Definite Integral to Infinity of Exponential of -a x by Sine of b x over x

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Theorem

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

where $a$ and $b$ are real number with $a > 0$.


Proof

Take $a$ constant and define:

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

We have:

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

so:

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

for some constant $C \in \R$.

We have:

\(\ds \map I 0\) \(=\) \(\ds \int_0^\infty \frac {e^{-a x} \sin 0} x \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty 0 \rd x\)
\(\ds \) \(=\) \(\ds 0\)

on the other hand:

\(\ds \map I 0\) \(=\) \(\ds \arctan 0 + C\)
\(\ds \) \(=\) \(\ds C\)

so:

$C = 0$

So we have:

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

as required.

$\blacksquare$


Sources

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