How to evaluate $$\int_{0}^{\frac{\pi}{2}} e^{-(\pi \tan(x) - 1)^2} \, dx$$ Source: I created this integral so I don’t know the closed form
I tried Wolfram Alpha, but Wolfram Alpha is unable to evaluate it.
Let $ u = \tan(x)$, then $du = \sec^2(x) dx$. When $x = 0$, $ u = \tan(0) = 0$, and when $x = \frac{\pi}{2}$, $u = \tan(\frac{\pi}{2}) = \infty $
Now, the integral becomes:
$$\int_{0}^{\infty} e^{-(\pi u - 1)^2} \frac{du}{\cos^2(x)}$$
Now, $\cos^2(x)$ in terms of $u$. Since $\tan^2(x) + 1 = \sec^2(x)$, we have $1 + u^2 = \sec^2(x)$, and thus $\cos^2(x) = \frac{1}{1+u^2}$.
Substituting;
$$\int_{0}^{\infty} e^{-(\pi u - 1)^2} \frac{du}{1+u^2}$$