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

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Theorem

$\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} }$

where:

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


Proof

Fix $p$ and set:

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

for all $\alpha \ge 0$.

Then:

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

We have:

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

so:

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

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

We then have:

\(\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x\) \(=\) \(\ds \map I a - \map I b\)
\(\ds \) \(=\) \(\ds \paren {-\frac 1 2 \map \ln {a^2 + p^2} + C} - \paren {-\frac 1 2 \map \ln {b^2 + p^2} + C}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\map \ln {b^2 + p^2} - \map \ln {a^2 + p^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} }\) Difference of Logarithms

$\blacksquare$


Sources