I know that $$ \Gamma(z) = \int_0^{\infty} x^{z-1}e^{-x}dx $$ is the Gamma function, but what about $$ \int_a^{\infty} x^{z-1}e^{-x}dx $$ for some $a > 0$? Does this function have a special name? I figure I'm okay with writing a function to evaluate this when $z \in \mathbb{Z}$ because if we take the change of variables $y = x-a$, then \begin{align*} \int_a^{\infty} x^{z-1}e^{-x}dx &= \int_0^{\infty}(y+a)^{z-1}e^{-(y+a)}dy \\ &= e^{-a} \int_0^{\infty} \sum_{k=0}^{z-1}{z-1 \choose k}y^ka^{z-1-k} e^{-y}dy \\ &= e^{-a} \sum_{k=0}^{z-1} {z-1 \choose k} a^{z-1-k} \int_0^{\infty} y^k e^{-y}dy \\ &= e^{-a} \sum_{k=0}^{z-1} {z-1 \choose k} a^{z-1-k} \Gamma(k+1), \end{align*} but this isn't enough for me because I am interested in $z \overset{\text{set}}{=} 3/2$ and considering it as a function of $a$.
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5https://en.wikipedia.org/wiki/Incomplete_gamma_function – carmichael561 Mar 16 '17 at 23:31
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1The particular thing you need is the error function for $z=3/2$, which is related to the incomplete gamma function. – Simply Beautiful Art Mar 16 '17 at 23:32
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oh I've heard of those. Thanks guys. Also related to the cumulative distribution function of a gamma random variable. – Taylor Mar 16 '17 at 23:43