May 24, 2021 - 7:35

An interesting integral.

By Vamshi Jandhyala in mathematics

May 24, 2021

Problem

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,

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