TAOCP 1.2.11.3 Exercise 12
We want to express the integral \[ \int_0^x e^{-t^2/2}\,dt \] in terms of the incomplete gamma function \[
Section 1.2.11.3: Some Asymptotic Calculations
Exercise 12. [HM20] The function $\int_0^x e^{-t^2/2},dt$ that appeared in connection with the normal distribution (see Section 1.2.10) can be expressed as a special case of the incomplete gamma function. Find values of $a$, $b$, and $y$ such that $b,\gamma(a,y)$ equals $\int_0^x e^{-t^2/2},dt$.
Verified: yes
Solve time: 5m01s
Solution
We want to express the integral [ \int_0^x e^{-t^2/2},dt ] in terms of the incomplete gamma function [ \gamma(a,y) = \int_0^y e^{-u},u^{a-1},du \qquad (a>0). ] Perform the substitution (u = t^2/2). Then (t = \sqrt{2u}), (dt = \frac{1}{\sqrt{2}} u^{-1/2},du). The limits become (u=0) when (t=0) and (u=x^2/2) when (t=x). Hence [ \int_0^x e^{-t^2/2},dt = \int_0^{x^2/2} e^{-u},\frac{1}{\sqrt{2}},u^{-1/2},du = \frac{1}{\sqrt{2}} \int_0^{x^2/2} e^{-u},u^{-1/2},du. ] Comparing with the definition of (\gamma(a,y)) we identify [ a-1 = -\frac{1}{2} \quad\Longrightarrow\quad a = \frac{1}{2}, ] [ y = \frac{x^2}{2}, ] and the constant factor is (b = \frac{1}{\sqrt{2}}).
Thus [ b,\gamma(a,y) = \frac{1}{\sqrt{2}},\gamma!\left(\frac{1}{2},,\frac{x^2}{2}\right) = \int_0^x e^{-t^2/2},dt. ]
\boxed{a = \frac{1}{2},\quad b = \frac{1}{\sqrt{2}},\quad y = \frac{x^2}{2}}