TITLE: How do we evaluate this general integral?
QUESTION [6 upvotes]: From convolving two chi-squared distributions, I find the general integral
$$\int _0^yx^{\frac m 2 -1} \cdot (y-x)^{\frac n 2 -1}dx$$
Is there a way to evaluate this for arbitrary integers $n, m$? We can of course assume without loss of generality that $n>m$ or $n<m$.

REPLY [11 votes]: Hint.  By the change of variable
$$
t=\frac xy,\qquad dx=y \,dt,\qquad 
$$one gets
$$
\int _0^yx^{\frac m 2 -1} \cdot (y-x)^{\frac n 2 -1}dx=y^{\frac m 2+\frac n 2 -1}\int _0^1t^{\frac m 2 -1} \cdot (1-t)^{\frac n 2 -1}dt
$$ then one may use the beta integral result
$$
\int _0^1t^{a -1} \cdot (1-t)^{b -1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0,\,b>0.
$$