What is the convolution of $ t^m \ast t^n $ where m and n are general positive integers?

Question

I want to work out a general form of the convolution $ t^m \ast t^n $. I started with:

$$ t^m \ast t^n = \int_0^t (t - \tau)^m \tau^n \ d \tau $$

Then using the binomial expansion:

$$ \int_0^t \tau^n(\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau)^k) \ d \tau $$

Integration by parts - let $ u = \sum (...)$ and $ v' = \tau^n \ d\tau $ :

$$ \frac{\tau ^{n+1}}{n+1}(\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau )^k) - \int \tau^n(\frac{\mathrm{d} }{\mathrm{d} \tau } (\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau )^k) ) $$

But I am unsure on how to differentiate the sum with the binomial in it.

Answer 1

\begin{eqnarray*} t^m \ast t^n = \int_0^t (t - \tau)^m \tau^n \ d \tau. \end{eqnarray*} Why not substitute $\tau = t \sigma $ and use the beta funcion \begin{eqnarray*} t^m \ast t^n = t^{n+m+1} \int_0^1 (1 - \sigma)^m \sigma^n \ d \sigma \\ = t^{n+m+1} \frac{m! n!}{(m+n+1)!}. \end{eqnarray*}

Edit: the beta function will give \begin{eqnarray*} \int_0^1 (1 - \sigma)^m \sigma^n \ d \sigma = \frac{\Gamma(m+1) \Gamma(n+1) }{\Gamma(m+n+2)} \end{eqnarray*} If $m$ and $n$ are not positive integers.

Answer 2

Putting $u=t\tau$ gives $du=t \, d\tau$, $$ \int_0^t (t-\tau)^m \tau^n \, d\tau = t^{m+n-1} \int_0^1 (1-u)^m u^n \, du. $$ The constant is given by the Beta-function, which for integers can be expressed in terms of factorials as $$ \int_0^1 (1-u)^m u^n \, du = \frac{m!n!}{(m+n+1)!}. $$