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

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Theorem

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

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


Proof

Note that the integrand is of the form:

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

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:

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

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

\(\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } x \rd x\) \(=\) \(\displaystyle \paren {\lim_{x \mathop \to \infty} e^{-x} - e^0} \ln \frac a b\)
\(\displaystyle \) \(=\) \(\displaystyle -\ln \frac a b\) Exponential of Zero
\(\displaystyle \) \(=\) \(\displaystyle \ln \frac b a\) Logarithm of Reciprocal

$\blacksquare$


Sources