# Partial Gamma Function expressed as Integral

## 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 $x > 0$.

$\blacksquare$