Why does this expression equal $\pi$? 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?
$$\int_{-\infty}^\infty\frac{1}{x^2+1}~dx=\pi$$

Here is another approach, which avoids taking $\arctan(\infty)$, etc.: let $u = \frac{1}{x}$, so that $\frac{du}{dx} = \frac{-1}{x^2}.$ Then we obtain $\int_{1}^{\infty} \frac{1}{1+u^{2}} du = \int_{0}^{1} \frac{1}{1+x^2}dx$. Hence $\int_{- \infty}^{\infty} \frac{1}{1+x^2} dx = 4\int_{0}^{1} \frac{1}{1+x^2}dx = 4 \arctan(1) = 4\frac{\pi}{4} = \pi$.
As is well-known, the last integral can also be evaluated by integrating $\frac{1}{1+x^2} = (1 - x^2 + x^4 -\ldots )$ term by term (this needs a little justification but is permissible here), to obtain the famous formula $\frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$. 
