As far as I can tell, the following infinite integral holds, $$ I(m,n) = \int_0^{\infty} \frac{1}{(1+x^n)^m} dx = \frac{\Gamma(1+1/n)\Gamma(m-1/n)}{\Gamma(m)}. $$ Here, $m \geq1, n \geq 2$ are integers. $\Gamma(x)$ is the well-known Gamma function.
When $m = 1$, it is shown in this link. Some other cases can be found here.
My Questions are that: (1) How can we proof the above integral identity nicely? (2) Is there a way to calculate the above integral by Residue theorem?
Firts of all, using the Gaussian hypergeometric function, the antiderivative is $$I(m,n) = \int \frac{dx}{(1+x^n)^m} = x\, \, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-x^n\right)$$ $$J(m,n) = \int_0^t \frac{dx}{(1+x^n)^m} =t \,\, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-t^n\right)$$ $$K(m,n) = \int_0^\infty \frac{dx}{(1+x^n)^m} =\frac{\Gamma \left(1+\frac{1}{n}\right) \Gamma \left(m-\frac{1}{n}\right)}{\Gamma (m)}$$
If you want to use residues, think about Barnes integrals.
We may use the generalized Newton's binomial and the Ramanujan Master Theorem (RMT).
The binomial theorem states the following (https://en.wikipedia.org/wiki/Binomial_theorem) $$\frac{1}{(1+y)^m}=\sum_{k=0}^\infty \binom{m+k-1}{k}(-x)^k=\sum_{k=0}^\infty \frac{(m+k-1)!}{(m-1)!}\frac{(-x)^k}{k!}$$
For convenience we may define $\phi(k)=\frac{(m+k-1)!}{(m-1)!}$.
Let's recall the RMT (https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem), which states that if a complex-valued function $f(x)$ has an expansion of the form $ f(x)=\sum _{k=0}^{\infty }{\frac {\,\varphi (k)\,}{k!}}(-x)^{k}$ then
$$\int _{0}^{\infty }x^{s-1}f(x)\,dx=\Gamma (s)\,\varphi (-s)$$
Now we may observe that the initial integral may be seen as a Mellin transform; applying it to the case at hand we obtain the following
$$I=\int_{0}^{\infty} \frac{dx}{(1+x^n)^m}=\frac{1}{n}\int_{0}^{\infty} \frac{y^{\tfrac{1}{n} -1}dy}{(1+y)^m}=\frac{1}{n}\Gamma{\left(\frac{1}{n}\right)}\frac{(m-1-\tfrac{1}{n})!}{(m-1)!} $$ the change of variable we made was $y=x^n$.
Solving further, we get the following $$I=\frac{\Gamma\left(1+\tfrac{1}{n}\right)\Gamma\left(m-\tfrac{1}{n}\right)}{\Gamma\left(m\right)}$$
