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 $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
 A: Without using the formula $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),$ I thought of the Gamma Function.
$$ \begin{aligned}
\text { Let } t &=\frac{1}{x^{n}+1}, \text { then } x=\left(\frac{1}{t}-1\right)^{\frac{1}{n}} \\
d x &=\frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1-n}{n}}\left(-\frac{1}{t^{2}} d t\right)=-\frac{(1-t)^{\frac{1}{n}-1}}{n t ^{\frac{1+n}{n }} }d t
\end{aligned}
$$
The integral is transformed into
$$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}&=\frac{1}{n} \int_{0}^{1} t^{m} \cdot \frac{(1-t)^{\frac{1-n}{n}}}{t^{\frac{1+n}{n}}} d t \\
&=\frac{1}{n} \int_{0}^{1} t^{\frac{m n-1}{n}-1}(1-t)^{\frac{1}{n} -1}dx \\
&=\frac{1}{n} B\left(m -\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Now applying the formula $$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
we can now conclude that
$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n(m-1)!}}$$
Letting $x\mapsto\frac{x}{\sqrt[n]a}$ yields  the general integral,
$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n a^{m-\frac{1}{n}}(m-1)!}}$$
Taking the same integral for example, we have
$$
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}= \frac{\Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)}{12 \Gamma(5)}= \frac{1}{288} \Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)\approx 0.845907, $$
which is checked by Wolframalpha.
However, I don’t know how to further simplify the formula
$$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right),$$
can you help?
$$\textrm{ ********} \tag*{} $$
Thanks to Mr Clathratus who gave me a nice simplification to $$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right).$$
Using the identity $$
\Gamma(x+1)=x \Gamma(x),
$$
we have $$
\begin{aligned}
\Gamma\left(m-\frac{1}{n}\right) &=\left(m-1-\frac{1}{n}\right) \Gamma\left(m-\left(-\frac{1}{n}\right)\right.\\
&=\left(m-1-\frac{1}{n}\right)\left(m-1-\frac{1}{n}\right) \Gamma\left(m-2-\frac{1}{n}\right) \\
& \qquad\qquad \vdots \\
&=\left(m-1-\frac{1}{n}\right)\left(m-2-\frac{1}{n}\right) \cdots\left(1-\frac{1}{n} \right)\Gamma\left(1-\frac{1}{n}\right)\\
&= \Gamma\left(1-\frac{1}{n}\right) \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \\
\Gamma\left(m-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right) &=\prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \Gamma\left(1-\frac{1}{n} \right)\Gamma\left(\frac{1}{n}\right)
\end{aligned}
$$
By the Euler's Reflection Theorem,
$$
\Gamma\left(1-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)=\pi \csc \left(\frac{\pi}{n}\right)
$$
Putting back, we now obtain the same formula as before
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! na^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k-\frac{1}{n} )}
$$
A: 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 $x=y^n$.
A: Glad to see that there are several nice solutions to the problem, now I want to add one more by discovering a reduction formula using integration by parts.
In order to use integration by parts, I applied the inversion substitution to the original integral.
For any fixed natural number $n$, we now define
$$
J_{m}:=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} \stackrel{x \rightarrow \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{m n-2}}{\left(x^{n}+1\right)^{m}} d x
$$
Noting that $$
\frac{d}{d x}\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)=-(m-1) \cdot \frac{nx^{n-1}}{\left(x^{n}+1\right)^{m}},
$$
we have
$$J_m=-\frac{1}{n(m-1)} \int_{0}^{\infty} x^{m n-n-1} d\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)
$$
By integration by parts, we get
$$
\begin{aligned}
J_{m} &\left.=-\frac{m-1}{n}\left[\frac{x^{m n-n-1}}{\left(x^{n}+1\right)^{m-1}}\right]_{0}^{\infty}+\frac{1}{n(m-1)} \int_{0}^{\infty} \frac{(m n-n-1) x^{m n-1-2}}{\left(x^{n}+1\right)^{m-1}}\right] \\
&=\frac{m n-n-1}{n(m-1)} \int_{0}^{\infty} \frac{x^{m n-n+2}}{\left(x^{n}+1\right)^{m-1}} d x\\&= \frac{\left(m-1-\frac{1}{n}\right)}{m-1} J_{m-1}
\end{aligned}
$$
Applying the reduction formula $m$ times yields inductively
\begin{aligned}
J_m&=\frac{m-1-\frac{1}{n}}{m-1} \cdot \frac{m-2-\frac{1}{n}}{m-2} \cdot \frac{m-3-\frac{1}{n}}{m-3} \cdot \frac{1-\frac{1}{n}}{1} J_{0} \\
& = \frac{1}{(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) J_{0}
\end{aligned}
Using the well-known formula,
$$
J_0=\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),
$$
we can conclude that
$$\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} =\frac{\pi \csc \left(\frac{\pi}{n}\right)}{n(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) $$
A: You could make it faster using
$$I=\int \frac{d x}{\left(x^{n}+a\right)^{m}}=x\, a^{-m} \, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-\frac{x^n}{a}\right)$$ which gives
$$J=\int_0^\infty \frac{d x}{\left(x^{n}+a\right)^{m}}=a^{\frac{1}{n}-m}\,\frac{\Gamma \left(1+\frac{1}{n}\right)  \Gamma
   \left(m-\frac{1}{n}\right)}{\Gamma (m)}$$ provided $\Re(m n)>1\land \Re(n)>0$.
