Partial Gamma Function expressed as Integral

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Theorem

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

Let $\map {\Gamma_m} x$ denote the partial Gamma function, defined as:

$\map {\Gamma_m} x := \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} }$


Then:

$\ds \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$

for $x > 0$.


Proof

Lemma

First we establish:


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

for $x > 0$.


The proof continues by induction on $m$.

For all $m \in \Z_{\ge 1}$, let $\map P m$ be the proposition:

$\ds \map {\Gamma_m} x = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$


Basis for the Induction

$\map P 1$ is the case:

\(\ds m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t\) \(=\) \(\ds \int_0^1 \paren {1 - t} t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \rd t - \int_0^1 t^x \rd t\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {t^x} x} 0 1 - \intlimits {\frac {t^{x + 1} } {x + 1} } 0 1\)
\(\ds \) \(=\) \(\ds \frac 1 x - \frac 1 {x + 1}\)
\(\ds \) \(=\) \(\ds \frac {x + 1 - x} {x \paren {x + 1} }\)
\(\ds \) \(=\) \(\ds \frac 1 {x \paren {x + 1} }\)
\(\ds \) \(=\) \(\ds \frac {1^x 1!} {x \paren {x + 1} }\)

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \map {\Gamma_k} x = k^x \int_0^1 \paren {1 - t}^k t^{x - 1} \rd t$


from which it is to be shown that:

$\ds \map {\Gamma_{k + 1} } x = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t$


Induction Step

This is the induction step:

With a view to expressing the primitive in the form:

$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \paren {1 - t}^{k + 1}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\rd u} {\rd t}\) \(=\) \(\ds -\paren {k + 1} \paren {1 - t}^k\) Power Rule for Derivatives


and let:

\(\ds \frac {\rd v} {\rd t}\) \(=\) \(\ds t^{x - 1}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {t^x} x\) Primitive of Power


Then:

\(\ds \) \(\) \(\ds \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^x \intlimits {-\paren {k + 1} \paren {1 - t}^k \frac {t^x} x} 0 1 - \paren {k + 1}^x \int_0^1 \paren {-\paren {k + 1} \frac {t^x} x} \paren {\paren {1 - t}^k} \rd t\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^{x + 1} } x \int_0^1 \paren {1 - t}^k t^x \rd t\) $u v$ term vanishes
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^{x + 1} } x \frac {\map {\Gamma_k} {x + 1} } {k^{x + 1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^{x + 1} } x \frac {k^{x + 1} k!} {k^{x + 1} \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + 1 + k} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^{x + 1} k!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^x \paren {k + 1}!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }\)
\(\ds \) \(=\) \(\ds \map {\Gamma_{k + 1} } x\)


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall m \in \Z_{\ge 1}: \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$

for $x > 0$.

$\blacksquare$


Sources