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

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Theorem

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

where:

$a$ and $b$ are non-negative real numbers
$p$ is a non-zero real number.


Proof

Fix $p$ and set:

$\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x$

for all $\alpha \ge 0$.

Then:

$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \map I a - \map I b$

We have:

\(\ds \map {I'} \alpha\) \(=\) \(\ds \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \csc p x} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds -\int_0^\infty e^{-\alpha x} \sin p x \rd x\) Derivative of $e^{a x}$, Definition of Cosecant
\(\ds \) \(=\) \(\ds -\frac p {\alpha^2 + p^2}\) Definite Integral to Infinity of $e^{-a x} \sin b x$

so:

\(\ds \map I \alpha\) \(=\) \(\ds -p \int \frac 1 {\alpha^2 + p^2} \rd \alpha\)
\(\ds \) \(=\) \(\ds -\arctan \frac \alpha p + C\) Primitive of $\dfrac 1 {x^2 + a^2}$

for all $\alpha \ge 0$ and constant $C \in \R$.

We therefore have:

\(\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x\) \(=\) \(\ds \map I a - \map I b\)
\(\ds \) \(=\) \(\ds \paren {-\arctan \frac a p + C} - \paren {-\arctan \frac b p + C}\)
\(\ds \) \(=\) \(\ds \arctan \frac b p - \arctan \frac a p\)

$\blacksquare$


Sources