Solution of equation $[1+\frac{x}{b}]e^{-x/b}=z$ Can we solve this equation
$$\left(1+\frac{x}{b}\right)e^{-x/b}=z$$
We have to determine value of $x$ in term of $z$.
Problem occur while calculating the following integral.
$$\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx $$
and subtituting
$$(1+\frac{x}{b})e^{-\frac{x}{b}}=z$$  
 A: As Famous Blue Raincoat commented, you can rewrite $$\left(1+\frac{x}{b}\right)e^{-x/b}=e\left(1+\frac{x}{b}\right)e^{-(1+x/b)}$$ and setting $y=-(1+\frac{x}{b})$, the equation write $$y e^{y}=-\frac{z}{e}$$ the solution of which being given by Lambert function $$y=W\left(-\frac{z}{e}\right)$$ and, back to $x$, $$x=-b \left(1+W\left(-\frac{z}{e}\right)\right)$$
In a more general manner, any equation which can be written as $$A+B x+C \log(D+Ex)=0$$ has solutions which can expressed in terms of Lambert function.
However, may I confess that I do not see how to compute the integral ? 
A: Your original problem is
$f(a, b, r)
=\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx
$.
I'll play with it and see what I get.
Letting
$x/b = y$,
so
$dx = b dy$,
$f(a, b, r)
=\frac{a}{b}\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}(by)^{r+1}e^{-y}dy
=ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy
$.
Since $e^y > 1+y$
for $y ? 0$,
$(1+y)e^{-y}
< 1$
so we could expand
$(1-(1+y)e^{-y})^{a-1}
$
with the binomial theorem.
This gives
$(1-(1+y)e^{-y})^{a-1}
=\sum_{n=0}^{\infty}
(-1)^n
\binom{a-1}{n}
((1+y)e^{-y})^n
$.
Integrating term-by-term,
$\begin{array}\\
f(a, b, r)
&=ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy\\
&=ab^r\int_{0}^{\infty}\sum_{n=0}^{\infty}
(-1)^n
\binom{a-1}{n}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\
&=ab^r\sum_{n=0}^{\infty}(-1)^n
\binom{a-1}{n}
\int_{0}^{\infty}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\
\end{array}
$
Looking at the inside integral,
$\begin{array}\\
\int_{0}^{\infty}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy
&=\int_{0}^{\infty}
(1+y)^ne^{-ny}y^{r+1}e^{-y}dy\\
&=\int_{0}^{\infty}
\sum_{k=0}^n \binom{n}{k}y^ke^{-(n+1)y}y^{r+1}dy\\
&=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty}
e^{-(n+1)y}y^{k+r+1}dy\\
&=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty}
e^{-z}\left(\frac{z}{n+1}\right)^{k+r+1}\frac{dz}{n+1}\quad (z = (n+1)y)\\
&=\sum_{k=0}^n \binom{n}{k}\frac{1}{(n+1)^{k+r+2}}\int_{0}^{\infty}
e^{-z}z^{k+r+1}dz\\
&=\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k+r+2}}\\
&=\frac1{(n+1)^{r+2}}\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k}}\\
\end{array}
$
There might be something
that can be done if
this is inserted into the sum above,
but I'll leave it at this.
