Existence of $E(X^2)$ when $X$ has the pdf $f(x)= \frac{1}{(2+x^2)^{3/2}}$ In a competitive exam, I came across an objective question which says

Let $X$ be a continuous random variable with the probability density function 
  $$f(x)= \frac{1}{(2+x^2)^{3/2}}\quad,\,-\infty<x<\infty$$ 
Find $E(X^2)$. And the options were
(a) $\quad0 \qquad$ (b) $\quad1 \qquad$ (c) $\quad2 \qquad$ (d) $\quad \text{ does not exist }$

My question is should I directly find out the $E(X^2)$ or should I first check whether it exists or not? Please suggest me something which takes less time as I am preparing for a competitive exam. If you suggest me to do the former please help me with the integration too. I wasn't able to solve it. It took me a lot of time. 
My attempt: I first checked whether the $E(X)$ exists or not, for that I evaluated $E(|X|)$ and found out that it doesn't exist, so $E(X)$ doesn't exist and hence $E(X^2)$ doesn't exist. Is it the correct way? 
 A: Here's one thing you can do that's quick if you recognize this. The t distribution has pdf
$$
f(x; \nu) = \frac{\Gamma\left(\frac{\nu + 1}2\right)}{\sqrt{\nu\pi}\Gamma\left(\frac \nu 2\right)} \left[1 + \frac 1\nu x^2\right]^{-\frac{\nu + 1}{2}}
$$
which suggests trying $\nu =2$. Plugging this in, this reduces to
$$
\frac{\Gamma(3/2)}{\sqrt{2\pi}\Gamma(1)} \left[1 + 2^{-1}x^2\right]^{-3/2}\\
= \frac{2^{-1}\sqrt\pi}{2^{1/2}\sqrt \pi}2^{3/2}(2 + x^2)^{-3/2} \\
= (2 + x^2)^{-3/2}
$$
which is exactly your pdf, so you've just got a t distribution with two degrees of freedom. This immediately tells me that $E(X^2)$ is infinite. But the mean does exist here and is $0$, which can be immediately known if you know a t distribution with two d.f. has a mean, or obtained directly via
$$
\text E(X) = \int_{0}^\infty \frac{x}{(2 + x^2)^{3/2}}\,\text dx + \int_{-\infty}^0\frac{x}{(2 + x^2)^{3/2}}\,\text dx \\
= \frac 12 \left(\int_2^\infty u^{-3/2}\,\text du - \int_2^\infty u^{-3/2}\,\text du\right) = 0
$$
because both integrals converge (if this was a Cauchy RV it'd be the indeterminate form $\infty - \infty$ which isn't equal to zero).
A: From the stated form of the density, $X$ has a Student's T distribution with two degrees-of-freedom.  We will show here that the second raw moment of this distribution does not to exist (i.e., it is infinite).  (You can find an answer to a more general question answered here that shows the derivation of all the raw moments of the T-distribution, but we will not need to go into this level of complexity here.)
When you are trying to prove that an integral is infinite, it is often easiest to do this by making liberal use of inequalities --- e.g., show that the integral of interest is larger than another integral, that is larger than another integral, that is infinite.  That is the method we will use here.  Let $y = x^2$ so that $dy = 2x \ dx$ and then use a change of variable to get:
$$\begin{equation} \begin{aligned}
\mathbb{E}(X^2) &= \int \limits_{-\infty}^\infty \frac{x^2}{(2+x^2)^{3/2}} \ dx \\[6pt]
&= 2 \int \limits_0^\infty \frac{x^2}{(2+x^2)^{3/2}} \ dx \\[6pt]
&= \int \limits_0^\infty \frac{x}{(2+x^2)^{3/2}} \cdot 2x \ dx \\[6pt]
&= \int \limits_0^\infty \frac{\sqrt{y}}{(2+y)^{3/2}} \ dy \\[6pt]
&= \int \limits_0^\infty \sqrt{\frac{y}{2+y}} \cdot \frac{1}{2+y} \ dy. \\[6pt]
\end{aligned} \end{equation}$$
Now, choose any value $0 < y_0 < \infty$ and you have:
$$\begin{equation} \begin{aligned}
\mathbb{E}(X^2) 
&= \int \limits_0^\infty \sqrt{\frac{y}{2+y}} \cdot \frac{1}{2+y} \ dy \\[6pt]
&> \int \limits_{y_0}^\infty \sqrt{\frac{y}{2+y}} \cdot \frac{1}{2+y} \ dy \\[6pt]
&> \sqrt{\frac{y_0}{2+y_0}} \int \limits_{y_0}^\infty \frac{1}{2+y} \ dy \\[6pt]
&= \sqrt{\frac{y_0}{2+y_0}} \Big[ \ln (2+y) \Big]_{y_0}^\infty \\[6pt]
&= \sqrt{\frac{y_0}{2+y_0}} \times \infty = \infty. \\[6pt]
\end{aligned} \end{equation}$$
