# 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!

• Apart from the obvious root $x=0$, are you sure there are others? – J. M. isn't a mathematician Apr 5 '13 at 5:57
• There are strong reasons to think so. This equation describes the radius of a black hole (x) in certain modified theory of gravity and I think there are some numerical solutions. Finding an expression for the black hole radius r is the reason why I need to solve this. – misi Apr 5 '13 at 6:12
• (But depending on A and B there could be no other solutions) – misi Apr 5 '13 at 6:13
• Okay, do you know of any particular values of $A,B$ where there are nonobvious solutions? – J. M. isn't a mathematician Apr 5 '13 at 6:16
• Just playing around plotting the equation it seems that A=2, B=.1 and r=2 work... – misi Apr 5 '13 at 6:25

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.
• Thanks for your thoughts. This is useful and might help me make numerical approximations. However, I'm still unable to find a closed expression, even an approximation, that is valid both when $x\rightarrow B$ and $B\rightarrow \infty$. I have reasons to think this might be impossible. It seems that any way I can come up with to effectively circumvent the gamma function ends up with something like $1+x e^x=x$ which I cannot solve. I'll open another question asking this. – misi Apr 6 '13 at 5:53
$$\gamma(2;x)=\int_0^x te^{-t}dt=-\left.e^{-t}(t+1)\right|_0^x=1-e^{-x}(x+1).$$
Then, $$\frac xA+e^{-x/B}(x+1)=1,$$ which doesn't seem to be expressible with the Lambert function.