<s>[INST] Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]} 
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26 
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
 [/INST] Proving $\displaystyle\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$:
Set $a=z$ and $b=1-z$ in the Beta function:
$$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\mathrm{d}x,$$
we obtain
\begin{gather}
\Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x\\
=\left(\int_0^1+\int_1^\infty\right) \frac{x^{z-1}}{1+x}\mathrm{d}x=\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\underbrace{\int_1^\infty \frac{x^{z-1}}{1+x}\mathrm{d}x}_{x\to 1/x}\\
=\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\int_0^1 \frac{x^{-z}}{1+x}\mathrm{d}x\\
\left\{\text{expand $\frac1{1+x}$ in series in both integrals}\right\}\\
=\int_0^1 x^{z-1}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x+\int_0^1 x^{-z}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x\\
\{\text{interchange integration and summation}\}\\
=\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n+z-1}\mathrm{d}x+\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n-z}\mathrm{d}x\\
=\sum_{n=0}^\infty\frac{(-1)^n}{n+z}+\sum_{n=0}^\infty\frac{(-1)^n}{n-z+1}\\
\{\text{seperate the first term of the first sum and shift the index of the second}\}\\
=\frac1z+\sum_{n=1}^\infty\frac{(-1)^n}{n+z}-\sum_{n=1}^\infty\frac{(-1)^n}{n-z}\\
=\frac1z-\sum_{n=1}^\infty\frac{2z(-1)^n}{n^2-z^2}\\
=\frac{\pi}{\sin(\pi z)}.
\end{gather}
To prove the last equality, set $x=0$ in the Fourier series of $\cos(zx)$:
\begin{equation}
\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}-\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2-z^2}\right],\quad z\notin\mathbb{Z}.\label{cos(zx)}
\end{equation}
Thus,
$$\Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x=\frac{\pi}{\sin(\pi z)}.$$
To complete the proof, replace $z$ by $1/n$ then let $x=y^n$.
</s>