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Let's say that we want to evaluate 011+x2dx\displaystyle\int_0^{\infty}\frac{1}{1+x^2}\,dx without knowing the anti derivative of 11+x2\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=rθ=θL=r\theta = \theta. We can equivalently write the arc length of the unit circle as L=0θdt=0θ1cos2(t)1cos2(t)dt=0θ1cos2(t)sin2(t)+cos2(t)cos2(t)dt=0θ1cos2(t)1+tan2(t)dt=0tan(θ)11+x2dx via x=tan(t). \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 θπ2\theta \to \frac{\pi}{2}, tan(θ)\tan(\theta) \to \infty. So recalling that L=θL = \theta: θ=0tan(θ)11+x2dxπ2=011+x2dx. \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 π4=0111+x2dxπ3=0311+x2dxπ2=011+x2dx. \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(θ)\tan(\theta) the anti-derivative in question. We know via the fundamental theorem of calculus θ=0tan(θ)11+x2dx=F(tan(θ))F(0),F(0)=0θ=F(tan(θ)).\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 f1(f(x))=xf^{-1}(f(x)) = x. So we can infer F(x)=arctan(x)+CF(x) = \arctan(x) + C, which leads us to the anti-derivative we assumed we didn't know at the beginning 11+x2dx=arctan(x)+C.\int\frac{1}{1+x^2}\,dx = \arctan(x) + C.

Note by Vincent Moroney
1 year, 1 month ago

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