Definite Integral to Infinity of Power of x over 1 + x/Proof 1
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Theorem
- $\ds \int_0^\infty \dfrac {x^{p - 1} \rd x} {1 + x} = \frac \pi {\sin \pi p}$
for $0 < p < 1$.
Proof
\(\ds \int_0^\infty \frac {x^{p - 1} \rd x} {1 + x}\) | \(=\) | \(\ds \int_0^\infty \frac 1 {p x^{p - 1} } \cdot \frac {x^{p - 1} \rd t} {1 + t^{1 / p} }\) | substituting $t = x^p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 p \int_0^\infty \frac {\rd t} {1 + t^{1 / p} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 p \cdot \frac \pi {\frac 1 p} \map \csc {\frac \pi {\frac 1 p} }\) | as $0 < p < 1$, $\dfrac 1 p > 1$, so we can apply Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \map \csc {\pi p}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi {\sin \pi p}\) | Definition of Cosecant |
$\blacksquare$