An Integral For Feynman Fans and Another for Math Lovers

Can you solve these integrals?

In this article we will look at two integrals; one focussing on Feynman’s technique and the other which uses infinite sums and leads to deriving the Gamma function.

Recently I posted an article about 2 very useful integration techniques, and I wanted to go even further using one of those techniques, the Feynman method. If you are not aware of this technique I suggest you read this article to familiarise yourself before coming back to give the first integral a go!

The First Integral

A tricky part when using Ft is knowing where to place our variable α. A few options jump out at us perhaps the power of the x in the denominator, the power of x in the numerator? I will save you the time and effort of exploring which placement of our parameter leads to a solution and tell you that the parameter will be the coefficient of the x in the ln term. So with this in mind, let us define our integral I(α) and quickly find the case of I(0).

Our target integral is I(1). We will need the case of I(0)=0 later on when we integrate our result for I’(α). So first to differentiate I(α) with respect to α. We should usually be very careful when we do this step, in particular, whether we can take the derivative inside the integral sign. We can do this when the integrand is continuous and converges, as well as the partial derivatives existing and themselves being continuous. But let us not worry about this too much for the time being!

Next the simplest way forward would to perform a partial fraction decomposition on the integrand.

Now we can rewrite our integrand using this partial fraction decomposition and from this we can integrate each fraction.

The next step in our method is to integrate both sides with respect to α. Sometimes we can simply do the indefinite integral on both sides and use our value I(0)=0 to find the constant of integration. However, for this integral we will integrate with the bounds 0 and 1. The motivation behind this is that notice for our result for I’(α) we have the original integral so to integrate indefinitely we would have to evaluate this integral, defeating the purpose of our method and going around in circles! This part was by far the most satisfying, so hopefully you too can appreciate this next step.

Personally, I love this method and love the crazy results we can obtain through using it. Additionally we generalise the result for our parameter α, hence we can find other integrals similar to this one. Have a go at quickly finding these two integrals!

The Second Integral

Some readers may recognise this is very similar to the Bernoulli integral but we have -x in the exponent rather than x.

How might we begin this integral? Our normal methods, like integration by parts or substitution seem far fetched and nothing immediately comes to mind. Perhaps I could nudge you in the right direction, then you can try yourselves. Our first step is to rewrite the integrand as exp(-xln(x)). So try this and see where this can take you before carrying on with our solution development.

We will now make use of the Maclaurin series for e^x. The motivation behind this is that we then get the integral of a sum. If we are careful with convergence, we can switch the summation and integration operator.

Next we can make a substitution, trying to guide the integrand to resemble the Gamma function.

Remembering the Gamma function

let us make the substitution (n+1)u=t to try and obtain the Gamma function.

I thought this would be a good opportunity to derive the Gamma function. The proof is not too long and involves integration by parts to obtain a recurrence formula, which nicely leads us to the result Γ(s+1).

With this result we can substitute this in to our sum so obtain our final result.

I find there is something fascinating about the integral, or continuous sum if you like, of x^(-x) from 0 to 1 is equal to the discrete sum of x^(-x) over all positive integers.

Thank you for reading. If you have any thoughts on these two integrals or figured out the answers to the two proposed at the end of the first integral, feel free to comment