August | 2020 | mathgarage

Archive for August, 2020

Finding the Integral of 1/(x^4+1) from 0 to infinity the EASY WAY

Posted: August 20, 2020 in Mathematics
Tags: math

Problem:

$\displaystyle \int_0^\infty \frac{1}{x^4+1} \text{ d}x$

Essentially there are two major ways of solving this integral, either by partial fraction (the most painful route), hint:

$\displaystyle x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2 - 2x^2$

or by Residue theorem. There is a nice video explained on youtube by Dr Peyam. I think he is a funny dude.

Of course there are many different methods out there such as the crazy approach here on this youtube video below:

But the problem is that these approaches require you to have some sorts of university math skill in order to do them. Today, I am going to present you a very simple and elegant way to do this integral. Even any highschool student can comprehend. All you will need is the technique of “Integration by substitution” aka “the u-sub method”. Here we go.

First, we substitute

$\displaystyle x = \frac1t, \qquad \text{d}x = -\frac{1}{t^2} \text{ d}t$

Then we get

$\displaystyle \int_0^\infty \frac{1}{x^4+1} \text{ d}x = \int_\infty^0 \frac{-1/t^2}{(1/t)^4+1} \text{ d}t = \int_0^\infty \frac{t^2}{t^4+1} \text{ d}t$

Here comes the tricky part, since x and t are just dummy variables. We can take the average of the two integrals that are equal as follow.

$\displaystyle \frac12 \int_0^\infty \frac{x^2+1}{x^4+1} \text{ d}x$

Then divides top and bottom by $x^2$ and using another substitution:

$\displaystyle w=x-\frac1x, \qquad \text{d}w = \left(1+\frac{1}{x^2}\right)\text{d}x$

$\displaystyle \frac12 \int_0^\infty \frac{x^2+1}{x^4+1} \text{ d}x = \frac12 \int_0^\infty \frac{1+1/x^2}{x^2+1/x^2} \text{ d}x = \frac12 \int_0^\infty \frac{1+1/x^2}{(x-1/x)^2+2} \text{ d}x$