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Each of the three following definite integrals are well known to have the same value of $\pi$: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x=2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x=\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\pi.$$

I like taking the first definite integral as the definition of $\pi$, since it represents half the circumference of the unit circle. The second integral obviously represents the area of the unit circle, but as an exercise I wanted to prove its equality with integral #1 using just elementary integration rules. I was successful once I tried integrating by parts:

$$\begin{align} \int\sqrt{1-x^2}\,\mathrm{d}x &=x\sqrt{1-x^2}-\int\frac{-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\sqrt{1-x^2}\,\mathrm{d}x\\ \implies2\int\sqrt{1-x^2}\,\mathrm{d}x&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\\ \implies2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x&=\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x. \end{align}$$

Having accomplished this much, I decided I'd like to demonstrate the equality of these two integrals to $\int_{-\infty}^{\infty}\frac{1}{1+x^2}\mathrm{d}x$ as well, in a similarly elementary manner, but I'm stumped as to what to try. Can anyone suggest a substitution or transformation that demonstrates their equality?

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  • $\begingroup$ I am not sure you can equate the last one to either of the first two since the bounds of integration are different. $\endgroup$ – Laars Helenius Dec 18 '13 at 20:56
  • $\begingroup$ On your note about complex analysis: this integral, if computed rigorously (of course, $\text{atan}(x) \rightarrow \pi/2$ as $x \rightarrow \infty$), is usually done with contour integration. So this would suggest complex analysis as a common approach. $\endgroup$ – Chris K Dec 18 '13 at 20:57
  • $\begingroup$ see Chebyshev polynomials:Orthogonality... $\endgroup$ – draks ... Dec 18 '13 at 21:15
  • $\begingroup$ @LaarsHelenius That can be avoided if the third one is replaced by $$\int_{-1}^1 \frac{2dx}{1+x^2} = \pi$$ $\endgroup$ – Jaume Oliver Lafont Apr 19 '16 at 8:42
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These are both immediate using two subsitutions. Everything is even, so split all of them at $0$.

$$\begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned}$$

And:

$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$

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    $\begingroup$ Whoa! Way better than my answer. Thanks! $\endgroup$ – John Hughes Dec 18 '13 at 21:25
  • $\begingroup$ Awesome. This is precisely the kind of answer I was looking for! Much appreciated. $\endgroup$ – David H Dec 18 '13 at 21:40
  • $\begingroup$ I can't see how these two substitutions would satisfy these two equations. Could someone clarify it a bit further? $\endgroup$ – Albert Netymk Mar 14 '18 at 23:24
  • $\begingroup$ @AlbertNetymk What do you mean? Applying those substitutions takes you from one integral to the other. Try it! $\endgroup$ – L. F. Jul 20 '18 at 15:54
  • $\begingroup$ @L.F. I think they are just wanting a bit more algebraic steps shown. $\endgroup$ – user64742 Jan 11 '19 at 1:50
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Let $x = \sin \theta, \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx = 2\int_{0}^1 \frac{1}{\sqrt{1-x^2}} dx = 2 \int_0^{\pi/2} \frac{1}{\cos\theta} d\sin\theta =\pi$

$\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx = \int_{-\infty}^{\infty} d \left(\tan^{-1}x \right) =\int_{-\pi/2}^{\pi/2} dy = \pi, where \quad y=\tan^{-1}x$

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    $\begingroup$ I think the OP's goal was to equate the two integrals without evaluating them explicitly. $\endgroup$ – Chris K Dec 18 '13 at 21:08
  • $\begingroup$ @ChrisK Indeed, as I would tend to take $\int_0^x\frac{1}{1+x^2}dx$ as the definition of $\arctan$, which would render this answer circular. Perhaps I should have made my question more explicit. $\endgroup$ – David H Dec 18 '13 at 21:12
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If in the third integral you let $x=\tan(t)$, you get

$$ I_3 = \int_{-\pi/2}^{\pi/2} 1 dt. $$

If, in the second integral, you do the substitution $x=\sin(u)$, you get $$ I_2 = 2\int_{-\pi/2}^{\pi/2} \cos^2 t ~ dt. $$

Alternatively, substitute $x = \cos(u)$ to get $$ I_2 = 2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt. $$

So combining these last two,

$$ I_2 = \frac{1}{2} \left (2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt + 2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt \right) \\ = \frac{1}{2} \left (2\int_{-\pi/2}^{\pi/2} \sin^2 t + \cos^2 t~ dt \right) \\ = \int_{-\pi/2}^{\pi/2} 1~ dt $$

hence the second and third integrals are equal, as you wanted.

Not completely satisfactory, but a shift from integrating from $-1$ to $1$ to integrating from $-\infty$ to $\infty$ is likely to involve something like a tangent substitution in general, so it's not too surprising to see this route work out.

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