Relationship between the incomplete gamma function of 2a and a

Question

If the gamma function is given by

$$\Gamma(\alpha) = \int_0^{+\infty}t^{\alpha-1}e^{-t}\text dt$$

and the lower incomplete gamma function by

$$\gamma(\alpha,x) = \int_{0}^{x}t^{\alpha-1}e^{-t}\text dt$$

Is it possible to derive $\gamma(2\alpha,x)$ from $\gamma(\alpha,x)$, $\Gamma(\alpha)$, and $\Gamma(2\alpha)$? The reason why I believe a relationship like that must hold is that, when adding two random variables distributed according to a truncated $\Gamma(\alpha,\beta)$ distribution with support on $[0,w)$

$$f(x;\alpha,\beta)=\mathbb{1}_{0\leq x < w}\frac{\beta^\alpha x^{\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)}$$

I get a term that goes like

$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)^2} \frac{\Gamma(\alpha)^2}{\Gamma(2\alpha)}$$

which looks a lot like the truncated gamma function with parametre $2\alpha$

$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(2\alpha,w\beta)}$$

Answer 1

Notice that

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

Now if $s = 2\alpha$...

Notice also that

$$\Gamma(2\alpha) = \frac{2^{2\alpha - 1}}{\sqrt{\pi}}\Gamma\left(\alpha + \frac{1}{2}\right)\Gamma(\alpha)$$