An Interesting Problem in Integral Calculus

Solved with the powerful Gamma function

Photo by Vitaly Gariev on Unsplash

Computing derivatives is a systematic procedure. The rules differentiation of most functions reduces the operation to a routine computation, with no guesswork.

Integration is a different kind of activity. Most functions have no elementary antiderivative, and there is no single recipe that fits every case. What we learn is really pattern recognition / systematic guesswork where you recognise the shape of an integrand, reach for a move that has worked on a similar integrand before, and see whether it works. Change of variables and integration by parts are some of these staple moves. One of these may solve the integral, or it may lead nowhere.

This blog is about one such problem. I assume you are comfortable with basic integration techniques. I do not assume that you already have the specific tools this problem needs. So rather than reach for them and directly solve the problem, I build them from scratch and only then apply them on the integral. Along the way I cover a bit more than necessary, and I hope you enjoy it.

Here is the integral we want to evaluate.

Result 1

Here, consider k to be a positive integer and the answer must be in terms of k and other constants. Give it a try before scrolling on.

Build up to the solution

We begin with the definition of the Gamma function and build everything from there.

The Gamma function

The Gamma function is defined by the following integral.

Result 2

Verify for yourselves using this definition that Γ(1) = 1. The definition also satisfies a recurrence relationship. Replace n by z + 1 in Result 2 and integrate by parts and simplify to obtain the following result.

Result 3

We can use this recurrence relation and Γ(1) = 1 to establish that: Γ(n+1) = n!

In the definition per Result 2, substitute x = t². This transforms the integral to the following:

Result 4

The product of two Gamma functions

Write two copies of Result 4, one in x and one in y, and multiply them. The product is a double integral over the first quadrant.

Result 5

Switch to polar coordinates by setting x = r cos θ and y = r sin θ. This transforms the integral to the following. Recall that the area element in polar coordinates is r dr dθ.

Result 6

Collect the powers of r and split the double integral into two single variable integrals. This is allowed because the integration limits have no variables in them.

Result 7

In Result 7, the integral involving r bears resemblance to Result 4, allowing us to write:

Result 8

Rearrange the above to obtain the key result:

Result 9

Now relabel the exponents. Set 2m — 1 = a and 2n — 1 = b. This transforms Result 9 to:

Result 10

This holds for a and b greater than -1, equivalently m and n positive, which is also what justifies the polar change above.

Computing Γ(1/2)

Set a = b = 0 in Result 10. The integral on the left hand side evaluates to π/2. The expression on the right hand side. Use the fact that Γ(1) = 1 and rearrange further to obtain:

Result 11

This result is used extensively in statistics. It shows up while studying the Gaussian distribution.

The Beta function and its link to the Gamma function

This is the bit beyond what the problem needs, promised at the start. The Beta function is defined as follows.

Result 12

Transform the above integral by substituting t = sin²(θ). This leads to:

Result 13

The right hand side resembles Result 10 allowing us to write:

Result 14

The double factorial

A double factorial, written with two exclamation marks. For an odd argument it is the product of the odd numbers. For any odd number:

Result 15

For an even argument it is the product of the even numbers. This can also be written in terms of the ordinary factorial:

Result 16

Together, Results 15 and 16 cover every integer from 1 to 2k, so an ordinary factorial is the product of its odd and even double factorials.

Result 17

Result 17 can be rearranged as follows by using Result 16:

Result 18

The Gamma function at the half-integers

The value of Γ(k + 1/2) can be found by first using Result 3.

Result 19

You can verify yourselves that the following result is true:

Result 20

Plugging into Result 19 and Using the value of Γ(1/2):

Result 21

Replace Result 18 into Result 21 to obtain:

Result 22

Finally, the Solution

We return to the integral in Result 1. Apply Result 10 with cosine power 0 and sine power 2k. The factor 2²ᵏ is a constant.

Result 23

Use the fact that Γ(k+1) = k! and Result 11 and Result 22 to simplify the above. This gives:

Result 24

A good eye will now spot the binomial coefficient. Recall that it is defined as:

Result 25

The binomial coefficient has two notations. I am using both for clarity. With this definition, Result 14 can be re-written as such:

Result 26

This is the final answer. Once all the necessary results were in hand, solving the problem is straightforward. For me, the build up to the solution is more interesting than the solution itself.

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