Consider the following integral :
$$\int_0^\infty\frac{(a+bx)^n}{e^{cx}-1}dx$$
Here a,b,c,n are constants
How to calculate above integral ?
Also, similarly:
$$\int_0^\infty\frac{(1+ix)^{2n-1}-(1-ix)^{2n-1}}{e^{2πx}-1}dx$$
How to calculate this for n=1,2..?
I know the following:
$$\int_0^\infty\frac{(1+ix)^{2n}-(1-ix)^{2n}}{e^{2πx}-1}dx=\frac{i (2n-1)}{2(2n+1)}$$
On
Assuming that the integral converges, you could write $$\frac{(a+bx)^n}{e^{cx}-1}=\sum_{k=0}^\infty e^{-c (k+1) x} \,(a+b x)^n$$ and, at the price of a simple change of variable, it could look that the sum of incomplete gamma function.
On
Notation: $C$ is the binomial coefficient, $$C(x,y)=\frac{\Gamma(x+1)}{\Gamma(x-y+1)\Gamma(y+1)}$$
You can use a binomial expansion and notice that the even terms cancel to obtain $$(1+ix)^{2n}-(1-ix)^{2n}=2\sum_{k=0}^{2n} C(2n,2k+1)(ix)^{2k+1}$$ And using an obvious property of the odd powers of the imaginary unit this is $$2i\sum_{k=0}^{2n} (-1)^{k}C(2n,2k+1)x^{2k+1}$$ So $$\int_0^\infty \frac{(1+ix)^{2n}-(1-ix)^{2n}}{e^{2\pi x}-1}\mathrm{d}x=2i\sum_{k=0}^{2n}(-1)^{k}C(2n,2k+1)\int_0^\infty \frac{x^{2k+1}}{e^{2\pi x}-1}\mathrm{d}x$$ But the final integral is very well known, in particular $$\mathcal{M}\left(x\mapsto \frac{1}{e^x-1}\right)(s)=\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\mathrm{d}x=\Gamma(s)\zeta(s)$$ Which I believe can be evaluated using Ramunajan's Master Theorem. Anyway this means that $$\int_0^\infty \frac{(1+ix)^{2n}-(1-ix)^{2n}}{e^{2\pi x}-1}\mathrm{d}x=2i\sum_{k=0}^{2n}(-1)^{k}\frac{(2n)!}{(2n-2k-1)!(2k+1)!}\frac{\Gamma(2k+2)\zeta(2k+2)}{(2\pi)^{2k+2}}$$ $$=2i(2n)!\sum_{k=0}^{2n}\frac{(-1)^{k}\zeta(2k+2)}{(2n-2k-1)!(2\pi)^{2k+2}}$$ But of course, the value of the zeta function at even integers is well known. We have $$\zeta(2(k+1))=(-1)^k\frac{(2\pi)^{2k+2}B_{2k+2}}{2(2k+2)!}$$ Therefore $$I_n=\int_0^\infty \frac{(1+ix)^{2n}-(1-ix)^{2n}}{e^{2\pi x}-1}\mathrm{d}x=i(2n)!\sum_{k=0}^{2n}\frac{B_{2k+2}}{(2n-2k-1)!(2k+2)!}$$ $$=i\sum_{k=0}^{2n}\frac{B_{2k+2}}{2k+2}C(2n,2k+1)$$ We can show that $$a_n=\sum_{k=0}^{2n}\frac{B_{2k+2}}{2k+2}C(2n,2k+1)=\frac{(2n-1)}{2(2n+1)}$$ Because, due to the poles of the Gamma function at the negative integers, letting $k\in\mathbb{N}_0$, $$C(n,k)=\frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)}=\begin{cases} \frac{n!}{(n-k)!k!} & k\leq n\\ 0 & k>n \end{cases}$$ Hence $$a_n=\sum_{k=0}^{n-1}\frac{B_{2k+2}}{2k+2}C(2n,2k+1)$$ Now we go by induction. The result is clearly true for $n=1$ since $$a_1=\frac{B_2}{2}C(2,1)=B_2=\frac{1}{6}=\frac{2\cdot 1-1}{2(2\cdot 1+1)}$$ Inductive step: $$a_{n+1}=\sum_{k=0}^{n}\frac{B_{2k+2}}{2k+2}C(2n+2,2k+1)$$ $$a_{n+1}=\sum_{k=0}^{n-1}\frac{B_{2k+2}}{2k+2}C(2n,2k+1)+2\sum_{k=0}^n\frac{B_{2k+2}}{2k+2}C(2n,2k)+\sum_{k=0}^n\frac{B_{2k+2}}{2k+2}C(2n,2k-1)$$ Where we have dropped the last term of the first sum due to the properties of the binomial coefficient mentioned before. Noting our defn. of $a_n$ and rearranging, $$a_{n+1}=a_n+\sum_{k=0}^n\left[\frac{B_{2k+2}}{2k+2}\bigg(2C(2n,2k)+C(2n,2k-1)\bigg)\right]$$ Again using the recursive properties of the binomial, $$a_{n+1}=a_n+\sum_{k=0}^n\left[\frac{B_{2k+2}}{2k+2}\bigg(C(2n,2k)+C(2n+1,2k)\bigg)\right]$$ And I must admit, I'm not sure where to go from here! Hopefully someone else can finish off the rest?
Hint: You should first try and figure out when the integral converges in the first place(looking at the limit in $x = 0$) and then maybe try per partes.