# How to prove following integral can be represented as lower incomplete gamma function?

In one of IEEE journal I have seen the following equation to be true $$2\pi \lambda\int_v^{\infty}r^{1-\beta}e^{-\pi\lambda r^2}dr=(\pi \lambda)^{\frac{\beta}{2}}\gamma\left(\pi\lambda v^2,1-\frac{\beta}{2}\right)$$when I use substitution $h=\pi \lambda r^2$ then my answer turns out to be as follows $$2\pi \lambda\int_v^{\infty}r^{1-\beta}e^{-\pi\lambda r^2}dr=(\pi \lambda)^{\frac{\beta}{2}}\Gamma\left(\pi\lambda v^2,1-\frac{\beta}{2}\right)$$ which is similar in appearance as the first equation except we now have Upper incomplete Gamma function. I want to know whether my answer is wrong or there is some pen mistake in the IEEE journal. Thanks in advance.

• I get your solution with the upper incomplete gamma function. It appears that there is a misprint in the journal article. – poweierstrass Oct 11 '16 at 11:26
• @poweierstrass I found that the solution with upper gamma function is also not applicable to all possible values of $\beta$. Just wanted to share it with you – Frank Moses Oct 17 '16 at 8:45
• I know. The incomplete gamma functions are tricky regarding allowable values of the argument, integration paths, and analytic continuation. See DLMF (dlmf.nist.gov/8.2) and Volume 1 of Higher Transcendental Functions (Bateman Manuscript) (resolver.caltech.edu/CaltechAUTHORS:20140123-104529738) for details. – poweierstrass Oct 17 '16 at 11:14
• yes exactly. in this dlmf page you can see that real part of $a$ should be greater than zero. that is what i was talking about – Frank Moses Oct 17 '16 at 11:24