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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.70$