TITLE: Compute $f'(t)=\int\limits_{-\infty}^{+\infty}e^{-tx^2}\mathrm dx$
QUESTION [0 upvotes]: For :$\int\limits_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$
Compute $f'(t)=\left(\int\limits_{-\infty}^{+\infty}e^{-tx^2}\mathrm dx\right)'$. With$t\in\mathbb{R}$.

REPLY [0 votes]: Going with what Ngua suggested, for t>0 $f'(t)$ will simply be equal to $\frac{\sqrt\pi}{\sqrt{t}}$.You can also check it on wolfram alpha.
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