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I noticed that the following expression equals $\pi$ and I was curious as to why. Is it just a coincidence, or is there a meaningful explanation?

$$\int_{-\infty}^\infty\frac{1}{x^2+1}~dx=\pi$$

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    It is susceptible to use of residues, with a semi-circle contour with diameter on $[-R,R]$, but the $\arctan$ method is much more straightforward.2011-07-29
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    @Geoff: I came up with a bell curve and decided to integrate it on my CAS calculator2011-07-29
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    Would it be perverse to factor the denominator (with complex coefficients) and integrate using partial fractions? Some knowledge of complex logarithms will get you a $\pi$ in the answer.2011-07-29
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    Related: http://math.stackexchange.com/questions/2899/can-this-standard-calculus-result-be-explained-intuitively/2905#29052011-07-29
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    In my experience, things in math are rarely coincidences. As Qiaochu points out, I asked a related question a while back, and he gives a *very* nice answer in that thread.2011-07-29
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    @Geoff the integrand is just the regular derivative of arctan(x) whereas arctan(x) is the antiderivative of the integrand... ;)2011-07-30
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    Oop! I think it was clear what I meant, though.2011-07-30

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