Definite integration includes Exponential and Logarithmic functions Does anyone have an idea of having solution for the following integration?
$$I=\int_0^\infty x^a e^{-b x} \log(1+cx)dx$$
where $a$ is a positive integer and $b,c>0$.
I found similar integral in eq. (2.6.23.4) in book " Integrals and Series: elementary functions. Vol.1", which is only valid for non-integer $a$.
Further, if I give same integral for known $a$ values such as $a=1,2,3,..$, then MATHEMATICA gives answers one by one. But it did not give for general $a$.
 A: Start with $x=\frac t c$ and $k=\frac b c$ to make
$$I=\int_0^\infty x^a e^{-b x} \log(1+cx)\,dx=\frac 1{c^{a+1}}\int_0^\infty t^a\, e^{-k t} \,\log(1+t)\,dt$$ Even specifying that $a$ is integer and $k>0$, you will recieve for
$$J=\int_0^\infty t^a\, e^{-k t} \,\log(1+t)\,dt$$
$$ k \left(-k^2\right)^a\, J=k (-k)^a \Gamma (a) \, _2F_2(1,1;2,1-a;k)-$$ $$\pi  k^a \csc (\pi  a) \Gamma (a+1,-k)+$$ $$a
   \Gamma (a) \left(\pi  k^a \csc (\pi  a)-(-k)^a \log (k)+(-k)^a \psi
  (a+1)\right)$$ which present problems of indeterminate forms when you try $a=1,2,\cdots$.
I must confess that I prefer the individual cases which show some interesting patterns.
$$J_a=\frac 1 {k^{a+1}} \Big[P_{a-1}+Q_a\,e^k\, \text{Ei}(-k) \Big]$$ where $P_n$ and $Q_n$ are simple polynomials of degree $n$ in $k$.
A: I was able to derive a recursive formula for the integral based purely on b in a simpler case, then expanded to the current problem.

Through IBP,

Where a is any positive integer and b is a positive number. Then, you can solve for the first two cases individually through IBP.


Where  is the Exponential Integral function
Your problem can be derived from this through substitution as the relation:

Where the b/c term replaces the b in the I integral.
I couldn't find an explicit formula for the integral, but the recursion works from what I could tell and try. Let me know if someone finds one or I did something wrong.
