Solution of definite integrals involving incomplete Gamma function The solution of the integral $$\int_0^{\infty}e^{-\beta x}\gamma(\nu,\alpha \sqrt x)dx $$ is given as $$2^{-\frac{1}{2}\nu}\alpha^{\nu}\beta^{\frac{1}{2}\nu-1}\Gamma(\nu)\exp(\frac{\alpha^2}{8\beta})D_{-\nu}(\frac{\alpha}{\sqrt{2\beta}})$$
[Re $\beta>0$, Re $\nu>0$]. I need to calculate the same integral for finite limit such as $0$ to $a$. So how can I calculate $\{\int_0^{a}e^{-\beta x}\gamma(\nu,\alpha \sqrt x)dx\}$. Is there any given form of solution for this definite integration?

$\int_0^ae^{-\beta x}\gamma(\nu,\alpha\sqrt x)~dx$
$=-\int_0^a\gamma(\nu,\alpha\sqrt x)~d\left(\dfrac{e^{-\beta x}}{\beta}\right)$
$=-\left[\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt x)}{\beta}\right]_0^a+\int_0^a\dfrac{e^{-\beta x}}{\beta}d\left(\gamma(\nu,\alpha\sqrt x)\right)$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^a\dfrac{\alpha^\nu x^{\nu-1}e^{-\beta x-\alpha\sqrt x}}{2\beta}dx$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-2}e^{-\beta x^2-\alpha x}}{2\beta}d(x^2)$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-1}e^{-\beta x^2-\alpha x}}{\beta}dx$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-1}e^{-\beta\left(x^2+\frac{\alpha x}{\beta}\right)}}{\beta}dx$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-1}e^{-\beta\left(x^2+\frac{\alpha x}{\beta}+\frac{\alpha^2}{4\beta^2}-\frac{\alpha^2}{4\beta^2}\right)}}{\beta}dx$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu e^\frac{\alpha^2}{4\beta}x^{2\nu-1}e^{-\beta\left(x+\frac{\alpha}{2\beta}\right)^2}}{\beta}dx$
$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_\frac{\alpha}{2\beta}^{\sqrt a+\frac{\alpha}{2\beta}}\dfrac{\alpha^\nu e^\frac{\alpha^2}{4\beta}\left(x-\dfrac{\alpha}{2\beta}\right)^{2\nu-1}e^{-\beta x^2}}{\beta}dx$
