# Integral of repeated convolution of the unit step function

## Background

Let $\theta$ be the unit step function: $$\theta(x) = \begin{array}{ll} \left\{ \begin{array}{ll} 0 & x \lt 0 \\ 1 & x\ge 0. \end{array}\right. \end{array}$$ Further, the convolution of two functions $f$ and $g$ is the function: $$(f \star g)(x) = \int_{-\infty}^{\infty} f(x-y)g(y) dy.$$ Let $\theta^{\star n}(x)$ denote the repeated convolution (and not pointwise multiplication) of the unit step function.

## Question

I know that, by simple integration, $$\theta^{\star n}(x) = \theta(x) \frac{x^{n-1}}{(n-1)!}$$ but I am unable to follow how this integral was calculated (the detailed calculations were not provided, just this result). Can anybody give me a hint? I guess my question is: If you didn't knew the result, how would you come up with it. Thank you.

• Let $f(x)=\theta^{\star n}(x)$ and $g(x)=\theta(x)$. Then both $f(x)$ and $g(x)$ are zero if $x<0$. So the integral is limited to values of $y$ for which $x-y$ and $y$ are both positive. – Empy2 Sep 27 '13 at 7:50

Firstly, $\theta^{\star1}(x)=\theta(x) = \theta(x)\frac{x^0}{0!}$.

Now, assume statement is true for some $k\in\Bbb N$. Then, for $x\ge0$,

\begin{align} \theta^{\star (k+1)}(x)=&\left(\theta\star\theta^{\star k}\right)(x)\\ =& \int_{-\infty}^{+\infty} \theta(x-y)\cdot\theta(y)\frac{y^{k-1}}{(k-1)!}dy\\ =& \int_{0}^{x} \frac{y^{k-1}}{(k-1)!}dy\\ =& \left[\frac{y^k}{k(k-1)!}\right]_0^x\\ =& \frac{x^k}{k!}\\ \end{align}

For $x<0$, $\theta^{\star (k+1)}(x)=0$ as $\theta(x-y)\cdot\theta(y)=0$ for all $y\in\Bbb R$. Therefore, combining two cases, $\theta^{\star (k+1)}(x) = \theta(x)\frac{x^k}{k!}$, and this finishes the induction.

Remaining to prove is that convolution is associative, and this repeated convolution can be performed in any order.

If you know Laplace transform, you can also show this result easily using that.

• Thanks peterwhy, that's very helpful. – COOLSerdash Sep 30 '13 at 8:54