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How can one calculate

$$ \int_{- \infty}^{\infty} \frac{dx}{x^{2n} + 1}, \;n \in \mathbb{N} $$

without using complex plane and Residue theorem?

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With $x=\tan^{1/n} t$ and some well-known properties of the Gamma function, you can prove the answer is $2/\operatorname{sinc}\frac{\pi}{2n}$. The case $n=1$ is an easy sanity check.

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Consider the integral $$F(a,b,c)=\int_0^\infty\frac{t^a}{(t^b+1)^c}\mathrm dt$$ Set $x^{1/b}=t$: $$F(a,b,c)=\frac1b\int_0^\infty\frac{x^{ \frac{a+1}b}}{(x+1)^c}\mathrm dx$$ Then $u=\frac1{x+1}$ gives $$\begin{align} F(a,b,c)&=\frac1b\int_0^1 \left[\frac{1-u}{u}\right]^{ \frac{a+1}b}u^c\frac{\mathrm du}{u^2}\\ &=\frac1b\int_0^1u^{c-2-\frac{a+1}{b}}(1-u)^{\frac{a+1}b}\mathrm du \end{align}$$ Then recall the definition of the Beta function: $$\mathrm{B}(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\mathrm dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ Where $\Gamma(s)$ is the gamma function. So of course we have $$F(a,b,c)=\frac1b\mathrm{B}\left(c-1-\frac{a+1}b,1+\frac{a+1}b\right)$$ We then have your integral as $$\int_0^\infty \frac{\mathrm dt}{t^{2n}+1}=F(0,2n,1)$$

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