I want to show that $\frac{i}{2\pi}\int^{2\pi}_0e^{(a+b)\cos\theta}\sin((a-b)\sin\theta-n\theta)d\theta = 0$

Question

So we have that $\theta\in\mathbb{R}$ and $a,b\in\mathbb{C}$, and $n\in\mathbb{N}$ and I have to show that $$ \frac{i}{2\pi}\int^{2\pi}_0e^{(a+b)\cos\theta}\sin((a-b)\sin\theta-n\theta)d\theta = 0.$$ I have the tools of complex analysis. So far I have tried using Euler's formula and gotten to the point where the above integral is equal to $$ \frac{1}{4\pi}\int^{2\pi}_0 e^{ae^{i\theta}+be^{-i\theta}}e^{-in\theta}-e^{ae^{-i\theta}+be^{i\theta}}e^{in \theta}.$$ But I can't really get much further. Some help would really be appreciated.

Thanks.

Answer 1

Note that the integrand $$ f(\theta) = e^{(a+b)\cos \theta} \sin\left((a-b)\sin\theta - n\theta\right) $$ is an odd function, and it is also periodic with period $2\pi$. In general, this always implies the integral over the whole period is $0$. Observe: $$ \int_0^{2\pi} f(\theta) d\theta =\int_0^\pi f(\theta)d\theta+\int_{\pi}^{2\pi}f(\theta)d\theta =\int_0^\pi f(\theta)d\theta + \int_{-\pi}^0 f(\theta+2\pi)d\theta = \int_0^\pi f(\theta)d\theta+\int_{-\pi}^0f(\theta)d\theta=\int_{-\pi}^\pi f(\theta)d\theta=0 $$ The integral of an odd function over a symmetric interval is always 0.

Answer 2

Use the property of definite integrals that: $$I=\int_{0}^{2a} f(x) dx=\int_{0}^{a} [f(2a-x)+f(x)] dx=2\int_{0}^{a} f(x) dx,~ if ~~ f(2a-x)=f(x);$$ $$\implies I= 0, ~if~~(2a-x)=-f(x).$$ So for $$J=\int_{0}^{2\pi} e^{p\cos x} \sin[q\sin x-nx]~dx$$ Here $$f(x)=e^{p\cos x} \sin[q\sin x-nx], f(2\pi-x)=e^{p \cos x}\sin [q \sin(2\pi-x)-n(2\pi-x)]$$ $$=e^{p\cos x}\sin [-q ~\sin x+nx-2n \pi]=-e^{p\cos x}\sin [q ~\sin x-nx] =-f(2\pi-x).$$ Hence the required $J$-integral vanishes.

Answer 3

Let $ n\in\mathbb{N} $, $ a,b\in\mathbb{C} \cdot $

Condider the function : $ f_{n}:\mathbb{C}^{*}\rightarrow\mathbb{C},\ z\mapsto\frac{1}{z^{n+1}}\,\mathrm{e}^{az+\frac{b}{z}} \cdot $

Denoting $ a_{p}=\left\lbrace\begin{aligned}\frac{a^{p}}{p!},\ \ \ &\textrm{If }p\geq0\\ 0,\ \ \ &\textrm{If }p<0\end{aligned}\right. $, and $ b_{p}=\left\lbrace\begin{aligned} 0,\ \ \ &\textrm{If }p>0\\\frac{b^{-p}}{\left(-p\right)!},\ \ \ &\textrm{If }p\leq0\end{aligned}\right. $, we have that :

$ \ \ \ \ \ \ \ \ f_{n}\left(z\right)=\frac{1}{z^{n+1}}\left(\sum\limits_{p=-\infty}^{+\infty}{a_{p}z^{p}}\right)\left(\sum\limits_{p=-\infty}^{+\infty}{b_{p}z^{p}}\right)=\sum\limits_{p=-\infty}^{+\infty}{c_{p}z^{p-n-1}} $, where $ c_{p}=\sum\limits_{k=-\infty}^{+\infty}{a_{k}b_{p-k}}=\sum\limits_{k=0}^{+\infty}{\frac{b^{k}a^{p+k}}{k!\left(p+k\right)!}} $

Considering the coefficient of $ z^{-1} $ in the previous expansion, we get that $ \mathrm{Res}\left(f_{n},0\right)=\sum\limits_{p=0}^{+\infty}{\frac{b^{p}a^{n+p}}{p!\left(n+p\right)!}} \cdot $

The Residue theorem allows us to write that $$ \oint_{\left|z\right|=1}{f_{n}\left(z\right)\mathrm{d}z}=2\pi\,\mathrm{i}\,\mathrm{Res}\left(f_{n},0\right) $$

Taking $ z=\mathrm{e}^{\mathrm{i}\,\theta} $, gives the following $$ \int_{0}^{2\pi}{f_{n}\left(\mathrm{e}^{\mathrm{i}\,\theta}\right)\mathrm{e}^{\mathrm{i}\,\theta}\,\mathrm{d}\theta}=2\pi\,\mathrm{Res}\left(f_{n},0\right) $$

Since \begin{aligned}\Large f_{n}\left(\mathrm{e}^{\mathrm{i}\,\theta}\right)\mathrm{e}^{\mathrm{i}\,\theta} =\mathrm{e}^{a\,\mathrm{e}^{\mathrm{i}\,\theta}+b\,\mathrm{e}^{-\mathrm{i}\,\theta}-\mathrm{i}\,n\theta}&\Large=\mathrm{e}^{\left(a+b\right)\left(\frac{\mathrm{e}^{\mathrm{i}\,\theta}+\mathrm{e}^{-\mathrm{i}\,\theta}}{2}\right)+\mathrm{i}\left(a-b\right)\left(\frac{\mathrm{e}^{\mathrm{i}\,\theta}-\mathrm{e}^{-\mathrm{i}\,\theta}}{2\,\mathrm{i}}\right)-\mathrm{i}\,n\theta}\\&\Large=\mathrm{e}^{\left(a+b\right)\cos{\theta}+\mathrm{i}\left(a-b\right)\sin{\theta}-\mathrm{i}\,n\theta} \end{aligned}

Taking the imaginary part, we get that $$ \int_{0}^{2\pi}{\mathrm{e}^{\left(a+b\right)\cos{\theta}}\sin{\left(\left(a-b\right)\sin{\theta}-n\theta\right)}\,\mathrm{d}\theta}=0 $$