Closed form of integral: $\displaystyle\int_{1}^{\infty}\frac{ax-b+1}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\ln\left(ax-b+1\right)\mathrm{d}x$

Question

Context
I was trying to calculate the entropy in the Benktander distribution of the second kind, where:
$$f_X(x)=\exp\left(\frac{a}{b}(1-x^b)\right)\cdot x^{b-2}\cdot(ax^b-b+1)\qquad x\geq 1$$ Where $a>0$ and $b\in(0,1]$
In this case the entropy is defined as: $$H[X]=-\int_1^\infty f(x)\cdot\ln(f(x))\mathrm{d}x$$ And after several steps I arrived at this point: $$H[X]=1+\frac{e^{\frac{a}{b}}}{b}\left(E_{\frac{1}{b}}\left(\frac{a}{b}\right)-\int_{1}^{\infty}\frac{ax-b+1}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\ln\left(ax-b+1\right)\mathrm{d}x\right)$$ Where $E_s(z)$ is the generalized exponential integral
What is the closed form of this integral? $$\displaystyle\int_{1}^{\infty}\frac{ax-b+1}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\ln\left(ax-b+1\right)\mathrm{d}x$$ I think it might be useful to consider the following function: $$E_s(z):=z^{s-1}\int_{z}^{\infty}e^{-t}t^{-s}\mathrm{d}t$$ Is integral representation of the exponential integral function, so we can define this other function: $$E^{(1,0)}_s(z):=\frac{\mathrm{d}}{\mathrm{d}s}E_s(z)=z^{s-1}\int_{z}^{\infty}e^{-t}t^{-s}\ln\left(\frac{z}{t}\right)\mathrm{dt}$$

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My approach is this: $$\int_{1}^{\infty}\frac{ax+b-1}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\ln\left(ax+b-1\right)\mathrm{d}x=\left.\frac{\partial}{\partial s}\int_{1}^{\infty}\frac{(ax+b-1)^s}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\mathrm{d}x\right|_{s=1}$$ Then I try to solve $$\int_{1}^{\infty}\frac{(ax+b-1)^s}{x^{\frac{1}{b}}}e^{-\frac{a}{b}x}\mathrm{d}x=\frac{\left(b-1\right)^{s+1-\frac{1}{b}}}{a^{1-\frac{1}{b}}}\int_{\frac{a}{b-1}}^{\infty}\frac{\left(x+1\right)^{s}}{x^{\frac{1}{b}}}e^{-\frac{b-1}{b}x}dx$$

Then I get stuck

Answer 1

This is just a partial answer given that the "closed form" result looks more complicated than the original integral.

Given the information in the first two comments (@TheSimpliFire and @FabioCaiazz0) and trying various integer values of $k$ using Mathematica for

$$\int_1^\infty x^k e^{-x} (x+a)^sdx$$

results in

$$\int_1^\infty x^k e^{-x} (x+a)^sdx=e^a \sum _{i=0}^k (-1)^i a^{k-i} \binom{k}{i} \Gamma (i+s+1,a+1)$$

Taking the derivative of $\Gamma (i+s+1,a+1)$ results in

$$G_{2,3}^{3,0}\left(a+1\left| \begin{array}{c} 1,1 \\ 0,0,i+s+1 \\ \end{array} \right.\right)+\log (a+1) \Gamma (i+s+1,a+1)$$

where $G$ is the Meijer G function. Evaluating that at $s=0$:

$$G_{2,3}^{3,0}\left(a+1\left| \begin{array}{c} 1,1 \\ 0,0,i+1 \\ \end{array} \right.\right)+\log (a+1) \Gamma (i+1,a+1)$$

Finally plugging that gives us

$$\int_1^\infty x^k e^{-x} \ln{(x+a)}dx=$$ $$e^a \sum _{i=0}^k (-1)^i a^{k-i} \binom{k}{i} \left(G_{2,3}^{3,0}\left(a+1\left| \begin{array}{c} 1,1 \\ 0,0,i+1 \\ \end{array} \right.\right)+\log (a+1) \Gamma (i+1,a+1)\right)$$

Answer 2

Assuming that $k$ is a positive integer, the integrals $$I_k=\int_1^\infty x^k\,\,e^{-x}\, \log (x+a)\,dx$$ show an interesting (but not yet identified) structure.

They write $$\color{blue}{I_k=\Gamma (k+1,1)\, \log(1+a)+}$$ $$\color{blue}{(-1)^{k+1}\,P_k(a)\,\,e^{a}\,\, \text{Ei}(-a-1)+(-1)^{k+1}\,\frac{Q_k(a)} e}\tag 1$$ The first polynomials $P_k(a)$ are $$\left( \begin{array}{cc} k & P_k(a) \\ 1 & a-1 \\ 2 & a^2-2 a+2 \\ 3 & a^3-3 a^2+6 a-6 \\ 4 & a^4-4 a^3+12 a^2-24 a+24 \\ 5 & a^5-5 a^4+20 a^3-60 a^2+120 a-120 \\ 6 & a^6-6 a^5+30 a^4-120 a^3+360 a^2-720 a+720 \\ 7 & a^7-7 a^6+42 a^5-210 a^4+840 a^3-2520 a^2+5040 a-5040 \\ \end{array} \right)$$

The first polynomials $Q_k(a)$ are

$$\left( \begin{array}{cc} k & Q_k(a) \\ 1 & 1 \\ 2 & a-4 \\ 3 & a^2-5 a+17 \\ 4 & a^3-6 a^2+25 a-84 \\ 5 & a^4-7 a^3+35 a^2-141 a+485 \\ 6 & a^5-8 a^4+47 a^3-226 a^2+911 a-3236 \\ 7 & a^6-9 a^5+61 a^4-345 a^3+1647 a^2-6703 a+24609 \\ \end{array} \right)$$

Edit

Thanks to @Martin R (have a look here), we now know that $$P_k(a) = \sum_{j=0}^k (-1)^{k-j} \frac{k!}{j!} a^j=(-1)^k\,e^{-a}\, \, \Gamma (k+1,-a)$$ which make simpler the second term of $(1)$.

Update

The problem of the $Q_k(a)$ is much more complex than I thought. If you look at @userrandrand's answer to this question of mine, it is obviously the result of a combination of two more than nasty recurrence equations. But the result (I shall not copy it here) is there.

Answer 3

Without any restriction beside $a\geq 0$, a series expansion gives for $$I_k=\int_1^\infty x^k\,\,e^{-x}\, \log (x+a)\,dx$$ $$I_k=G_{2,3}^{3,0}\left(1\left| \begin{array}{c} 1,1 \\ 0,0,k+1 \\ \end{array} \right.\right)-\sum_{n=1}^\infty (-1)^n \,\frac{ \Gamma (k-n+1,1)}{n}\,a^n$$

A quick and dirty nonlinear regression (data generated for $1\leq k \leq 200$) gives with $(R^2=0.9999940)$ $$\log \left(G_{2,3}^{3,0}\left(1\left| \begin{array}{c} 1,1 \\ 0,0,k+1 \\ \end{array} \right.\right)\right)=-a+b\,k^c$$

$$\begin{array}{|llll} \hline \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ a & 6.9200879 & 0.2693403 & \{6.3889276,7.4512482\} \\ b & 1.2690172 & 0.0074564 & \{1.2543126,1.2837218\} \\ c & 1.2333239 & 0.0010909 & \{1.2311725,1.2354753\} \\ \end{array}$$