Problem of the Week17 November – 21 November 2014

Week #12

The following quote is taken from Richard Feynman’s memoir Surely You’re Joking, Mr. Feynman!1:

“. . .Advanced Calculus, by Woods. . . It had Fourier series, Bessel functions, determinants,elliptic functions – all kinds of wonderful stuff that I didn’t know anything about. Thatbook also showed how to differentiate parameters under the integral sign – it’s a certainoperation. It turns out that’s not taught very much in the universities; they don’t emphasizeit. But I caught on how to use that method, and I used that one damn tool again and again.So because I was self-taught using that book, I had peculiar methods of doing integrals.”

See if you can channel your inner Feynman and evaluate the following integral from Woods’s book withoutthe aid of technology : ∫ π

0log(1− 2α cosx+ α2)dx, for |α| > 1.

Solution: The hint was to differentiate with respect to a parameter. This requires one to use Leibnizintegral rule. The obvious parameter to use in the differentiation is α, so if we let

I(α) =

∫ π

0log(1− 2α cosx+ α2)dx

1Feynman, Richard P., Ralph Leighton, and Edward Hutchings. Surely You’re Joking, Mr. Feynman!: Adventures of aCurious Character. New York: W.W. Norton, 1985.

2

then

dI

dα=

d

dα

∫ π

0log(1− 2α cosx+ α2)dx

=

∫ π

0

d

dαlog(1− 2α cosx+ α2)dx

=

∫ π

0

−2 cosx+ 2α

1− 2α cosx+ α2dx

=1

α

∫ π

0

(1− 1− α2

1− 2α cosx+ α2

)dx

=π

α− 1− α2

α

∫ π

0

1

1− 2α cosx+ α2dx.

It may seem that we have gone from a very difficult integral to. . . a very difficult integral. But this onecan be computed using everyone’s favorite technique from freshman calculus: a trig substitution! If we setθ = tan x

2 , we can use a little trigonometry to conclude that

cosx = cos(2 arctan θ) = cos2(arctan θ)− sin2(arctan θ)

=1

1 + θ2− θ2

1 + θ2

=1− θ2

1 + θ2.

Also, dx =2

1 + θ2dθ, so – after a little bit of algebra – the integrand becomes (in terms of θ)

dx

1− 2α cosx+ α2= 2

dθ

(1 + θ2)(

1− 2α1−θ21+θ2

+ α2) =

2dθ

(1 + 2α+ α2)θ2 + (1− 2α+ α2)

=2dθ

(1 + α)2θ2 + (1− α)2

We then have

dI

dα=π

α− 2

α

∫ θ(π)

θ(0)

(1− α)2

(1− α)2 + (1 + α)2θ2dθ

=π

α− 2

α

∫ ∞0

1

1 +(1+α1−αθ

)2dθ=π

α− 2

αarctan

(1 + α

1− αθ

)∣∣∣∣∞0

=π

α+

2

α· π

2=

2π

α.

This follows from the fact that 1+α1−α < 0 for |α| > 1. Now, we need to compute I(α):

dI

dα=

2π

α=⇒ I(α) = 2π

∫dα

α= π logα2 + C.

3

We need only determine the constant C to finish the job. There are many ways to do this. One way is toobserve that

π logα2 + C =

∫ π

0log(1− 2α cosx+ α2)dx

=

∫ π

0log

(α2

(1− 2

αcosx+

1

α2

))dx

= π logα2 +

∫ π

0log

(1 +

1

α2− 2

αcosx

)dx,

which reduces to C =∫ π0 log

(1 + 1

α2 − 2α cosx

)dx. We leave the details as an exercise to the interested

reader, but it turns out that

log

(1 +

1

α2− 2

αcosx

)→ 0 uniformly as α→∞.

It follows that, if we take limα→∞

of both sides of the above equality, we can bring the limit inside the integral

to conclude that

C = limα→∞

∫ π

0log

(1 +

1

α2− 2

αcosx

)dx =

∫ π

0limα→∞

log

(1 +

1

α2− 2

αcosx

)dx = 0.

This means that C = 0, giving us that I(α) = π logα2 for |α| > 1. Phew!