Having Fun with an Integral


The main point is not quantity or speed—the main point is quality of thought.

–- Jane Gilman

I=02πecosθcos(sinθ) dθ I = \int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta)\; d\theta

is an integral that WolframAlpha cannot YET compute analytically as of today (December 8, 2018). Here is the link.


I think that there are two ways to tackle this problem by exploiting the analyticity of the integrand ecosθcos(sinθ)e^{\cos \theta} \cos(\sin \theta). We can calculate it in the real domain by Taylor series or in the complex domain by Cauchy's integral formula.

Solution by Taylor Series

First, let us introduce the function

f(r)=02πercosθcos(rsinθ) dθ, f(r) = \int_{0}^{2\pi} e^{r \cos \theta} \cos(r \sin \theta)\; d\theta,

so we have the equation

I=f(1)I = f(1)

. By Taylor expansion, we have

f(r)=n=0f(n)(0)n!rn. f(r) = \sum_{n=0}^{\infty} \frac {f^{(n)}(0)}{n!} r^{n}.

It is easy to show that all the coefficients of this Taylor expansion vanish except for the first one. Namely, f(0)=02πe0cosθcos(0sinθ) dθ=2πf(0) = \int_{0}^{2\pi} e^{0 \cos \theta} \cos(0 \sin \theta)\; d\theta = 2\pi. The nn-th derivatives of ff at 00 must be in the form of

f(n)(0)=02πk even  0kn(1)k2(nk)cosnkθsinkθ dθ=02πcos(nθ) dθ=0. f^{(n)}(0) = \int_0^{2\pi} \sum_{k{\text{ even }}\wedge\; 0\le k\le n}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta \; d\theta = \int_{0}^{2\pi} \cos(n \theta) \; d\theta = 0.

Hence, we have f(r)=2π=If(r) = 2\pi = I.

Solution by Cauchy's Integral Formula

First, let us introduce the same function

f(r)=02πercosθcos(rsinθ) dθ. f(r) = \int_{0}^{2\pi} e^{r \cos \theta} \cos(r \sin \theta)\; d\theta.

The integrand can be converted to a complex exponential,

ercosθcos(rsinθ)={ercosθcos(rsinθ)+isin(rsinθ)}={ercosθ+irsinθ}={ereiθ}. e^{r \cos \theta} \cos(r \sin \theta) = \Re\{e^{r \cos \theta} \cos(r \sin \theta) + i \sin(r \sin \theta)\} = \Re\{e^{r\cos \theta + ir\sin \theta}\} = \Re\{e^{re^{i\theta}}\}.

Cauchy's Integral formula at 00 is

g(0)=12πiγg(z)z dz. g(0)=\frac{1}{2\pi i} \oint_\gamma \frac{g(z)}{z} \,dz.

If we take g(z)=ezg(z) = e^z, and γ:reiθ\gamma: re^{i\theta}, we have

g(0)=12πi02πg(reiθ)ireiθreiθ dθ=12π02πg(reiθ) dθ. g(0)=\frac{1}{2\pi i} \int_{0}^{2\pi} \frac{g(re^{i\theta}) ire^{i\theta}}{re^{i\theta}} \,d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} g(re^{i\theta}) \,d\theta.

Note that we have

{g(reiθ)}={ereiθ}=ercosθcos(rsinθ). \Re\{g(re^{i\theta})\} = \Re\{e^{re^{i\theta}}\} = e^{r \cos \theta} \cos(r \sin \theta).


g(0)=e0=1g(0)=12πf(r)f(r)=2πI=f(1)=2π.\begin{aligned} g(0) &= e^0 = 1 \\ g(0) &= \frac{1}{2\pi} f(r) \\ f(r) &= 2\pi \\ I &= f(1) = 2\pi. \end{aligned}