# Jun 6, 2021 - 15:34

#### An interesting integral.

By Vamshi Jandhyala in mathematics

June 7, 2021

## Problem

— 級数bot (@infseriesbot) June 7, 2021

## Solution

Define

$$
f(k) = \int_0^{\infty} \frac{1-e^{-kx^2}}{x^2} dx
$$

Differentiating both sides wrt $k$ and using the well known result $\int_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$, we get

$$
\begin{aligned}
f'(k) &= \int_0^{\infty} e^{-kx^2} dx = \frac{1}{\sqrt{k}} \frac{\sqrt{\pi}}{2}\\

\implies f(k) &= \sqrt{\pi k} + c
\end{aligned}
$$

We also have $f(\frac{1}{2}) = \frac{1}{\sqrt{2}} f(1)$ which gives us $c=0$.

Therefore,

$$ \int_0^{\infty} \frac{1-e^{-x^2}}{x^2} dx = f(1) = \sqrt{\pi} $$