# May 24, 2021 - 7:35

#### An interesting integral.

By Vamshi Jandhyala in mathematics

May 24, 2021

## Problem

— 級数bot (@infseriesbot) May 24, 2021

## Solution

$$ \text{Define } f(a) = \int_0^\infty \frac{e^{-ax}}{x} dx $$

Differentiating under the integral sign, we get

$$
\begin{aligned}
f'(a) &= - \int_0^\infty e^{-ax} dx = -\frac{1}{a} \\

\implies f(a) &= -\ln{a} + c
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
\int_0^\infty \frac{e^{-ax}-e^{-bx}}{x} dx &= f(a) - f(b) \\

&= -\ln{a} + \ln{b} \\

&= \ln \frac{b}{a}
\end{aligned}
$$