Definite Integral to Infinity of Power of x over 1 + x

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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 1

\(\ds \int_0^\infty \frac {x^{p - 1} \rd x} {1 + x}\) \(=\) \(\ds \int_0^\infty \frac 1 {p x^{p - 1} } \cdot \frac {x^{p - 1} \rd t} {1 + t^{1 / p} }\) 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

$\blacksquare$


Proof 2


This theorem requires a proof.
In particular: We can probably use Reduction Formula for Primitive of Power of x by Power of a x + b but I can see it being a long slog
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Sources

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