I'm trying to calculate the following definite integral
$$\int_{-\infty }^{b} \frac{2(y^2)^a}{\sqrt{e^{2(b-y)}-1}} \, dy=\int_0^{\infty}\frac{\left(\left(b-\frac{y}{2}\right)^2\right)^a}{\sqrt{e^y-1}}$$
with $a>0$ and $0\leq b<1$.
Wolfram Mathematica don't give me the result. I wonder if it will be solved using complex analysis (contour integral).
Any help will be wellcome.
Equivalently, you are asking for analytical solutions to integrals of the form
$$I_n(\mu) \equiv \int\limits_0^\infty \frac{(x-\mu)^n}{\sqrt{e^x-1}}\,\mathrm dx$$
But rather than consider $\mu\in[0,2), n\in\mathbb R^+$, $\mu$ can be treated as any number in $\mathbb R$ without affecting convergence, and for my own convenience I will restrict $n$ to the nonnegative integers $\mathbb Z^*$.
Now, if $\mu\neq0$ we can expand out the numerator in terms of the binomial theorem to find that
$$I_n(\mu)=\sum\limits_{k=0}^n (-1)^k {n \choose k}\mu^{n-k}I_k(0) \tag1$$
and so instead of evaluating $I_n(\mu)$ for $\mu\in\mathbb R$, it is sufficient to only evaluate those cases where $\mu=0$ and utilize $(1)$. Therefore, let us define and seek solutions to
$$I_n \equiv I_n(0)=\int\limits_0^\infty \frac{x^n}{\sqrt{e^x-1}}\,\mathrm dx$$
To do so, define a generating function
$$\begin{align} F(z) &\equiv \sum\limits_{n=0}^\infty I_n\frac{z^n}{n!} \\ &= \int\limits_0^\infty \left(\sum\limits_{n=0}^\infty \frac{x^n z^n}{n!}\right) \frac{\mathrm dx}{\sqrt{e^x-1}} \\ &= \int\limits_0^\infty \frac{e^{xz}}{\sqrt{e^x-1}}\,\mathrm dx \end{align}$$
And by making the substitution $e^x \mapsto \frac{1}{1-t}$, we see that the integral can be expressed in terms of the beta function as
$$F(z) = \operatorname{B}\left(\frac{1}{2},\frac{1}{2}-z\right) \tag2$$
And since by definition of the functional equation,
$$I_n = F^{(n)}(0)$$
we have reduced the task of solving the integrals in question to computing partial derivatives of the beta function and evaluating at $z=0$.
As an example, let us evaluate $I_2(\pi)$. First we utilize $(1)$ to find that
$$\begin{align} I_2(\pi) &= 1\cdot\pi^2 I_0(0) - 2\cdot\pi^1 I_1(0) + 1\cdot\pi^0 I_2(0) \\ &= \pi^2I_0 - 2\pi I_1 + I_2 \end{align}$$
Now, we utilize $(2)$ to evaluate the remaining integrals.
Trivially,
$$I_0=\operatorname{B}\left(\frac{1}{2},\frac{1}{2}\right)=\pi$$
For $I_1$, we will need to first find
$$\frac{\mathrm d}{\mathrm dz}\operatorname{B}\left(\frac{1}{2},\frac{1}{2}-z\right)$$
which we can determine by writing the beta function in terms of the gamma function and applying derivative rules as normal. Then by using Gauss's Digamma Theorem we can evaluate at $z=0$ to conclude that
$$I_1 = 2\pi\ln(2)$$
Taking the derivative with respect to $z$ a second time, evaluating at $z=0$, and simplifying the resulting polygamma functions, we can similarly conclude that
$$I_2 = \frac{\pi^3}{3}+4\pi\ln^2(2)$$
and finally, plugging these values back in, we find that
$$I_2(\pi) = \frac{4\pi^3}{3}-4\pi^2\ln(2)+4\pi\ln^2(2)$$