What is the result of $\int_{0}^{+a} \sqrt{z} e^{-z^2/2}\,dz$?

Question

I encounter this integration: $$\int_{0}^{+a} \sqrt{z} e^{-z^2/2}\,dz$$ I think that it is a special function, but I don't remember now. Can anyone give me a pointer?

Answer 1

For the sake of the answer, I will replace $\sqrt{z}\to\sqrt{|z|}$ to take care of the ambiguity of the branch of the square root function. Your integral is then, after the change of variable $z\to\sqrt{z}$ followed by $z\to 2z$, $$2\int_0^a \sqrt{z}e^{-z^2/2}dz=\sqrt{2}\int_0^{a^2/2} z^{-1/2} e^{-z}dz.$$ This is the lower incomplete Gamma function: $$\sqrt{2}\,\gamma\left(\frac{1}{2}, \frac{a^2}{2}\right).$$ You have that $$\gamma\left(\frac{1}{2}, u\right)\uparrow \sqrt{\pi}\qquad \text{ as }u\to\infty.$$ The error is given by $$0<\sqrt{\pi}-\gamma\left(\frac{1}{2}, u\right)=\int_u^{\infty} z^{-1/2} e^{-z}dz:=R(u).$$ We have the simple estimate $$0<R(u)<\frac{e^{-u}}{\sqrt{u}}.$$ An integration by parts will also show you that $$0<\frac{e^{-u}}{\sqrt{u}} - R(u)<\frac{e^{-u}}{2\,u^{3/2}}.$$ Continuing, we can derive the full asymptotic expansion of $\gamma(1/2,u)$ as $u \to\infty$. For all $N\in \mathbb{N}_0$, we have that $$\gamma(1/2, u)=\sqrt{\pi}-\sum_{k=0}^N (-1)^k \frac{e^{-u}}{u^{k+1/2}}\frac{(2k-1)!!}{2^k} +O\left(\frac{e^{-u}}{u^{N+3/2}}\right).$$ with $(2k-1)!!=(2k-1)(2k-3)\cdots 1$ the double-factorial, satisfying $(-1)!!=1$.