Convolution problem involving binomial

Question

Do I have to use binomial theorem? Also how does 1 get in the limit of integrand?

Answer 1

$$\begin{array}{rcl} (f * g) (t) &=& \displaystyle \int_0^t f(a) g(t-a) \ \mathrm da \\ &=& \displaystyle \int_0^t a^m (t-a)^n \ \mathrm da \\ &=& \displaystyle \int_0^t (ut)^m (t-ut)^n \ \mathrm d(ut) \\ &=& \displaystyle t \int_0^1 (ut)^m (t-ut)^n \ \mathrm du \\ &=& \displaystyle t \int_0^1 t^m u^m t^n (1-u)^n \ \mathrm du \\ &=& \displaystyle t^{m+n+1} \int_0^1 u^m (1-u)^n \ \mathrm du \\ \end{array}$$

where $a = ut$.

Note: The limits are $0$ and $t$ instead of $-\infty$ and $\infty$, because $f$ and $g$ are assumed to be only defined on $\Bbb R_{>0}$.