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

Question

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.