Taking the Derivative Under the Integral, or, First Post

Well, this is my first post. I'm just making this blog to record thoughts I have about science, school, my research, and anything else. I don't really expect anyone to read it, except maybe future me.

The first thing I'm going to post is a math trick that I really like. To me it's called 'taking a derivative under the integral.' I'm going to show how it works in the context of integrating the Planck function. The Planck function is the 'spectral radiance' as a function of frequency (or wavelength), and it describes the radiation a blackbody at a certain temperature will emit. 'Spectral radiance' is a measure of energy, in units of W/m^2/Hz/sr (watts/meters^2/hertz/steradians, or, energy per time per area per frequency per solid angle). Here is the function:

Integrating this function over all frequencies will give the total power per area per solid angle emitted by a blackbody at a certain temperature. h*nu and k*T are both energies, and I'm going to assume that h*nu is much greater than k*T, so that (h*nu)/(k*T) is a large number, so that exp(h*nu/k/T) is a very large number, so that exp(h*nu/k/T) - 1 ~ exp(h*nu/k/T). Now we have:

Since we don't care about the constants until the end, let's arbitrarily have 2*h/c^2 = N and h/k/T = M. Now we have:

So, the integral we are concerned with is this:

Since N is a constant, we can take it out of the integral, and just add it back in at the end. Now we have:

So, the idea is to add a variable in that we will make equal to 1 at the end, and take the derivative with respect to that variable so that the integral is simplified to something we can easily do. I will add the variable alpha into the exponential, and keep in mind it will ultimately be equal to 1 so I'm not really changing anything:

Now, we want to take the derivative of this with respect to alpha so that it can replace what we currently have in the integral. Here are the steps:

So, we're getting close to having what is in the integral. To fine tune it:

(Remember, alpha=1) So there we have it. Now we can substitute this into the integral:

We can take the -1/M^3 out of the integral and put it back in later. We can also take the derivative out of the integral since it doesn't depend on nu. Now we have:

This integral is much simpler:

Now, we do the derivative:

So, I will put the N and the -1/M^3 back in:

Now, we set alpha=1 and plug in N=2*h/c^2 and M=h/k/T, and get our final result!

There you have it. This seems like an overly long process when I explain it all like this, but once you get the hang of it you can do it all pretty quickly. You could also have done 'integration by parts,' but personally I find this method easier and more straightforward.

It also works for arbitrary exponentials:

I'll leave it as an exercise to work it out yourself (heh ... it's fun, trust me!), but the solution is:

This is just one case, too. I bet there are tons of situations where it would be possible to 'differentiate under the integral.' It's just nice to keep in mind if you have a tough integral in front of you.

That's all for my first post :)