Given \[ \int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt\pi, \] define for \(t>0\) \[ f(t)=\int_{-\infty}^{\infty}e^{-t x^2}\,dx. \] Find \(f'(t)\) explicitly.
Solution
1. For every \(t>0\),\[ f(t)=\frac{\sqrt\pi}{\sqrt t}. \]
Set \[ u=\sqrt t\,x. \] Since \(t>0\), this substitution gives \[ x=\frac{u}{\sqrt t}, \qquad dx=\frac{du}{\sqrt t}, \] and therefore \[ \begin{aligned} f(t) &= \int_{-\infty}^{\infty}e^{-t x^2}\,dx\\ &= \frac{1}{\sqrt t} \int_{-\infty}^{\infty}e^{-u^2}\,du\\ &= \frac{\sqrt\pi}{\sqrt t}, \end{aligned} \] using the given Gaussian integral.
2. For every \(t>0\),\[ f'(t) = \boxed{-\frac{\sqrt\pi}{2t^{3/2}}}. \]
By step 1, \[ f(t)=\sqrt\pi\,t^{-1/2}. \] Differentiating this explicit formula gives \[ f'(t) = -\frac12\sqrt\pi\,t^{-3/2} = -\frac{\sqrt\pi}{2t^{3/2}}. \]
3. Q.E.D.
Step 2 is the requested derivative.