I am having trouble proving the following inequality.
$$ \iint_{x^2+y^2 \leq 1} e^{x^2+y^2}\mathrm{d}x\mathrm{d}y \leqslant {\left(\int_{\sqrt{\pi}/2}^{\sqrt{\pi}/2} e^{x^2} \mathrm{d}x\right)}^2 $$
I have tried the following approaches but all failed.
- $$0\leq x\leq 1\implies e^{x^2}\geq xe^{x^2}=\left(\frac{1}{2} e^{x^2}\right)'$$
this failed because $xe^{x^2}$ is too small
sage: numerical_integral(x*exp(x^2),0,sqrt(pi)/2)
(0.5966400253690076, 6.6240349357760975e-15)
sage: numerical_integral(exp(x^2),0,sqrt(pi)/2)
(1.1849166170926333, 1.3155217105571918e-14)
- $${\left(\int_{-L}^L e^{x^2}\mathrm{d}x\right)}^2=\iint_{\begin{subarray}{l}{-L\leq x\leq L\\ -L\leq y\leq L}\end{subarray}} e^{x^2+y^2}\mathrm{d}x\mathrm{d}y\leq \iint_{x^2+y^2\leq L^2} e^{x^2+y^2}\mathrm{d}x\mathrm{d} y$$
The area $x^2+y^2\leq L^2$ is too small.
- partial(prefix) sum taylor series of $e^{x^2}$, theoretically this can lead to a big enough estimation, however it involves too much calculation.
so: $$I_1=\iint\limits_{x^2+y^2\le 1}e^{x^2+y^2}\,dxdy=\int\limits_0^{2\pi}\int\limits_0^1re^{r^2}\,drd\theta$$
$$I_2=\left(\int\limits_{-\sqrt{\pi}/2}^{\sqrt{\pi}/2}e^{x^2}\,dx\right)^2=\int\limits_{-\sqrt{\pi}/2}^{\sqrt{\pi}/2}\int\limits_{-\sqrt{\pi}/2}^{\sqrt{\pi}/2}e^{x^2+y^2}\,dxdy=\int\limits_0^{2\pi}\int\limits_0^{\sqrt{\pi/2}}re^{r^2}\,drd\theta$$ now since the domains for $\theta$ are the same we can ignore those, and since the functions of $r$ are the same we only really care about comparing the domains, since we know that $\exp(r^2)>0$ and so it comes down to $\sqrt{\pi/2}>1$ so we are integrating a strictly positive function over a larger domain and so the integral will be greater.