Integral to Infinity of Sine p x over x/Proof 1
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Theorem
- $\ds \int_0^\infty \frac {\sin p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ 0 & : p = 0 \\ -\dfrac \pi 2 & : p < 0 \end {cases}$
Proof
Let $p > 0$.
We have:
\(\ds \int_0^\infty \frac {\sin p x} x \rd x\) | \(=\) | \(\ds \frac 1 p \int_0^\infty \frac {\sin t} { \frac 1 p t} \rd t\) | substituting $t = p x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac {\sin t} t \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2\) | Dirichlet Integral |
Then:
\(\ds \int_0^\infty \frac {\sin \left({- p x}\right)} x \rd x\) | \(=\) | \(\ds -\int_0^\infty \frac {\sin p x} x \rd x\) | Sine Function is Odd | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac \pi 2\) | per above computation |
For $p = 0$, we have:
\(\ds \int_0^\infty \frac {\sin 0 x} x \rd x\) | \(=\) | \(\ds \int_0^\infty \frac 0 x \rd x\) | Sine of Zero is Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Hence the result.
$\blacksquare$