Partial Gamma Function expressed as Integral/Lemma

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Theorem

Let $m \in \Z_{\ge 1}$.

Then:

$(1): \quad \ds \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$

for $x > 0$.


Proof

Let:

\(\ds z\) \(=\) \(\ds \frac t m\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d t} {\d z}\) \(=\) \(\ds m\)

Recalculating the limits:

\(\ds t\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds 0\)
\(\ds t\) \(=\) \(\ds m\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds 1\)


Hence:

\(\ds \paren {1 - \frac t m}^m\) \(=\) \(\ds \paren {1 - z}^m\)
\(\ds t^{x - 1}\) \(=\) \(\ds \paren {m z}^{x - 1}\)
\(\ds \) \(=\) \(\ds m^{x - 1} z^{x - 1}\)


Thus $(1)$ can be written:

\(\ds \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t\) \(=\) \(\ds m^x \int_0^1 \paren {1 - z}^m z^{x - 1} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t\) changing the name of the dummy variable

$\blacksquare$


Sources

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