Integral to Infinity of Sine p x over x

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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 1

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.

$\blacksquare$


Proof 2


This theorem requires a proof.
In particular: Use Primitive of Sine of a x over x, and also the analytic solution as found in Integration, 2nd ed.
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Sources

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