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 $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$

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
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! na^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k-\frac{1}{n} )}
$$
