# Old School Cool

Let's say that we want to evaluate $\displaystyle\int_0^{\infty}\frac{1}{1+x^2}\,dx$ without knowing the anti derivative of $\frac{1}{1+x^2}$. How could we do this? To start let's consider the unit circle, we know the unit circle has arc-length $L=r\theta = \theta$. We can equivalently write the arc length of the unit circle as \begin{aligned} L = & \int_0^{\theta} \,dt \\ = & \int_0^{\theta}\frac{\frac{1}{\cos^2(t)}}{\frac{1}{\cos^2(t)}}\,dt \\ = & \int_0^{\theta} \frac{\frac{1}{\cos^2(t)}}{\frac{\sin^2(t)+\cos^2(t)}{\cos^2(t)}}\,dt\\ = & \int_0^{\theta} \frac{\frac{1}{\cos^2(t)}}{1+\tan^2(t)}\,dt\\ = & \int_0^{\tan(\theta)} \frac{1}{1+x^2}\,dx \color{#3D99F6} \,\,\text{ via } x =\tan(t). \end{aligned} Notice when $\theta \to \frac{\pi}{2}$, $\tan(\theta) \to \infty$. So recalling that $L = \theta$: \begin{aligned} \theta = & \int_0^{\tan(\theta)} \frac{1}{1+x^2}\,dx \\ \frac{\pi}{2} = & \int_0^{\infty} \frac{1}{1+x^2}\,dx .\\ \end{aligned} We can also use this to discover many other relations. For example \begin{aligned} \frac{\pi}{4} = & \int_0^{1}\frac{1}{1+x^2}\,dx \\ \frac{\pi}{3} = & \int_0^{\sqrt{3}}\frac{1}{1+x^2}\,dx\\ -\frac{\pi}{2} = & \int_0^{-\infty}\frac{1}{1+x^2}\,dx. \end{aligned} This was the method of Italian mathematician Giulio Fangnano.

We can of course immediately infer via the inverse function of $\tan(\theta)$ the anti-derivative in question. We know via the fundamental theorem of calculus \begin{aligned} \theta =& \int_0^{\tan(\theta)} \frac{1}{1+x^2}\,dx = F(\tan(\theta)) - F(0), \, \,F(0) = 0\\ \theta = &F(\tan(\theta)) . \end{aligned} And we know this is in the form $f^{-1}(f(x)) = x$. So we can infer $F(x) = \arctan(x) + C$, which leads us to the anti-derivative we assumed we didn't know at the beginning $\int\frac{1}{1+x^2}\,dx = \arctan(x) + C.$ Note by Vincent Moroney
1 year, 5 months ago

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