Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x

Theorem

$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } x \rd x = \ln \frac b a$

where $a$ and $b$ are strictly positive real numbers.

Proof

Note that the integrand is of the form:

$\ds \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x$

where:

$\map f x = e^{-x}$

We have, by Derivative of Exponential Function:

$\map {f'} x = -e^{-x}$

which is continuous on $\R$.

We also have, by Exponential Tends to Zero and Infinity:

$\ds \lim_{x \mathop \to \infty} \map f x = \lim_{x \mathop \to \infty} e^{-x} = 0$

As $f$ is continuously differentiable, and $\ds \lim_{x \mathop \to \infty} \map f x$ exists and is finite, we may apply Frullani's Integral, giving:

\(\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } x \rd x\)

\(=\)

\(\ds \paren {\lim_{x \mathop \to \infty} e^{-x} - e^0} \ln \frac a b\)

\(\ds \)

\(=\)

\(\ds -\ln \frac a b\)

Exponential of Zero

\(\ds \)

\(=\)

\(\ds \ln \frac b a\)

Logarithm of Reciprocal

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.71$