How do we evaluate this general integral?

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 $$ \int _0^1t^{a -1} \cdot (1-t)^{b -1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0,\,b>0. $$

Hint. By letting $t=x/y$, we have that $$\int _0^yx^{\frac m 2 -1} \cdot (y-x)^{\frac n 2 -1}dx= y^{\frac m 2 +\frac n 2-1} \int _0^1t^{\frac m 2 -1 } \cdot (1-t)^{\frac n 2 -1}dt=y^{\frac m 2 +\frac n 2-1}B\left(\frac{m}{2},\frac{n}{2}\right)$$ where $B(x,y)$ is the Beta function.