Today, we check how we may exploit some interesting symmetries when evaluating definite integrals. The problem is from 1987 Putnam. I personally think this question is so elegant as a nice symmetric translation of coordinates hugely simplify the calculation. Imagine what would happened if we tried to approach this by directly calculating its antiderivative!

Problem.

Find the exact value of

$$ I:=\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log (x+3)}}dx $$

Solution.

Introducing a change in variables

$$ y=x-6 $$

shifts the domain of integration to (-4,-2). Note that the above change of coordinates is not random. It is intuitively a result of endeavor to somehow ‘maintain’ symmetry (note that the integral domain has just been ‘reflected’ about the y-axis), and exploit how the integrand changes. Then we have

$$ I=\int_{-4}^{-2}\frac{\sqrt{\log(3-y)}}{\sqrt{\log(3-y)}+\sqrt{\log(9+y)}}dy. $$

now, another motivation is to actually make the integral domain same by additionally using the inversion

$$ z=-y $$

Then we have

$$ I=\int_{-4}^{-2}\frac{\sqrt{\log(3-y)}}{\sqrt{\log(3-y)}+\sqrt{\log(9+y)}}dy=\int_2^4\frac{\sqrt{\log(3+z)}}{\sqrt{\log(3+z)}+\sqrt{\log{(9-z)}}}dz $$

This is a marvelous result. Some may question why it is useful to ‘manipulate’ the integral that we want to find, because it seems unsolved in either way. However, it is magnificent in the sense that we have obtained two different expressions for the same integral with same integral domain! .

We can add the two.

$$ 2I=I+I\\=\int_2^4\frac{\sqrt{\log(3+z)}}{\sqrt{\log(3+z)}+\sqrt{\log{(9-z)}}}dz+\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log (x+3)}}dx\\=\int_2^4\frac{\sqrt{\log(9-x)}+\sqrt{\log(x+3)}}{\sqrt{\log(9-x)+}+\sqrt{\log (x+3)}}dx=\int_2^4dx=2. $$

Therefore, we conclude by dividing,

$$ I=1 $$