Integral of incomplete gamma function I am trying to integrate this:
$$\int_0^\infty z^{-|M|-1}\,\Gamma(A,z)\;dz$$
where $A$ is a real positive, and note that the power of $z$ is $-|M|-1$, i.e., is forced to be negative real.
 A: First: I fear that your initial integral is ill defined at $0$ since the integrant will be equivalent to $z^{-|M|-1}\,\Gamma(A)$ there (for $A>0$ ; it looks even worse for other values of $A$...).
Since you asked at least two other questions about this kind of integrals let's see how far we can go :
The integral would be straightforward without the $|M|$ sign constraint since for $B:=-|M|$ (with $B$ supposed positive!) we would have :
\begin{align}
\int_0^\infty z^{B-1}\,\Gamma(A,z)\;dz&=\int_0^\infty \int_z^\infty z^{B-1}\,e^{-t}\,t^{A-1}\,dt\;dz\\
&=\int_0^\infty \int_0^t z^{B-1}\,dz\;e^{-t}\,t^{A-1}\;dt\\
&=\int_0^\infty \left.\frac{z^B}B\right|_0^t\,dz\;e^{-t}\,t^{A-1}\;dt\\
&=\frac 1B\int_0^\infty e^{-t}\,t^{A+B-1}\;dt\\
&=\frac {\Gamma(A+B)}B,\quad\text{for}\;\Re(A+B)>0\\
\end{align}
The problem for $B$ negative comes from the lower bound $0$ of $\int_0^t z^{B-1}\,dz$ which will give infinity.
The same problem appears using integration by parts :
\begin{align}
\int_0^\infty z^{B-1}\,\Gamma(A,z)\;dz&=\left.\frac{z^B}B\Gamma(A,z)\right|_0^\infty+\int_0^\infty \frac{z^B}B\,z^{A-1}\,e^{-z}\;dz\\
&=\frac {\Gamma(A+B)}B,\quad\text{for}\;\Re(B)>0\;\text{and}\;\Re(A+B)>0\\
\end{align}
I'll add that Gradshteyn and Ryzhik's Table contains the entry $(6.455)$ from Erdelyi's book H.T.F. $2$ chap. $9.3$ (for $\;\Re(b)>0,\;\Re(a+b)>0,\;\Re(s)>-\frac 12$) :
$$\int_0^\infty e^{-sz}\,z^{b-1}\,\Gamma(a,z)\,dz=\frac{\Gamma(a+b)}{b\,(1+s)^{a+b}} \,_2F_1\left(1,a+b;b+1;\frac s{1+s}\right)$$
The substitutions $\;a:=A,\;b:=B=-|M|,\;s=0\;$ would give the previous result (because the hypergeometric function $_2F_1$ takes the value $1$ when the rightmost term is $0$) but of course the Erdelyi formula requires the same $\;\Re(-|M|)>0\,$ and $\,\Re(A)>\Re(|M|)$ we had earlier. Another idea would be to use integration under the integral sign relatively to $s$ to obtain negative powers of $z$ but the hypergeometric function is not encouraging...).
Perhaps that 'the physics' of your problem allows to use analytic continuation for negative values of $B$ so that the answer will still be $\displaystyle \frac{\Gamma(A-|M|)}{-|M|}$ (or replace the lower bound $0$ by $\epsilon$ in my attempts) I don't know...
Hoping this helped anyway,
