# 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{align*} 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{blue} \,\,\text{ via } x =\tan(t). \end{align*} Notice when $$\theta \to \frac{\pi}{2}$$, $$\tan(\theta) \to \infty$$. So recalling that $$L = \theta$$: \begin{align*} \theta = & \int_0^{\tan(\theta)} \frac{1}{1+x^2}\,dx \\ \frac{\pi}{2} = & \int_0^{\infty} \frac{1}{1+x^2}\,dx .\\ \end{align*} We can also use this to discover many other relations. For example \begin{align*} \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{align*} 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{align*} \theta =& \int_0^{\tan(\theta)} \frac{1}{1+x^2}\,dx = F(\tan(\theta)) - F(0), \, \,F(0) = 0\\ \theta = &F(\tan(\theta)) . \end{align*} 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
3 months, 2 weeks ago

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