Compute $$\iint_D(x^2-y^2)e^{2xy}dxdy,$$ where $D=\{(x,y):x^2+y^2\leq 1, \ -x\leq y\leq x, \ x\geq 0\}.$

The area is a circlesector disk with radius $1$ in the first and fourth quadrant. Going over to polar coordinates I get

$$\left\{ \begin{array}{rcr} x & = & r\cos{\theta} \\ y & = & r\sin{\theta} \\ \end{array}, \ \ \implies E:\left\{ \begin{array}{rcr} 0 \leq r\leq 1 \\ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \\ \end{array} \right. \right.$$

and $$J(r,\theta)=\frac{d(x,y)}{d(r,\theta)}=r.$$

So $$\iint_D(x^2-y^2)e^{2xy}r \ dxdy=\iint_Er^3(\cos^2{\theta}-\sin{2\theta})e^{r^22\cos{\theta}\sin{\theta}}drd\theta= \\ =\iint_Er^3\cos{2\theta} e^{r^2\sin{2\theta}}drd\theta = 2\int_0^{4/\pi}\cos{2\theta}\cdot\left(\int_0^1 r^3e^{r^2\sin{2\theta}}dr\right)d\theta.$$

I have no idea how to compute the inner integral. I seem to get quite complicated integrals everytime I do this.

  • $\begingroup$ You might be able to do it by parts -- but you'd probably be even better off switching the order of integration and doing the $\theta$ integral first. $\endgroup$ – Y. Forman Feb 11 '18 at 17:36
  • $\begingroup$ Isn't that equally difficult? I've tried it. $\endgroup$ – Parseval Feb 11 '18 at 17:50
  • $\begingroup$ The $\theta$ integral is equally difficult, or the resulting $r$ integral is equally difficult? $\endgroup$ – Y. Forman Feb 11 '18 at 17:58
  • $\begingroup$ The $\theta$ integral. I still get the integrand to $\cos{2\theta}e^{r^2\sin{2\theta}}$. $\endgroup$ – Parseval Feb 11 '18 at 18:08
  • 1
    $\begingroup$ You can now substitute $u = \sin 2 \theta$. Remember $r$ is constant with respect to $\theta$. $\endgroup$ – Y. Forman Feb 11 '18 at 18:15

Call $\alpha = \sin(2\theta)$ for simplicity.

$$\int_0^1 r^3 e^{\alpha r^2}\ dr = \int_0^1 \frac{d}{d\alpha} r e^{\alpha r^2} = \frac{d}{d\alpha} \int_0^1 r e^{\alpha r^2}\ dr$$

The latter can easily be done by parts once to get

$$\int r e^{\alpha r}\ dr = \frac{e^{a r} (a r-1)}{a^2}$$

Hence with the extrema it becomes

$$\frac{e^a (a-1)+1}{a^2}$$


$$\frac{d}{d\alpha} \left(\frac{e^a (a-1)+1}{a^2}\right) = \frac{\left(a^2-2 a+2\right) e^a-2}{a^3}$$

Getting back $\alpha = \sin(2\theta)$ and thou hast

$$\frac{\left(\sin^2(2\theta)-2 \sin(2\theta)+2\right) e^{\sin(2\theta)}-2}{\sin^3(2\theta)}$$

  • $\begingroup$ And integrating $\cos{2\theta}\cdot\text{your answer}$ is simple? $\endgroup$ – Parseval Feb 11 '18 at 18:10
  • $\begingroup$ @Parseval Not really, but this is what comes out by HIS final integral. I didn't check if the whole procedure up to that is correct. $\endgroup$ – Von Neumann Feb 11 '18 at 18:14

By letting $x=\frac{u+v}{\sqrt{2}},y=\frac{u-v}{\sqrt{2}}$ the given integral becomes

$$\begin{eqnarray*} \iint_{\substack{u^2+v^2\leq 1 \\ u,v\geq 0}} 2uv\,e^{u^2-v^2}\,du\,dv &=& \int_{0}^{1}\int_{0}^{\pi/2}\rho^3\sin(2\theta)e^{\rho^2\cos(2\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}\int_{0}^{\pi}\rho^3\sin(\theta)e^{\rho^2\cos(\theta)}\,d\theta\,d\rho\\&=&\frac{1}{2}\int_{0}^{1}2\rho \sinh(\rho^2)\,d\rho\\&=&\frac{1}{2}\left[\cosh(\rho^2)\right]_{0}^{1}=\frac{\cosh(1)-1}{2}=\color{red}{\sinh^2\left(\tfrac{1}{2}\right)}.\end{eqnarray*}$$


Switch the order of integration to do the $\theta$ integral first. Substitute $w = r^2\sin2\theta$, $dw = 2r^2\cos2\theta$.

$$\begin{align} \int_0^1 \int_0^\frac\pi4 2r^3 \cos2\theta e^{r^2\sin2\theta}d\theta dr &= \int_0^1 \int_0^{r^2} re^w dwdr \\ &= \int_0^1 r(e^{r^2}-1)dr \\ &= \frac12\int_0^12re^{r^2}dr - \int_0^1rdr \\&= \frac12(e-1)-\frac12 \\&=\frac e2 -1 \end{align}$$


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