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

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