# Polynomials arising from residue of $\frac{z^m}{(1+z^N)^k}$ at $e^{\frac{i\pi}N}$

I have been trying to find a general expression for $${I=\int_0^{\infty}}\frac{x^m}{(1+x^N)^k}dx$$ where $$m,N,k\in\mathbb{N}_0$$ , $$m\leq{Nk-2}$$ , $$N\geq2$$ and $$k\geq1$$. To do so I have been using residue theorem from complex analysis. In doing this, the residue at $$e^{\frac{i\pi}N}$$ needs to be calculated and this being a k-th order pole, the calculation is quite complicated. Using the formula described here, I created a script in MATLAB to symbolically differentiate $$k-1$$ times and take the limit and upon doing this several patterns emerged. When $$m=0$$, the residue seems to resolve to $$\frac{(-1)^k}{(k-1)!}\frac{e^{\frac{i\pi}N}}{N^k}\prod_{j=1}^{k-1}(jN-1)$$ though I do not have a rigourous proof for this. Otherwise, it seems the residue takes the form $$\frac{(-1)^k}{(k-1)!}\frac{e^{\frac{i\pi{(m+1)}}N}}{N^k}(g_k(N,m)+(m+1)^{k-1})$$ Where $$g_k$$ is a polynomial in N and m of degree $$k-1$$. Again, I have no rigourous proof for this, but it seems the pattern holds. Particular values of $$g_k$$ are as follow: $$g_1=0\\g_2=-N\\g_3=2N^2-3Nm-3N\\g_4=-6N^3+11N^2m+11N^2-6Nm^2-12Nm-6N\\g_5=24N^4-50N^3m-50N^3+35N^2m^2+70N^2m+35N^2-10Nm^3-30Nm^2-30Nm-10N$$ To me, it looks like there is a whole lot of relationships between the coefficients of these polynomials and that there might be a closed form expression for $$g_k$$, but other than the highest degree coefficient being $$(-1)^{k-1}(k-1)!$$ for $$k\gt1$$, I have no ideas about the nature of these polynomials, nor if there actually is a closed form expression for them. One thing to note is the separation of the $$(m+1)^{k-1}$$ may be making things harder to resolve, as including the one with the terms with only N makes the terms with only N equal to the product of factors seen in the case where $$m=0$$. The polynomials do not seem to factor. Are these polynomials a known series (i.e. are these a named group of polynomials). If not does anyone have any insights into the nature of these polynomials? Is there anyway to discern whether or not a closed form expression can be found? If there is and if these polynomials do have a closed form expression, what is it?

Edit: To be clear, I am looking for insights into the described polynomials, not the integral itself.

As one could expect the antiderivative is given by a gaussian hypegeomtric function $$I=\int\frac{x^m}{(1+x^n)^k}dx$$ Let $$x^n=t$$ to make $$I=\frac 1n\int \frac {t^{\frac{m+1-n}{n}} } {(1+t)^k }dt=\frac{t^{\frac{m+1}{n}}}{m+1}\,\, _2F_1\left(k,\frac{m+1}{n};\frac{m+n+1}{n};-t\right)$$ Back to $$x$$ $$I=\frac{x^{m+1}}{m+1}\,\, _2F_1\left(k,\frac{m+1}{n};\frac{m+n+1}{n};-x^n\right)$$ $$J=\int_0^\infty\frac{x^m}{(1+x^n)^k}dx=\frac{\Gamma \left(\frac{m+1}{n}\right) \Gamma \left(k-\frac{m+1}{n}\right)}{n \, \Gamma (k)}$$ provided that $$\Re(n)>0\land \Re(m-k n)<-1\land \Re(m)>-1$$
Hope this will help. My approach is using the Beta function. By substitute $$t = x^n$$: $${I=\int_0^{\infty}}\frac{x^m}{(1+x^N)^k}dx=\frac{1}{n}\int_0^{\infty}\frac{t^{\frac{m+1}{n}-1}}{(1+t)^k}dt=\frac{\Gamma \left(\frac{m+1}{n}\right) \Gamma \left(k-\frac{m+1}{n}\right)}{n \, \Gamma (k)}$$