Community Problem Contributed By debroy0611

image containing math problem statement
A sleek integration problem from the 2024 CalTech Math Meet. Watch out for potholes.
Problem contributed by anonymous user
Let:
$I = \displaystyle \int \limits_0^{\infty} \dfrac{dx}{\left(x + \dfrac{1}{x}\right)^2}$       $...(i)$

-----------book page break-----------
Let $x = \dfrac{1}{u} \Rightarrow dx = -\dfrac{1}{u^2} du$  the limits become: $x = 0, u = \infty$ and $x = \infty, u = 0$

Therefore, we get:
$I =  \displaystyle \int \limits_{\infty}^0 \dfrac{-\dfrac{1}{u^2} du}{\left(u + \dfrac{1}{u}\right)^2}$

$\Rightarrow I = \displaystyle \int \limits_0^{\infty} \dfrac{\dfrac{1}{u^2} du}{\left(u + \dfrac{1}{u}\right)^2}$

Switching variable names from $u$ to $x$, we get:
$I = \displaystyle \int \limits_0^{\infty} \dfrac{\dfrac{1}{x^2} dx}{\left(x + \dfrac{1}{x}\right)^2}$       $...(ii)$

Adding equations $(i)$ and $(ii)$, we get:
$2I = \displaystyle \int \limits_0^{\infty} \dfrac{\left(1 + \dfrac{1}{x^2} \right) dx}{\left(x + \dfrac{1}{x}\right)^2}$

-----------book page break-----------
$\Rightarrow 2I = \displaystyle \int \limits_0^{\infty} \dfrac{\left(1 + \dfrac{1}{x^2} \right) dx}{\left(x - \dfrac{1}{x}\right)^2 + 4}$

Let $u = x - \dfrac{1}{x}$
$\therefore du = \left(1 + \dfrac{1}{x^2}\right) dx$, when $x = 0, u = -\infty$, and $x = \infty, u = \infty$

We get the integral as:
$2I = \displaystyle \int \limits_{-\infty}^{\infty} \dfrac{du}{u^2 + 4}$

$\Rightarrow 2I = \displaystyle \int \limits_{-\infty}^{\infty} \dfrac{du}{u^2 + 2^2}$

$\Rightarrow 2I = \dfrac{1}{2} tan^{-1}\left(\dfrac{u}{2}\right) \Biggr | _{-\infty}^{\infty}$

$\Rightarrow 2I = \dfrac{1}{2} \left[ \dfrac{\pi}{2} - \left(-\dfrac{\pi}{2}\right)\right]$

$\Rightarrow I  = \dfrac{\pi }{4} = \dfrac{3.14}{4} = 0.785 = 0.79$