3
$\begingroup$

Here, I have a little unpleasant way to calculate the following double integral

$$\iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A$$

where $D$ is the square area $D=\{(x,y)\in\mathbb{R}^2: 0 \le x \le \pi/2, 0 \le y \le \pi/2\}$

my attempt:

with the symmetry of the area we know

$$I=\iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A=\iint_{D}{\sin(y)e^{\sin(x)\sin(y)}}\text{d}A$$

thus

$$I=\frac1{2}\iint_{D}{(\sin(x)+\sin(y))e^{\sin(x)\sin(y)}}\text{d}A$$

with the substitution $u=\sin(x)$ and $v=\sin(y)$, we have

$$I=\frac1{2}\iint_{D^{*}}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A$$

and now the integral area is $D^{*}=\{(u,v)\in\mathbb{R}^2: 0 \le u \le 1, 0 \le v \le 1\}$

notice that the function $\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus

$$I=\iint_{D^{*}\cap(u\le v)}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A=\iint_{D^{*}\cap(u\ge v)}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A$$

so I can only foucs on area $D^{*}\cap(u\ge v)$, set another substitution

$$u+v=\alpha$$

$$uv=\beta$$

under the condition $u\ge v$, the determinant of Jacobian matrix writes

$$|J(\alpha,\beta)|=\frac1{\sqrt{\alpha^2-4\beta}}$$

and

$$\frac1{\sqrt{(1-u^2)(1-v^2)}}=\frac1{\sqrt{(1+\beta)^2-\alpha^2}}$$

the integral area change to $\{(\alpha,\beta)\in\mathbb{R}^2: 2\sqrt{\beta} \le \alpha \le 1+\beta, 0 \le \beta \le 1\}$, so

$$\begin{align} I&=\int_{0}^{1}\int_{2\sqrt{\beta}}^{1+\beta}{\frac{\alpha}{\sqrt{((1+\beta)^2-\alpha^2)(\alpha^2-4\beta)}}e^{\beta}}\text{d}\alpha \text{d}\beta \\&=\int_{0}^{1}\left(-\tan^{-1}\sqrt{\frac{(1+\beta)^2-\alpha^2}{\alpha^2-4\beta}}\right)\biggr|_{\alpha=2\sqrt{\beta}}^{1+\beta} e^{\beta}\text{d}\beta \\&=\frac{\pi}{2}(e-1) \end{align}$$

obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.

thanks in advance for any suggestion!

$\endgroup$
2
$\begingroup$

Using the Taylor expansion of the exponential $$ \iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A=\sum_{k\geq 0}\frac{1}{k!}\int_0^{\pi/2}\mathrm{d}x~(\sin x)^{k+1}\int_0^{\pi/2}\mathrm{d}y~(\sin y)^{k}=\sum_{k\geq 0}\frac{1}{k!}I(k)I(k+1)\ , $$ where $$ I(k)=\int_0^{\pi/2}\mathrm{d}x~(\sin x)^{k}=\int_0^1\mathrm{du}~u^k(1-u^2)^{-1/2}=\frac{\sqrt{\pi}~\Gamma \left(\frac{k+1}{2}\right)}{2 \Gamma \left(\frac{k}{2}+1\right)}\ . $$ Therefore $$ \frac{1}{k!}I(k)I(k+1)=\frac{\left(\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right)\right) \left(\sqrt{\pi } \Gamma \left(\frac{k}{2}+1\right)\right)}{k! \left(2 \Gamma \left(\frac{k}{2}+1\right)\right) \left(2 \Gamma \left(\frac{k+3}{2}\right)\right)}=\frac{\pi }{2 \Gamma (k+2)} $$ after simplifications, which then leads to $$ \sum_{k\geq 0}\frac{1}{k!}I(k)I(k+1)=\frac{1}{2} (e-1) \pi\ . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.