chi-square random variable pdf

Question

Hi could anyone please help me in computing the bellow integral $$\int_{0}^{\tau}\frac{y^{-1/2}e^{-y/2}}{2^{1/2}\gamma(1/2)}dy$$

Answer 1

For starters, notice that both $2^{0.5}$ and $\Gamma(1/2)$ are constants, so they can safely be moved in front of

the integral: $\displaystyle\int_0^\tau\frac{y^{-0.5}e^{y/2}}{2^{0.5}\Gamma(1/2)}dy=\frac1{2^{0.5}\Gamma(1/2)}\int_0^\tau y^{-0.5}e^{y/2}dy$ . Now, we know that $a^{1/b}=\sqrt[b]a$ ,

so $2^{0.5}=2^{1/2}=\sqrt2$ . Also, $\Gamma(1/2)=\sqrt\pi$ . All that's now left to do is to let $y=t^2$, and express the

remaining integral in terms of the Gaussian error function: $\displaystyle\int_0^\tau y^{-0.5}e^{y/2}dy=\int_0^\sqrt\tau t^{-1}e^{t^2/2}2t\,dt$

$=\displaystyle2\int_0^\sqrt\tau e^{t^2/2}dt=2\frac{\sqrt2}i\int_0^{i\sqrt\frac\tau2} e^{-x^2}dx=-2i\sqrt2\frac{\sqrt\pi}2\text{erf}\bigg(i\sqrt\frac\tau2\bigg)=\sqrt{2\pi}\cdot\text{erfi}\bigg(\sqrt\frac\tau2\bigg)$. The

second substitution was $\dfrac{t^2}2=-x^2$. Now all that's left to do is to divide this result through the two constant terms above, finally arriving at $\text{erfi}\bigg(\sqrt{\dfrac\tau2}\bigg)$.