Calculate $I=\int_0^{2\pi}dx \int_0^{\pi} e^{\sin y(\cos x-\sin x)}\sin y\,dy$

Question

I'm not quite sure how to calculate this integral:

$$I=\int_0^{2\pi}dx \int_0^{\pi} e^{\sin y(\cos x-\sin x)}\sin y\,dy$$

With Mathematica I see the result is $2\pi \sqrt 2 \sinh \sqrt 2$.

Is there any special technique for calculating this integral?

Any help will be appreciated.

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Thank you for these brilliant solutions using Bessel function or complex analysis.

After looking into this integral for a long time I found another "primary" solution:

Rewrite it as

$$I=\int_0^{2\pi}d\theta \int_0^{\pi} e^{\sin \varphi(\cos \theta-\sin \theta)}\sin \varphi\,d\varphi$$

By polar coordinates it becomes

$$I=\int_{u^2+v^2+w^2=1} e^{u-v}\,dS$$

since $dS=\sin\varphi d\theta d\varphi$. Then choose an orthogonal transform such that $z=\frac{u-v}{\sqrt 2}$, we get

$$I=\int_{x^2+y^2+z^2=1} e^{\sqrt 2 z}\,dS$$

Going back to polar coordinates it becomes

$$I=\int_0^{2\pi}d\theta \int_0^{\pi} e^{\sqrt 2 \cos\varphi}\sin \varphi\,d\varphi$$

And finally we get

$$I=2\pi \int_{-1}^1 e^{\sqrt 2 t}\,dt$$

Answer 1

I see a solution using the modified Bessel function (not in the result, however) $$I_0(z)=\sum_{n=0}^\infty\frac{1}{n!^2}\left(\frac{z}{2}\right)^{2n}=\frac1\pi\int_0^\pi e^{z\cos t}\,dt.$$ Since $\cos x-\sin x=\sqrt2\cos(x+\pi/4)$ being integrated over its period, we may drop the $\pi/4$: $$\int_0^{2\pi}e^{r(\cos x-\sin x)}\,dx=\int_0^{2\pi}e^{r\sqrt2\cos x}\,dx=2\pi I_0(r\sqrt2),$$ and using the power series, we can compute \begin{align*} \int_0^\pi I_0(a\sin y)\sin y\,dy &=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\int_0^\pi\sin^{2n+1}y\,dy \\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\mathrm{B}\left(\frac12,n+1\right) \\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n} n!^2}\frac{\sqrt\pi\cdot n!}{2^{-n-1}(2n+1)!!\sqrt\pi} \\&=2\sum_{n=0}^\infty\frac{a^{2n}}{(2n+1)!}=\frac2a\,\sinh a. \end{align*} This gives the result you've got.

Answer 2

Denoting : $ f : \mathbb{C}\rightarrow\mathbb{C},\ z\mapsto\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z} $, where $ a\in\mathbb{R} \cdot $

$ \left(\forall n\in\mathbb{Z}\right),\ a_{n}=\left\lbrace\begin{matrix}\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!},\ \text{If }n\geq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n<0\end{matrix}\right. $, and $ \left(\forall n\in\mathbb{Z}\right),\ b_{n}=\left\lbrace\begin{matrix}\frac{a^{-n}\left(1-\mathrm{i}\right)^{-n}}{2^{-n}\left(-n\right)!},\ \text{If }n\leq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n>0\end{matrix}\right. $

$$ \oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a}{2}\left(z+\frac{1}{z}\right)-\frac{a}{2\,\mathrm{i}}\left(z-\frac{1}{z}\right)}}{\mathrm{i}\,z}\,\mathrm{d}z}=\oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z}\,\mathrm{d}z} $$

\begin{aligned} f\left(z\right)&=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!}z^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1-\mathrm{i}\right)^{n}}{2^{n}n!}z^{-n}}\right)\\ &=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=-\infty}^{+\infty}{a_{n}z^{n}}\right)\left(\sum_{n=-\infty}^{+\infty}{b_{n}z^{n}}\right)\\ f\left(z\right) &=\frac{1}{\mathrm{i}\,z}\sum_{n=-\infty}^{+\infty}{c_{n}z^{n}} \end{aligned}

Where $$ \left(\forall n\in\mathbb{Z}\right),\ c_{n}=\sum_{k=-\infty}^{+\infty}{a_{k}b_{n-k}}=\sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}} $$

Setting $ n $ to $ 0 $, we'll get : $$ \sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}}=\sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$

Thus, applying the residue theorem : $$ \int_{0}^{2\pi}{\mathrm{e}^{a\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}=2\pi \sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$

For $ y \in\mathbb{R} $, setting $ a=\sin{y} $, then integrating with respect to $ y $, we get : $$ \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}=4\pi\sum_{n=0}^{+\infty}{\frac{W_{2n+1}}{2^{n}\left(n!\right)^{2}}} $$

Where $ W_{n} $ for $ n\in\mathbb{N} $, is the Wallis integral $ W_{n}=\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} \cdot $

Note that since we're integrating on a segment, switching the integral and the infinite sum, which we've just done, is in deed possible.

Since $ \left(\forall n\in\mathbb{N}\right),\ W_{2n+1}=\frac{2^{2n}\left(n!\right)^{2}}{\left(2n+1\right)!} $, we have : \begin{aligned} \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}&=4\pi\sum_{n=0}^{+\infty}{\frac{2^{n}}{\left(2n+1\right)!}} \\ &=2\pi\sqrt{2}\sinh{\sqrt{2}}\end{aligned}