Integral of $\int_{-\infty}^\infty\frac{1}{t}e^{-\frac{t^2}{4}}\int_0^{\frac{t}{2}}e^{u^2}dudt$ How to calculate the following integral:
$$\int_{-\infty}^\infty\frac{1}{t}e^{-\frac{t^2}{4}}\int_0^{\frac{t}{2}}e^{u^2}dudt\ \ ?$$
I don't understand how to calculate. Please help. Thank you.


I thought it might be instructive to apply Feynman's trick of differentiating under the integral to evaluate the integral of interest.  To that end, we now proceed.  


Let $I(a)$, $a> 1$, be the integral given by
$$\begin{align}
I(a)&=\int_{-\infty}^\infty \frac{e^{-at^2/4}}{t}\int_0^{t/2}e^{u^2}\,du\,dt\\\\
&\overbrace{=}^{t/2\,\mapsto \,t}\int_{-\infty}^\infty \frac{e^{-at^2}}{t}\int_0^{t}e^{u^2}\,du\,dt\tag1
\end{align}$$ 

Differentiating $(1)$ reveals
$$I'(a)=- \int_{-\infty}^\infty te^{-at^2} \int_0^{t}e^{u^2}\,du\,dt\tag2 $$

Next, integrating by parts the (outer) integral in $(2)$ with $u=\int_0^{t}e^{u^2}\,du$ and $v=-\frac2a e^{-at^2}$, we find that 
$$\begin{align}
I'(a)&=-\frac1{2a}\int_{-\infty}^\infty e^{-(a-1)t^2}\,dt\\\\
&=-\frac{\sqrt\pi}{2a\sqrt{a-1}}\tag3
\end{align}$$

Integrating $(3)$ and using $\lim_{a\to \infty}I(a)=0$, we arrive at  
$$\begin{align}
I(a)&=\frac{\sqrt\pi}2 \int_a^\infty \frac{1}{x\sqrt{x-1}}\,dx\\\\
&=\frac{\pi^{3/2}}{2}-\sqrt{\pi}\arctan(\sqrt{a-1})\tag 4
\end{align}$$

Finally, taking the limit as $a\to 1^+$ yields the coveted integral
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{e^{-t^2/4}}{t}\int_0^{t/2}e^{u^2}\,du\,dt=\frac{\pi^{3/2}}{2}}$$
And we are done!

