What substitution (to be used in integration) can proof the following equality?

Question

I have seen following to be true in one of the research paper $$2\pi \lambda\int_v^{\infty}r^{1-\beta}e^{-\pi\lambda r^2}dr=(\pi\lambda)^{\frac{\beta}{2}}\gamma\left(\pi\lambda v^2,1-\frac{\beta}{2}\right)$$ where $\beta>2$ $v$ is some positive value and $\lambda>0$ ($\gamma$ is the lower incomplete Gamma function). I can show it to be true (with upper gamma function instead of lower gamma function, and that too for some specific range of $\beta$) but I do not know how above equation is valid for all values of $\beta >2$. I need your help in this regard. Many thanks in advance.

Answer 1

Concerning the antiderivative $$I=2\pi \lambda\int r^{1-\beta}e^{-\pi\lambda r^2}\,dr=-(\pi \lambda)^{\frac \beta 2}\,\,\Gamma \left(1-\frac{\beta }{2},\pi \lambda r^2\right)$$ it seems that the limit at $\infty$ is equal to $0$ if $\lambda >0$ without any other restriction.

Then, if $v>0$, $$2\pi \lambda\int_v^\infty r^{1-\beta}e^{-\pi\lambda r^2}\,dr=(\pi \lambda)^{\frac \beta 2}\,\,\Gamma \left(1-\frac{\beta }{2},\pi \lambda v^2\right)$$ But, as far as I remember,

$$\gamma(s,x) + \Gamma(s,x) = \Gamma(s)$$

In any manner, it seems that, using the upper gamma function, $\beta$ could be whatever you wish.