Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$ $$R(s;n)= \int^{\infty}_0 \frac{x^n}{x^s+1}dx$$
From a previously asked question, I know:
$$R(s;0)=\frac{1}{s} \varGamma\left(\frac{1}{s}\right) \varGamma\left(1-\frac{1}{s}\right)$$
The obvious approach is to do integration by parts but I did not manage to find it using that approach, can any of you provide hints or solutions?
 A: Set $y = x^{n+1}$, then $dy = (n+1)x^n dx$, and $ x^s  = y^{ s/({n+1})}$, so
$$R(s;n)=\frac1{n+1}R\left(\frac{s}{n+1};0\right)$$
A: 
Not an answer, but three worked out cases for $\beta=0$, $\beta=1$, and $\beta=2$.

Well, we have the following integral:
$$\mathcal{I}_\text{n}\left(\beta\right):=\int_0^\infty\frac{x^\text{n}}{x^\beta+1}\space\text{d}x\tag1$$
Now, we can use the 'evaluating integrals over the positive real axis' property of the Laplace transform in order to write:
$$\mathcal{I}_\text{n}\left(\beta\right)=\int_0^\infty\mathcal{L}_x\left[x^\text{n}\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{x^\beta+1}\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$
Using the table of selected Laplace transforms, we can find:


*

*$$\mathcal{L}_x\left[x^\text{n}\right]_{\left(\text{s}\right)}=\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\tag3$$

*When $\beta=0$:
$$\mathcal{L}_x^{-1}\left[\frac{1}{x^0+1}\right]_{\left(\text{s}\right)}=\frac{\delta\left(\text{s}\right)}{2}\tag4$$
Where $\delta\left(x\right)$ is the Dirac delta function.

*When $\beta=1$:
$$\mathcal{L}_x^{-1}\left[\frac{1}{x^1+1}\right]_{\left(\text{s}\right)}=\exp\left(-\text{s}\right)\tag5$$

*When $\beta=2$:
$$\mathcal{L}_x^{-1}\left[\frac{1}{x^2+1}\right]_{\left(\text{s}\right)}=\sin\left(\text{s}\right)\tag6$$
So, we can see the three cases:


*

*When $\beta=0$:
$$\mathcal{I}_\text{n}\left(0\right)=\int_0^\infty\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\cdot\frac{\delta\left(\text{s}\right)}{2}\space\text{ds}=\frac{\Gamma\left(1+\text{n}\right)}{2}\int_0^\infty\frac{\delta\left(\text{s}\right)}{\text{s}^{1+\text{n}}}\space\text{ds}=$$
$$\frac{\Gamma\left(1+\text{n}\right)}{2}\cdot\lim_{\text{k}\to0}\frac{1-\theta\left(\text{k}\right)}{\text{k}^{1+\text{n}}}\tag7$$
Where $\theta\left(x\right)$ is the Heaviside theta function and we can use the fact that $\int_0^\infty\frac{\delta\left(x\right)}{\text{y}\left(x\right)}\space\text{d}x=\frac{1-\theta\left(0\right)}{\text{y}\left(0\right)}$.

*When $\beta=1$:
$$\mathcal{I}_\text{n}\left(1\right)=\int_0^\infty\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\cdot\exp\left(-\text{s}\right)\space\text{ds}=\Gamma\left(1+\text{n}\right)\int_0^\infty\frac{\exp\left(-\text{s}\right)}{\text{s}^{1+\text{n}}}\space\text{ds}=$$
$$\Gamma\left(1+\text{n}\right)\Gamma\left(-\text{n}\right)=-\pi\csc\left(\text{n}\pi\right)\tag8$$
To see why that is true you can look at this answer.

*When $\beta=2$:
$$\mathcal{I}_\text{n}\left(2\right)=\int_0^\infty\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\cdot\sin\left(\text{s}\right)\space\text{ds}=\Gamma\left(1+\text{n}\right)\int_0^\infty\frac{\sin\left(\text{s}\right)}{\text{s}^{1+\text{n}}}\space\text{ds}=$$
$$-\Gamma\left(1+\text{n}\right)\Gamma\left(-\text{n}\right)\sin\left(\frac{\text{n}\pi}{2}\right)=\frac{\pi}{2}\cdot\csc\left(\frac{\text{n}\pi}{2}\right)\tag9$$
To see why that is true you can look at this answer.

