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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.33$