Solve equation with lower gamma function: $A \gamma(2;x/B)=x$ for $x$ I need to find an expression for $x$ given:
$A \gamma(2;x/B)=x$
where $\gamma(a,x)=\int\limits_0^x t^{a-1} e^{-t} \mathrm{d}t$ is the lower incomplete gamma function. $A$ and $B$ are real, positive constants and $x$ should also be real and positive.
Mathematica is unable to solve this equation directly (neither Solve nor Reduce work). I have tried a Taylor expansion around $r=B$, but I don't get the desired convergence when $B \rightarrow 0$. This is important because eventually I'd like to get an expression valid when $r$ is of the order of magnitude of $B$, but that also reduces to $r=A$ when $B$ vanishes.
I've also tried writing the gamma function in terms of modified Bessel functions (http://dlmf.nist.gov/8.7), but that didn't work either.
Would an asymptotic expansion work? In that case, what kind and how would I implement it? Also, where would it converge?
A second order approximation would be fine, but a numerical answer would not be ideal.
Any suggestions? Thanks!
 A: Well, hope this will be of any help (not solution, just thoughts).
$$A \gamma\bigg(2;\frac{x}{B}\bigg)=x$$
I'll call $a=\frac{A}{B}, \ z=\frac{x}{B}$, so  $a \gamma(2;z)=z$. Then I use the relation between incomplete gamma and confluent hypergeometric function $c z^c \gamma(c,z)=\mathrm{M}(c,c+1,-z)$, so, if we consider that $z\ne0$:
$$z \ \mathrm{M}(2,3,-z)=\frac{2}{a}$$.
$$\mathrm{M}(2,3,-z)=\frac{2 e^{-z} \left(-z+e^z-1\right)}{z^2}$$
$$\frac{2 e^{-z} \left(-z+e^z-1\right)}{z}=\frac{2}{a}$$
So (if I didn't loose anything)  I get the equation like this:
$$\frac{1}{a \ z}\bigg((1+z)e^{-z}+\bigg(\frac{z}{a}-1\bigg)\bigg)=0 $$
I guess starting here you can use series expansion or different approximations in cases of big/small arguments.
A rough plotting(for example when $a=5$) shows that there should be some real positive roots. 
A: The integral is elementary:
$$\gamma(2;x)=\int_0^x te^{-t}dt=-\left.e^{-t}(t+1)\right|_0^x=1-e^{-x}(x+1).$$
Then, $$\frac xA+e^{-x/B}(x+1)=1,$$
which doesn't seem to be expressible with the Lambert function.
