MIT Integration Bee 2023 – Finals Problem 1 – An Ode to Some Beautiful Integrals

The MIT Integration Bee (which started at MIT in 1981) is the first of many integral calculus competitions that are held at educational institutions all over the world. Interested participants first take a qualifying exam, after which the successful candidates are drawn up against each other, tournament style, in brackets. Generally, the knockout rounds involve participants solving integrals in front of a chalkboard, often with a time limit. The champion gets the title of “Grand Integrator”, along with prizes in the form of cash, vouchers, or books.

The 2023 MIT Integration Bee was held Thursday, January 26, 2023. So over the next couple of weeks, I’ll try and showcase some of my solutions to the problems in the finals of this year’s bee. Without further ado, let’s begin! 🙂

Problem 1 of this year’s integration bee finals asks us to evaluate

$\displaystyle I = \int_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{(\sin x + \cos x)^2} \mathrm{d}x.$

We first factor out $\cos x$ from the denominator. What we have then is

$\displaystyle I = \int_0^{\pi/2} \frac{\sqrt[3]{\tan x}}{(\tan x + 1)^2} \sec^2 x \: \mathrm{d}x.$

This looks neat and we can substitute $\tan x = z$ , which means that $\displaystyle \mathrm{d}z = \sec^2 x \: \mathrm{d}x$ . Our integral reduces to

$\displaystyle I = \int_0^{\infty} \frac{z^{1/3}}{(1+z)^2} \: \mathrm{d}z.$

After a little bit of head scratching at this point, I’ll admit I looked up the definition of the beta function I needed to tackle this integral. The requisite relation turns out to be

$\displaystyle \mathrm{B} (m,n) = \int_0^{\infty} \frac{z^{m-1}}{(1+z)^{m+n}} \: \mathrm{d}z.$

One comparison with our integral tells us we need to solve $m-1 = 1/3$ and $m+n = 2$ and so, $m=4/3$ and $n=2-4/3= 2/3$ .

Therefore,

$\displaystyle I = \mathrm{B} \left( \frac{4}{3}, \frac{2}{3} \right).$

We now need two properties. The first is the relationship of the beta function to the gamma function

$\displaystyle \mathrm{B} (m,n) = \frac{\Gamma (m) \Gamma (n)}{\Gamma (m+n)},$

and the second is the fundamental recurrence relation for the gamma function

$\Gamma (y+1) = y \Gamma (y)$

on an appropriate domain. We can then write

$\displaystyle I = \mathrm{B} \left( \frac{4}{3}, \frac{2}{3} \right) = \frac{\Gamma \left( \frac{4}{3} \right) \Gamma \left( \frac{2}{3} \right)}{\Gamma \left( \frac{4}{3} + \frac{2}{3} \right)} = \frac{\Gamma \left( \frac{4}{3} \right) \Gamma \left( \frac{2}{3} \right)}{\Gamma (2)} = \frac{\Gamma \left( 1 + \frac{1}{3} \right) \Gamma \left( \frac{2}{3} \right)}{\Gamma (2)} = \frac{\frac{1}{3} \Gamma \left( \frac{1}{3} \right) \Gamma \left( \frac{2}{3} \right)}{\Gamma (2)}.$

For positive integers $n$ , $\Gamma (n) = (n-1)!$ and so, $\Gamma (2) = 1! = 1$ . Therefore,

$I = \frac{1}{3} \Gamma \left( \frac{1}{3} \right) \Gamma \left( \frac{2}{3} \right) = \frac{1}{3} \Gamma \left( \frac{1}{3} \right) \Gamma \left( 1-\frac{1}{3} \right).$

We can then use the Euler reflection formula for the gamma function, which states that for non-integral $y$ ,

$\displaystyle \Gamma (1-y) \Gamma (y) = \frac{\pi}{\sin \pi y}.$

We can then wrap things up by writing

$\displaystyle I = \frac{1}{3} \cdot \frac{\pi}{\sin \left( \frac{\pi}{3} \right)} = \frac{1}{3} \cdot \frac{\pi}{\frac{\sqrt{3}}{2}} = \frac{2 \pi}{3 \sqrt{3}} = \boxed{\mathbf{\frac{2\sqrt{3} \pi}{9}}}.$

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