Answer: Frullani Integral: (e^{-x}-e^{-2x})/x

Key Idea / Intuition

The integrand looks like a single ugly fraction, but hidden inside it is a Frullani integral: the formula that says $\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx = (f(0)-f(\infty))\ln(b/a)$ whenever $f$ is continuous and the limits at $0$ and $\infty$ exist. Here $f(x)=e^{-x}$ , so $f(0)=1$ , $f(\infty)=0$ , and the ratio of the two "speeds" is $2/1$ . The answer pops out in one line.

Alternatively, one can differentiate under the integral sign: introduce a parameter, swap differentiation and integration, then integrate back. Both routes lead to the same beautiful answer $\ln 2$ .

Formal Proof / Solution

Method 1: Frullani's Integral

Frullani's theorem. If $f:[0,\infty)\to\mathbb{R}$ is continuous and $f(0)$ , $f(\infty):=\lim_{x\to\infty}f(x)$ both exist, then for $0 < a < b$ ,

$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx = \bigl(f(0)-f(\infty)\bigr)\ln\!\frac{b}{a}.$

Write our integral as

$I = \int_0^\infty \frac{e^{-x} - e^{-2x}}{x}\,dx = \int_0^\infty \frac{f(1\cdot x)-f(2\cdot x)}{x}\,dx,\quad f(t)=e^{-t}.$

Then $f(0)=1$ , $f(\infty)=0$ , $a=1$ , $b=2$ , so Frullani gives

$\boxed{I = (1-0)\ln\!\frac{2}{1} = \ln 2.}$

Method 2: Differentiation Under the Integral Sign (Feynman)

Define

$I(t) = \int_0^\infty \frac{e^{-x} - e^{-tx}}{x}\,dx, \quad t > 0.$

Differentiate with respect to $t$ (justified by dominated convergence, since $|{-xe^{-tx}}/{x}| = e^{-tx}$ is integrable):

$I'(t) = \int_0^\infty \frac{\partial}{\partial t}\!\left(\frac{e^{-x}-e^{-tx}}{x}\right)dx = \int_0^\infty e^{-tx}\,dx = \frac{1}{t}.$

Integrate back: $I(t) = \ln t + C$ .

Boundary condition: $I(1) = \int_0^\infty \frac{e^{-x}-e^{-x}}{x}\,dx = 0$ , so $0 = \ln 1 + C = C$ .

Therefore $I(t) = \ln t$ , and

$I = I(2) = \ln 2.$

Quick sanity check