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Integral of $r frac{2^{r-1} log (2) e^{-frac{sqrt{2^r-1}}{b}} left(2^r-1right)^{frac{d}{2}-1}}{b^d Gamma (d)}$ with Mathematica

Mathematica Asked on May 14, 2021

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

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

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

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

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}]

One Answer

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

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

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

Integrate[f, {r, 0, Infinity}, 
 Assumptions -> 
  d ∈ Reals && b ∈ Reals && d > 0 && b > 0]

then, the solution to this integral is

$$frac{b^{-d} left(frac{2 pi csc left(frac{pi d}{2}right) , _1F_2left(frac{d}{2};frac{1}{2},frac{d}{2}+1;-frac{1}{4 b^2}right)}{d}+frac{2 left(-pi b sec left(frac{pi d}{2}right) , _1F_2left(frac{d}{2}+frac{1}{2};frac{3}{2},frac{d}{2}+frac{3}{2};-frac{1}{4 b^2}right)+(d+1) b^d Gamma (d-2) , _2F_3left(1,1;2,frac{3}{2}-frac{d}{2},2-frac{d}{2};-frac{1}{4 b^2}right)+2 left(d^3-2 d^2-d+2right) b^{d+2} Gamma (d-2) (log (b)+psi ^{(0)}(d))right)}{b^2 (d+1)}right)}{log (4) Gamma (d)}$$

Correct answer by Felipe Augusto de Figueiredo on May 14, 2021

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