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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.

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  • $\begingroup$ Hint: $$\int_0^\infty z^{-|M|-1}\,\Gamma(A,z)\;dz=\int_0^\infty \int_z^\infty z^{-|M|-1}\,e^{-t} t^{A-1}\,dt\;dz$$ Change order of integration $\;\displaystyle \int_0^\infty \int_z^\infty f(t,z)\,dt\;dz=\int_0^\infty \int_0^t f(t,z)\,dz\;dt$ (you'll get a gamma function but with a possibly negative parameter). $\endgroup$ – Raymond Manzoni Mar 16 '14 at 18:26
  • $\begingroup$ Can you please complete the integration? $\endgroup$ – kazekage Mar 16 '14 at 19:13
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    $\begingroup$ When $\displaystyle{\large z \sim 0}$, the integrand is $\displaystyle{\large\sim{\Gamma(a,0) \over z^{\vert M\vert + 1}}}$. So ?. $\endgroup$ – Felix Marin Mar 16 '14 at 20:43
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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,

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  • $\begingroup$ Thank you for your answer. Actually, the integral: $${\int_0^\infty z^{B-1}\,\Gamma(A,z)\;dz}$$ is part of a Taylor series expansion. That Taylor series can be considered for values of |z|<1 and |z|>1. For one case, M tends to be strictly positive and is working fine. However, for the other case, it tends to be strictly negative and this where my problem comes from. As you just showed, the derivation is straightforward for the positive case, but I don't know how I can deal with the negative case. $\endgroup$ – kazekage Mar 17 '14 at 8:48
  • $\begingroup$ @user84310: the additional infinite term at the lower bound is problematic. If you replace the lower bound $0$ by $\epsilon>0$ in my derivations you'll get instead of $\infty$ the additional term $\dfrac{\epsilon^{-|M|}\,\Gamma(A)}{|M|}$. Perhaps that your (not shown) divergent integrals may be 'regularized' the same way so that the $\epsilon$ contributions disappear at the end. If not your initial problem should be ill defined (an infinite contribution missing?). $\endgroup$ – Raymond Manzoni Mar 17 '14 at 12:45
  • $\begingroup$ It seems not working, I wrote a MATLAB code to test it and did not work. $\endgroup$ – kazekage Mar 18 '14 at 6:26

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