I am asked to find E(Y) of the density function:
$f(y) = \frac{y^{\frac{3}{2}}}{\frac{3}{4}\sqrt\pi}e^{-y},$ for $0<y<\infty$
So, by definition
E(Y)= $\int_0^\infty yf(y)dy = \int_0^\infty \frac{y^{\frac{5}{2}}}{\frac{3}{4}\sqrt\pi}e^{-y}$
How would I compute that integral without using "erf" (I do not even know what that is, so I doubt that in my evaluation we use it).
I can see the similarity on the gamma function, but I cannot solve the integral. Like, how can I even start?
On
The PDF $f(y)$ is of a $\text{Gamma}(\alpha = 5/2, \beta=1)$ distribution, as you've mentioned in the comments. Note that $\Gamma(5/2) = (3/4)\sqrt{\pi}$.
Without resorting to the shortcut formulas, here's another way you can approach the problem.
Observe $$\mathbb{E}[Y] = \int_{0}^{\infty}\dfrac{y^{5/2}}{(3/4)\sqrt{\pi}}e^{-y}\text{ d}y\text{.}$$ Take into account that the integrand looks very, very similar to a $\text{Gamma}(\alpha= 7/2, \beta = 1)$ PDF, in which case
$$\begin{align}\int_{0}^{\infty}\dfrac{y^{5/2}}{\Gamma(7/2)}e^{-y}\text{ d}y &= \int_{0}^{\infty}\dfrac{y^{5/2}}{(5/2)\Gamma(5/2)}e^{-y}\text{ d}y \\ &= \int_{0}^{\infty}\dfrac{y^{5/2}}{(5/2)(3/4)\sqrt{\pi}}e^{-y}\text{ d}y\\ &=\dfrac{2}{5}\int_{0}^{\infty}\dfrac{y^{5/2}}{(3/4)\sqrt{\pi}}e^{-y}\text{ d}y \\ &= 1 \end{align}$$ because we are integrating a density function. From simple algebra, we find that $$\mathbb{E}[Y] = \int_{0}^{\infty}\dfrac{y^{5/2}}{(3/4)\sqrt{\pi}}e^{-y}\text{ d}y = \dfrac{5}{2}\text{.}$$
To calculate the integral without using the formulae for the distribution simply notice that
$$E(Y)=\frac{1}{\frac{3}{4}\sqrt\pi}\int_0^\infty y^{7/2-1}e^{-y}\mathrm{d}y=\frac{4}{3\sqrt{\pi}}\Gamma(7/2)=\frac{4}{3}\frac{15}{8}=\frac{5}{2} $$
because $\Gamma(1/2)=\sqrt\pi$ and $\Gamma(x+1)=x\Gamma(x)$