Hard looking yet beautiful integral

Never judge a book by its cover: this is the case. This integral requires a bit of work and knowledge, as you need to know about the Gaussian integral (I covered it in these two posts: https://atomic-temporary-167386734.wpcomstaging.com/gaussian-integral-using-feynmans-technique/https://atomic-temporary-167386734.wpcomstaging.com/integral-of-e-xt-using-feynmans-technique/) and the most basic method of integration, u-substitution (https://addjustabitofpi.wordpress.com/2019/10/05/integration-techniques-u-substitution/ & https://addjustabitofpi.wordpress.com/2019/10/13/integration-techniques-u-substitution-part-2/). So let’s start!

Look at the Gaussian integral:

Instead of having –x², we have -(a-x)²/4kt. If you think about it though, we can write this as

so that, when substituting, we will get the form we know. So,

As you can see, we can do some simplifications:

but there is still something to do: since we substituted, we are now integrating with respect to u, therefore all x‘s must be written in terms of u; also, the bounds change, since those are for x.

I stopped at -x because we have e^-x inside the integral, therefore there’s no point in finding x and then inverting the sign again.

Let’s put everything together:

Since we have the – sign in front of the integral, we can swap the bounds:

There is still one thing we can do. e is raised to a sum, so we can write it as a product:

e^-a is a constant (with respect to u), meaning we can move it outside the integral:

Here instead, we can do the opposite:

We have u², we have u… it looks like the expansion of the square of a binomial! To make things easier, let’s factor out the – sign:

Now we are going to “complete the square”: say you have x²-2x; you can see that the missing term in order to make a square is 4, so you can write it as x²-2x+4-4= (x-2)²-4. (-4 because if you simply add 4, the two things won’t be equivalent anymore). This is exactly what we are doing here. First though, let’s take that 4 out of the square root:

The missing term is kt, so

At this point, we can perform another substitution:

This time the bounds stay the same, since nothing changes when you plug infinity and -infinity into u-sqrt(kt).

This is the Gaussian integral! And we saw here that this is equal to sqrt(pi).



And that’s it!

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