calculus - Proving of Integral $int_{0}^{infty}frac{e^{-bx}-e^{-ax}}{x}dx = lnleft(frac{a}{b}right)$

Prove that

$$\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right)$$

My Attempt:

Define the function $I(a,b)$ as

$$I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx$$

Differentiate both side with respect to $a$ to get

$$\begin{align} \frac{dI(a,b)}{da} &= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &= \int_{0}^{\infty}e^{-ax}\,dx\\ &= -\frac{1}{a}(0-1)\\ &= \frac{1}{a} \end{align}$$

How can I complete the proof from here?

Answer

A problem-specific solution is as follows:

$$\begin{align*} \int_{0}^{\infty} \frac{e^{-bx} - e^{-ax}}{x} \, dx &= - \int_{0}^{\infty} \int_{a}^{b} e^{-xt} dt \, dx \\ &= - \int_{a}^{b} \int_{0}^{\infty} e^{-xt} dx \, dt \\ &= - \int_{a}^{b} \frac{dt}{t} = - \left[ \log x \right]_{a}^{b} = \log\left(\frac{a}{b}\right). \end{align*}$$

Interchanging the order of integration is justified either by Fubini's theorem or Tonelli's theorem.