<s>[INST] $\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$ similar identities I recently found that 
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$$
iff
$$b^2-4ac=-4.$$
I found it by integrating
$$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}.$$
If the reciprocal of the function is to be integrated over the entire real line, then the function must not have any real zeros. This implies that $$b^2-4ac<0$$
With this in mind, we complete the square:
$$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}}.$$
Then setting $g=c-\frac{b^2}{4a}$, and $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$,
$$I=\sqrt{\frac{g}{a}}\int_{-\pi/2}^{\pi/2}\frac{\sec^2u\ \mathrm{d}u}{g\tan^2u+g}=\frac{\pi}{\sqrt{ag}}.$$
So our identity holds for 
$$ag=1.$$
And with a little algebra, 
$$b^2-4ac=-4.$$
So my question is, are there any other similar identities involving other famous constants? Cheers!
 [/INST] First if we consider the integral:
\begin{equation}
 I(a) = \int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx = \frac{\pi}{a}
\end{equation}
Then we can see:
\begin{equation}
 I(a) + cI(b) = \left[\int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx + c\int_{-\infty}^{\infty}\frac{1}{x^2 + b^2}\:dx \right] = \frac{\pi}{a} + c\frac{\pi}{b} = \pi\left[\frac{1}{a} + c\frac{1}{b} \right]
\end{equation}
We can see that this can be expanded to any number of $I(a)$ terms and any sequence of $c_i$ values, i.e. 
\begin{equation}
 \sum_{i = 1}^{\infty} (-1)^{i + 1} \frac{1}{i + 1}I(i) = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
Thus, 
\begin{equation}
 \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty}  \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
Or as another example, 
\begin{equation}
 \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty}  \frac{1}{x^2 + i^4} \:dx = \frac{ \pi^3}{6}
\end{equation}
What's even better (IMO) is that we can take each integral spoken to and apply (through Glasser Master Theorem)
$$x = t - \sum_{i = 1}^{n - 1}\frac{\left| d_{i}\right|}{ t - e_{i}}$$
Where $d_i, e_i \in \mathbb{R}$ and $n \in N$ that the value of the integrals remain unchanged!! 
As an addendum to the example:
\begin{equation}
 \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty}  \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
If we call on this integral as addressed here we see that
\begin{equation}
 2\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = 2\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty}  \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx =\pi\ln(2)
\end{equation}
Or, 
\begin{equation}
 \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty}  \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \frac{\pi\ln(2)}{2}
\end{equation}
</s>