Definite Integral to Infinity of Sine m x over x by x Squared plus a Squared
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Theorem
- $\ds \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$
where:
- $a$ is a positive real number
- $m$ is a non-negative real number.
Proof
Fix $a$ and set:
- $\ds \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x$
for $m \ge 0$.
We have:
\(\ds \map {I'} m\) | \(=\) | \(\ds \frac \d {\d m} \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac \partial {\partial m} \paren {\frac {\sin m x} {x \paren {x^2 + a^2} } } \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x\) | Derivative of $\sin a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi {2 a} e^{-m a}\) | Definite Integral to Infinity of $\dfrac {\cos m x} {x^2 + a^2}$ |
So by Primitive of $e^{a x}$:
- $\map I m = -\dfrac \pi {2 a^2} e^{-m a} + C$
for some constant $C \in \R$.
We have:
\(\ds \map I 0\) | \(=\) | \(\ds \int_0^\infty \frac {\sin 0 x} {x^2 + a^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty 0 \rd x\) | Sine of Zero is Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
On the other hand:
\(\ds \map I 0\) | \(=\) | \(\ds -\frac \pi {2 a^2} e^0 + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac \pi {2 a^2} + C\) | Exponential of Zero |
So:
- $C = \dfrac \pi {2 a^2}$
giving:
- $\ds \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.42$