Solución de integrales definidas que involucran la función Gamma incompleta

Question

La solución de la 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 $\nu>0$]. Necesito calcular la misma integral de límite finito como$0$$a$. Entonces, ¿cómo puedo calcular el $\{\int_0^{a}e^{-\beta x}\gamma(\nu,\alpha \sqrt x)dx\}$. Es allí cualquier forma de solución para esta integración numérica?

Answer 1

$\int_0^ae^{-\beta x}\gamma(\nu,\alpha\sqrt x)~dx$

$=-\int_0^a\gamma(\nu,\alpha\sqrt x)~d\left(\dfrac{e^{-\beta x}}{\beta}\right)$

$=-\left[\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt x)}{\beta}\right]_0^a+\int_0^a\dfrac{e^{-\beta x}}{\beta}d\left(\gamma(\nu,\alpha\sqrt x)\right)$

$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^a\dfrac{\alpha^\nu x^{\nu-1}e^{-\beta x-\alpha\sqrt x}}{2\beta}dx$

$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-2}e^{-\beta x^2-\alpha x}}{2\beta}d(x^2)$

$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-1}e^{-\beta x^2-\alpha x}}{\beta}dx$

$=-\dfrac{e^{-\beta x}\gamma(\nu,\alpha\sqrt a)}{\beta}+\int_0^\sqrt a\dfrac{\alpha^\nu x^{2\nu-1}e^{-\beta\left(x^2+\frac{\alpha x}{\beta}\right)}}{\beta}dx$

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

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

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