Solve a definite integral $\int_0^\pi e^{2\cos(\theta)}\cos(2\sin(\theta) - \theta)\,d\theta$

Question

My friend gave me this integral to solve but I am unable to even start. I even suspect even if it possible to solve analytically. Please help me. You are welcome to solve with whatever method you want.
I checked the answer using online integral solver and it comes out to be $2\pi$ (if it is solvable analytically). $$\int_0^\pi e^{2\cos(\theta)}\cos(2\sin(\theta) - \theta)\,d\theta$$

Thank you in advance!

Answer 1

It's an interesting integral. I think you are stuck on how to tackle $\theta$ sitting in the $cos(2\sin(\theta) - \theta)\,$. I will use complex integration to solve it.

Put $cos(2\sin(\theta) - \theta)\, = \Re\; e^{i(2\sin(\theta) - \theta)}$. Then the integral $I$ becomes $$I = \Re\int_0^\pi \,e^{2\cos(\theta)}e^{i(2\sin(\theta) - \theta)}\,\mathrm{d}\theta$$ Now I want to convert $I$ into a closed line integral and for that, I try the substitution $$\theta \rightarrow 2\pi-\theta$$ in the original integral. It is easy to see then that integral remains unchanged so, by the property of definite integrals for real variables, $$I = \frac{1}{2}\int_0^{2\pi} e^{2\cos(\theta)}\cos(2\sin(\theta) - \theta)\,d\theta$$

Put $$\int_0^{2\pi} e^{2\cos(\theta)}\cos(2\sin(\theta) - \theta)\,d\theta =\Re\int_0^{2\pi} \,e^{2\cos(\theta)}e^{i(2\sin(\theta) - \theta)}\,\mathrm{d}\theta = J$$ Then $I = J/2$. Now substitute $\cos(\theta) + i \sin(\theta) = z\;$ in $J\;$. Then clearly, $$\int_0^{2\pi}\,e^{2\cos(\theta)}e^{i(2\sin(\theta) - \theta)}\,\mathrm{d}\theta = -i\oint \limits_{|z|=1}\frac{e^{2z}}{z^2}\mathrm{d}z$$

Now, $0$ is the only point of singularity of the integrand in the complex integral. Further, it is a pole of order $2$ and its residue is given by $$Res(0)= \frac{2e^{2\times0}}{1!} = 2$$ Then, by applying Cauchy's Residue theorem, we get $$\oint_{|z|=1}\frac{e^{2z}}{z^2}\mathrm{d}z = 2\pi i \times (1 \times 2)= 4\pi i$$

Therefore, $I = J/2 = \frac{\Re(4\pi i \times (-i))}{2} = \frac{\Re(4\pi)}{2} = 2\pi.$

Answer 2

For $k\in\Bbb Z$, $\int_0^\pi\exp(ik\theta)\mathrm{d}\theta$ is $\pi$ if $k=0$, and otherwise $\frac{(-1)^k-1}{ik}$, i.e. $0$ for even $k$ and $\frac{2i}{k}$ for odd $k$. Hence $\Re\int_0^\pi\exp(ik\theta)\mathrm{d}\theta=0$ for any nonzero integer $k$. Your integral is$$\Re\int_0^\pi\exp(2\exp i\theta-i\theta)\mathrm{d}\theta=\sum_{n\ge0}\frac{2^n}{n!}\Re\int_0^\pi\exp(n-1)i\theta\mathrm{d}\theta=2\pi.$$