It is well-known that $$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.$$
I noticed that $y = \frac{1}{x^n + 1}$ for even $n$ has a similar shape to the bell curve. Let $$A(n) = \int_{-\infty}^\infty \frac{1}{x^{2n} + 1}, \qquad n \in \mathbb{N}.$$ Then, $A(1) = \pi$ using the fact that $\frac{1}{x^2 + 1} = \frac{d}{dx}(\tan^{-1}x)$. Also, $A(2) = \frac{\pi}{\sqrt{2}}$ using partial fractions. Now, for me at least, it starts to get tedious. Through WolframAlpha, some results including the first ones are the following:
| $n$ | $A(n)$ |
|---|---|
| $1$ | $\pi$ |
| $2$ | $\frac{\pi}{\sqrt{2}}$ |
| $3$ | $\frac{2\pi}{3}$ |
| $4$ | $\frac{1}{4}\pi\csc(\pi/8)$ |
| $5$ | $\frac{4\pi}{5(\sqrt{5} - 1)}$ |
When finding $A(6)$ through WolframAlpha, the result given was $\frac{1}{12}\Gamma(\frac{1}{12})\Gamma(\frac{11}{12})$. For me, I think it is unusual for the Gamma function to show up.
I saw this answer for $A(2)$ where it uses the Gamma function to solve. Let $k = 2n$. By generalizing this, \begin{align} \int_0^\infty \frac{1}{x^k + 1} \, dx &= \int_0^\infty\int_0^\infty e^{-(x^k + 1)t} \, dx\, dt \\ &= \int_0^\infty e^{-t}\int_0^\infty e^{-x^k t} \, dx\, dt \\&=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^k t}\, dx\right)\,dt \\&=\frac{1}{k}\int_0^\infty t^{-\frac{1}{k}}e^{-t}\left(\int_0^\infty u^{\frac{1}{k}-1} e^{-u}\,du\right)\,dt \\&=\frac{1}{k}\left(\int_0^\infty t^{-\frac{1}{k}}e^{-t}\,dt\right)\left(\int_0^\infty u^{\frac{1}{k}-1} e^{-u}\,du\right) \\&=\frac{1}{k} \Gamma\left(1-\frac{1}{k}\right)\Gamma\left(\frac{1}{k}\right) \end{align}
Is this the simplest closed form for the area under $\frac{1}{x^{2n} + 1}$?
