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I'm trying to find the integral given below with Mathematica

$\int_0^{\infty } r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)} \, dr$

However, it takes too long for it to return something and when it returns it outputs the same integral.

$\int_{0}^{\infty } \frac{2^{r-1} r \log (2) b^{-d} e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{\Gamma (d)} \, dr$

I'd like to figure out the solution for this integral.

Mathematica code

y = (2^(-1 + r)*(-1 + 2^r)^(-1 + d/2)*Log[2])/(E^(Sqrt[-1 + 2^r]/b)*(Gamma[d]*b^d)); 
Integrate[r*y, {r, 0, Infinity}]
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    $\begingroup$ Please include your Mathematica code so we don't have to type in your integral. $\endgroup$ – JimB Jul 10 '20 at 1:59
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    $\begingroup$ Even without multiplying by $r$ it doesn't appear that the integral converges on {$0,\infty$}. What values of $b$ and $d$ do you think there is finite result? $\endgroup$ – JimB Jul 10 '20 at 2:12
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    $\begingroup$ So you want the expectation of a random variable $R$ with a pdf of $\frac{2^{r-1} \log (2) \left(2^r-1\right)^{\frac{d}{2}-1} \exp \left(-\frac{\sqrt{2^r-1}}{b}\right)}{b^d \Gamma (d)}$ with $b>0$ and $0<d<1$? If so, adding in such information would be helpful. $\endgroup$ – JimB Jul 10 '20 at 3:51
  • $\begingroup$ If you do a change of variables technique with $x=2^r-1$, then Mathematica will give you a result with the assumptions of $b>0$ and $0<d<1$. Sorry, I won't have time to write up an answer with that until sometime tomorrow. $\endgroup$ – JimB Jul 10 '20 at 5:31
  • $\begingroup$ @JimB, I've just added the Mathematica code. Thanks! $\endgroup$ – Felipe Augusto de Figueiredo Jul 10 '20 at 12:24
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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

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

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