Jacobian factor dt = iReiudu.). Change variables to s = pu:

=iR

p

∫ π2

0

eixRpeiseis/pds (52)

Write eis = cos s+ i sin s in the exponent:

=iR

p

∫ π2

0

ei[xRp cos s−s/p]e−xR

p sin s ds. (53)

In the limitR→∞, this integral—which involves a rapidly decaying exponential—may be analyzed using the methods discussed earlier in these notes. The max-imum value of the argument of the exponential (−xRp sin s) occurs at the leftendpoint, s = 0; expanding around that point yields

=iReixR

p

p

∫ π2

0

e−xRps

[1− is

p+O(s2)

] [1 +O(s3)

]ds (54)

→ ieixRp

pxRp−1+O

(1

R2p−1

)(55)

so the integral vanishes as R→∞ for any p > 1.


C Delta sequences

A delta sequence is a sequence of functions that tends toward a delta functionin a certain limit. Examples include

δ(x) = limL→∞

1

πL

sin(xL

)(xL

)1

Le−|x|/L

1

L√πe−x

2/L2

1

2π

∫ L

−Leiλxdλ


D The curious case of the sinc integrals

Everything we have said in these notes was well known to the classical analystsby the late 19th century if not earlier. On the other hand, there exists a range ofphenomena that eluded detection until the advent of the computer age. Promi-nent among these is the story of the sinc integral, which was only discoveredvery recently (in the 2000s).

The story goes like this: Consider the sinc function sinc(x) ≡ sin xx , which

arises (among other places) in sampling theory, and also furnishes one exampleof a delta sequence, as discussed in the previous section. The total area underthe curve of sinc(x) is area of the unit circle:∫ ∞

∞

sinx

xdx = π

This is perhaps not especially surprising. What is somewhat surprising is thatthe total area under the curve of the product function sinc(x)sinc

(x3

)is also

the area of the unit circle:∫ ∞−∞

sinx

x· sinx/3

x/3dx = π

Further investigation reveals a similar pattern for the next several sinc-productfunctions: ∫ ∞

−∞

(sinx

x

)(sinx/3

x/3

)(sinx/5

x/5

)dx = π∫ ∞

−∞

(sinx

x

) (sinx/3

x/3

) (sinx/5

x/5

) (sinx/7

x/7

)dx = π

and so on, up to ∫ ∞−∞

(sinx

x

) (sinx/3

x/3

)· · ·(

sinx/13

x/13

)dx = π.

Since the pattern holds with no modification whatsoever for the first 7 instances,we might expect it to continue in this way forever. But this expectation isconfounded by the next instance, which reads∫ ∞−∞

(sinx

x

) (sinx/3

x/3

)· · ·(

sinx/15

x/15

)dx =

(467 807 924 713 440 738 696 537 864 469

935 615 849 440 640 907 310 521 750 000

)·2π

Who ordered that?!In contrast to the first 7 cases considered above, the numerical value here is

not π—but it is very close, differing from π in the 11th decimal place. I findthis to be totally weird.

This bizarre behavior of sinc integrals was discovered by the father-and-sonteam of experimental mathematicians David and Jonathan Borwein, and youcan read about it here:


• D. Borwein and J. Borwein, “Some remarkable properties of sinc and re-lated integrals,” Ramanujan Journal 5 73 (2001), http://link.springer.com/article/10.1023%2FA%3A1011497229317

I am also a big fan of David Borwein’s book Computational Excursionsin Analysis and Number Theory, available online (with MIT certificates) here:http://link.springer.com/book/10.1007%2F978-0-387-21652-2

http://link.springer.com/article/10.1023%2FA%3A1011497229317

http://link.springer.com/article/10.1023%2FA%3A1011497229317

http://link.springer.com/book/10.1007%2F978-0-387-21652-2

18.305 lecture notes: power tools for de nite integrals...

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