The Strange Integral

$\begin{aligned} I & = \int_{-\infty}^\infty \frac 1{1+x^2} dx & \small \color{#3D99F6} \text{Since the integrand is even.} \\ & = 2 \int_0^\infty \frac 1{1+x^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \\ & = 2 \int_0^\frac \pi 2 \frac {\sec^2 \theta}{1+\tan^2 \theta} d\theta & \small \color{#3D99F6} \text{Note that }1 + \tan^2 \theta = \sec^2 \theta \\ & = 2 \int_0^\frac \pi 2 d\theta \\ & = 2 \theta\ \bigg|_0^\frac \pi 2 = \boxed \pi \end{aligned}$

We will first prove that:

$\int\frac{1}{1+x^2} dx = tan^{-1}(x) + C$

If we define f(x) as:

$\frac{d}{dx} tan(x)$

Then by the inverse function rule:

$\frac{d}{dx} tan^{-1}(x) = \frac{1}{f(tan^{-1}(x))}$

By the quotient rule:

$f(x) = \frac{d}{dx} \frac{sin(x)}{cos(x)} = \frac{cos^2(x)-(-sin^2(x))}{cos^2(x)} = \frac{1}{cos^2(x)} = sec^2x$

It follows that:

$\frac{d}{dx} tan^{-1}(x) = \frac{1}{sec^2(tan^{-1}(x))} = \frac{1}{1+tan^2(tan^{-1}(x))} = \frac{1}{1+x^2}$

Now that we have determined the indefinite integral:

$\int_{-\infty}^\infty\frac{1}{1+x^2} dx = (lim_{x\rightarrow\infty}tan^{-1}(x)) - (lim_{x\rightarrow-\infty}tan^{-1}(x)) = \frac{\pi}{2} - \frac{-\pi}{2} = \pi$

Upon first inspection of the integral, we notice that this is a definite integral form of the identity $\\ \int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx = arctan(x)$ However, because the bounds are infinities, this definite integral is improper and therefore must be evaluated via limits $\Rightarrow \lim _{ t\rightarrow \infty }{ \int _{ -t }^{ t }{ \frac { 1 }{ 1+{ x }^{ 2 } } dx } } \\ \Rightarrow \lim _{ t\rightarrow \infty }{ \arctan { (t) } -\arctan { (-t) } } \\ =\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \quad =\quad \pi \\$