# Define I(x)=int_x^infty e^(-y^2)dy. Evaluate int_0^infty e^(x^2)[I(x)]^2dx

Define

Evaluate
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To wit
$I\left(x\right)=x{\int }_{1}^{\mathrm{\infty }}du\phantom{\rule{mediummathspace}{0ex}}{e}^{-{x}^{2}{u}^{2}}$
which means that
$I\left(x{\right)}^{2}={x}^{2}{\int }_{1}^{\mathrm{\infty }}du\phantom{\rule{mediummathspace}{0ex}}{\int }_{1}^{\mathrm{\infty }}dv\phantom{\rule{mediummathspace}{0ex}}{e}^{-{x}^{2}\left({u}^{2}+{v}^{2}\right)}$
Now we can put this into the desired integral and reverse the order of integration to get
$\begin{array}{r}{\int }_{0}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}{e}^{{x}^{2}}I\left(x{\right)}^{2}={\int }_{1}^{\mathrm{\infty }}du\phantom{\rule{mediummathspace}{0ex}}{\int }_{1}^{\mathrm{\infty }}dv\phantom{\rule{mediummathspace}{0ex}}{\int }_{0}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}{x}^{2}{e}^{-{x}^{2}\left({u}^{2}+{v}^{2}-1\right)}\end{array}$
The inner integral converges because ${u}^{2}+{v}^{2}-1\ge 0$. We evaluate the inner integral and reduce the integral to a double integral:
${\int }_{0}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}{e}^{{x}^{2}}I\left(x{\right)}^{2}=\frac{\sqrt{\pi }}{4}{\int }_{1}^{\mathrm{\infty }}du\phantom{\rule{mediummathspace}{0ex}}{\int }_{1}^{\mathrm{\infty }}dv\phantom{\rule{mediummathspace}{0ex}}\left({u}^{2}+{v}^{2}-1{\right)}^{-3/2}$
The integral over v may be attacked by a trig substitution: $v=\sqrt{{u}^{2}-1}\mathrm{tan}t$, $dv=\sqrt{{u}^{2}-1}{\mathrm{sec}}^{2}tdt$; the result is
${\int }_{1}^{\mathrm{\infty }}dv\phantom{\rule{mediummathspace}{0ex}}\left({u}^{2}+{v}^{2}-1{\right)}^{-3/2}=\left(1-\frac{1}{u}\right)\frac{1}{{u}^{2}-1}=\frac{1}{u\left(u+1\right)}$
The problem now reduces to the evaluation of
${\int }_{1}^{\mathrm{\infty }}\frac{du}{u\left(u+1\right)}=\underset{r\to \mathrm{\infty }}{lim}\mathrm{log}\left(\frac{r}{r+1}\right)+\mathrm{log}2=\mathrm{log}2$
Therefore, the desired integral has the value
${\int }_{0}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}{e}^{{x}^{2}}I\left(x{\right)}^{2}=\frac{\sqrt{\pi }}{4}\mathrm{log}2$