What is the integral of $r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)}$?

Question

I have been trying to solve a research problem for a while now and in doing so, I stumbled upon the following integral: $$\int_0^{\infty } r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)} \, dr.$$ However, I've got no idea how to solve that. Therefore, I'd like to figure out the solution for this integral.

Answer 1

An solution for this integral can be found at Mathmatica.SE, which is reproduced next.

After applying the change of variable technique with $x=2^r-1$ we get

$$f=\frac{e^{-\frac{\sqrt{r}}{b}} r^{\frac{d}{2}-1} \log _2(r+1)}{2 \left(b^d \Gamma (d)\right)} $$

$$\text{Integrate}[f,\{r,0,\infty \},\text{Assumptions}\to d\in \mathbb{R}\land b\in \mathbb{R}\land d>0\land b>0]$$

then, the solution to this integral is

\begin{align} \frac{1}{\log(4)b^d\Gamma(d)} \left[\frac{2\pi}{d} \csc\left(\frac{\pi d}{2}\right) \, {}_1F_2\left(\frac{d}{2};\frac{1}{2},\frac{d}{2}+1;-\frac{1}{4 b^2}\right)\\ + \frac{1}{b^2 (d+1)} \left(2 \left\{-\pi b \sec \left(\frac{\pi d}{2}\right) \, {}_1F_2\left(\frac{d}{2}+\frac{1}{2};\frac{3}{2},\frac{d}{2}+\frac{3}{2};-\frac{1}{4 b^2} \right) \\ +\,(d+1) b^d \Gamma (d-2)\, {}_2F_3\left(1,1;2,\frac{3}{2}-\frac{d}{2},2-\frac{d}{2};-\frac{1}{4 b^2}\right) \\ + 2(d^3-2 d^2-d+2) b^{d+2} \Gamma (d-2) (\log (b)+\psi ^{(0)}(d)) \right\}\right)\right] \end{align}