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.

With the substitution $u\to\frac{t}2x$, the inner integral becomes
$$\int_0^{\frac{t}{2}}e^{u^2}\,du=\frac{t}2\int^1_0e^{\frac{t^2}{4}x^2}\,dx,$$
implying
$$\int_{-\infty}^\infty\frac{1}{t}e^{-\frac{t^2}{4}}\int_0^{\frac{t}{2}}e^{u^2}\,du\,dt=\frac12\int_{-\infty}^\infty\int^1_0 e^{-\frac{t^2}{4}(1-x^2)}\,dx\,dt=\int_{-\infty}^\infty\int^1_0 e^{-s^2(1-x^2)}\,dx\,ds.$$
Since the integrand is positive, Mr. Fubini encourages us to change the order of integration, and
$$\int_{-\infty}^\infty e^{-s^2(1-x^2)}\,ds=\frac{\sqrt{\pi}}{\sqrt{1-x^2}}.$$ Now clearly $$\int^1_0\frac{dx}{\sqrt{1-x^2}}\,dx=\frac{\pi}2,$$
so we arrive at $$\int_{-\infty}^\infty\frac{1}{t}e^{-\frac{t^2}{4}}\int_0^{\frac{t}{2}}e^{u^2}\,du\,dt=\frac{\pi^{3/2}}2.$$
