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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $19$