I gave a challenge question at the end of class a week or so ago. Here I will give the solution and show that it gives us something rather strange and surprisingly useful.

I wrote down the following, and asked you to prove it:

$\int_0^\infty e^{-t} t^N dt=N!$

For $N\ge 0, N\in \mathbb{Z}$. Now, N! can be thought of as the number of different orderings of pulling N objects out of a bag (without replacement) when they are all different. If you have N things in a bag, then there are N possible things that you can pull out first. There are then N-1 ways of pulling out the next object, N-2 ways of pulling out the next, etc. and finally, when you’ve pulled out N-1 objects there’s only a single possibility of pulling out the last. So:

$N!=N(N-1)(N-2)(N-3)...3.2.1$

And the number of ways of pulling no objects out of a bag is 1, because you just don’t pull anything out. So, 0!=1.

ok, so let’s see how we can prove the above integral. Well, whenever we want to prove something for an infinite sequence of integers, we should think of induction. The first step here is thus to prove that the expression holds true for the base case, which in this case is 0:

$\int_0^\infty e^{-t} t^0 dt=\int_0^\infty e^{-t} dt=\left.-e^{-t}\right|_0^\infty=0-(-1)=1$

which is indeed =0!, so the base case holds.

Now, let’s assume that it holds true for some integer $N=k$. This is our inductive hypothesis:

$\int_0^\infty e^{-t} t^k dt=k!$

Now, we want to check that so long as the hypothesis is true, the expression holds true for $N=k+1$. In trying to prove this we will be allowed to use our inductive hypothesis. So, let’s look at:

$\int_0^\infty e^{-t} t^{k+1} dt$

We somehow want to manipulate this into a form which includes the case $N=k$ above. To do this, we would need to reduce the power of t, from k+1 to k. We know that in general, integrating by parts can be used to do such a trick, because we will choose a part of the product in the integral to integrate, and a part to differentiate. So, let’s choose:

$t^{k+1}=u$ and $e^{-t}dt=dv$

This means that:

$dt (k+1)t^k=du$ and $-e^{-t}=v$

Now we plug this into our integration by parts formula $\int u dv=uv-\int v du$ and we get:

$\int_0^\infty e^{-t} t^{k+1} dt=\left.-t^{k+1}e^{-t}\right|_0^\infty+\int_0^\infty e^{-t}(k+1)t^k dt$

The first term vanishes when we plug in either of the limits, so we can write:

$\int_0^\infty e^{-t} t^{k+1} dt=(k+1)\int_0^\infty e^{-t}t^k dt$

But the last part is precisely what we had on the LHS of our inductive hypothesis, so we can simply substitute in the RHS of the inductive hypothesis, which gives:

$\int_0^\infty e^{-t} t^{k+1} dt=(k+1)k!=(k+1)!$

Which is what we wanted to prove. Thus, by the principle of mathematical induction, we have proved the original statement to be true.

So, why is this interesting? Well, one particularly interesting thing to note is that the factorial function in general is only applicable to positive integers. However, the integral expression doesn’t seem to specify integers. We can actually use the integral expression to define a function which is a bit more general than factorial, which, for positive integers is equal to our usual factorial, but is valid for all real numbers. We call this function the Gamma function and define it through a very similar integral:

$\Gamma(z)=\int_0^\infty e^{-t} t^{z-1} dt$

Such that $\Gamma(z+1)=(z)!$ for $z\ge 0, z\in\mathbb{Z}$.

We can plot both $\Gamma(z+1)$ and z! for positive integers and see that it does indeed coincide. For $z\ge 0$ the Gamma function looks all nice and smooth, but you can see that for negative z, something very funny is going on!

In fact, the Gamma function seems to be doing something funky for negative integers only. There is a huge amount of interesting mathematics behind this, but most of it requires understanding complex numbers. Let’s see if we can discuss this later when we get onto this amazing topic!

 How clear is this post?