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

Question

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 1

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.

Answer 2

Note that the following is a general technique that can handle much harder problems. Recalling the Laplace transform

$$ F(s) = \int_{0}^{\infty} f(x) e^{-sx} dx. $$

Consider the more general integral

$$ F(s) = \int_{0}^{\infty} \frac{e^{-bx}-e^{-ax}}{x} e^{-sx} dx \implies F'(s) = -\int_{0}^{\infty} ({e^{-bx}-e^{-ax}}) e^{-sx} dx .$$

Now, it is just a matter of evaluating the last integral and integrating the answer with respect to $s$ and then taking the limit as $s\to 0$ to find the desired value.

Note: When you integrate with respect to $s$ do not forget the constant of integration. To find it use the fact that

$$ \lim_{s\to \infty} F(s) = 0. $$