# Definite Integral from -a to a of Power of a plus x by Power of a minus x

## Theorem

$\ds \int_{-a}^a \paren {a + x}^{m - 1} \paren {a - x}^{n - 1} \rd x = \paren {2 a}^{m + n - 1} \frac {\map \Gamma m \map \Gamma n} {\map \Gamma {m + n} }$

where:

$\Gamma$ denotes the Gamma function
$a$, $m$ and $n$ are positive real numbers.

## Proof

Note the resemblance of this result to the integral defining the beta function.

In view of this, we apply the substitution:

$u = \dfrac {a + x} {2 a}$

We then have, by Derivative of Power:

$\dfrac {\d u} {\d x} = \dfrac 1 {2 a}$

and:

 $\ds 1 - u$ $=$ $\ds 1 - \frac {a + x} {2 a}$ $\ds$ $=$ $\ds \frac {2 a - a - x} {2 a}$ $\ds$ $=$ $\ds \frac {a - x} {2 a}$

so that:

$a + x = 2 a u$

and:

$a - x = \paren {2 a} \paren {1 - u}$

and:

$\dfrac {\d x} {\d u} = 2 a$

We therefore have:

 $\ds \int_{-a}^a \paren {a + x}^{m - 1} \paren {a - x}^{n - 1} \rd x$ $=$ $\ds 2 a \int_0^1 \paren {2a u}^{m - 1} \paren {2 a \paren {1 - u} }^{n - 1} \rd u$ Integration by Substitution $\ds$ $=$ $\ds \paren {2 a}^{m + n - 2 + 1} \int_0^1 u^{m - 1} \paren {1 - u}^{n - 1} \rd u$ $\ds$ $=$ $\ds \paren {2 a}^{m + n - 1} \map \Beta {m, n}$ Definition of Beta Function: Definition 1 $\ds$ $=$ $\ds \paren {2 a}^{m + n - 1} \frac {\map \Gamma m \map \Gamma n} {\map \Gamma {m + n} }$ Definition of Beta Function: Definition 2

$\blacksquare$