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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