As the variable X is from t-distribution, so,\(E\left( X \right) = 0\)
Let us consider the square of the expectation of the random variable\(E\left( {{X^2}} \right)\)
So,
\(\begin{align}E\left( {{X^2}} \right) &= c\int_{ - \infty }^\infty {{x^2}{{\left( {1 + \frac{{{x^2}}}{m}} \right)}^{\frac{{ - \left( {m + 1} \right)}}{2}}}dx} \\ &= 2c\int_{ - \infty }^\infty {{x^2}{{\left( {1 + \frac{{{x^2}}}{m}} \right)}^{\frac{{ - \left( {m + 1} \right)}}{2}}}dx} \end{align}\)
Where\(c = \frac{{\Gamma \left( {\frac{{m + 1}}{2}} \right)}}{{{{\left( {m\pi } \right)}^{\frac{1}{2}}}\Gamma \left( {\frac{m}{2}} \right)}}\)
Now, there is provided that,
\(\begin{align}y &= \frac{{\frac{{{x^2}}}{m}}}{{1 + \frac{{{x^2}}}{m}}}\\x &= {\left( {\frac{{my}}{{1 - y}}} \right)^{\frac{1}{2}}}\\\frac{{dx}}{{dy}} &= \frac{{\sqrt m }}{2}{y^{ - \frac{1}{2}}}{\left( {1 - y} \right)^{ - \frac{3}{2}}}\end{align}\)
Now by substituting the value of x and dx in \(E\left( {{X^2}} \right)\), we get,
\(\begin{align}E\left( {{X^2}} \right) &= 2c\int_{ - \infty }^\infty {{x^2}{{\left( {1 + \frac{{{x^2}}}{m}} \right)}^{\frac{{ - \left( {m + 1} \right)}}{2}}}dx} \\ &= \sqrt m c\int_0^1 {\frac{{my}}{{\left( {1 - y} \right)}}{{\left( {1 + \frac{y}{{1 - y}}} \right)}^{\frac{{ - \left( {m + 1} \right)}}{2}}}{y^{ - \frac{1}{2}}}{{\left( {1 - y} \right)}^{ - \frac{3}{2}}}dy} \\ &= {m^{\frac{3}{2}}}c\frac{{\Gamma \left( {\frac{3}{2}} \right)\Gamma \left( {\frac{{\left( {m - 2} \right)}}{2}} \right)}}{{\Gamma \left( {\frac{{\left( {m + 1} \right)}}{2}} \right)}}\\ &= m{\pi ^{ - \frac{1}{2}}}\Gamma \left( {\frac{3}{2}} \right)\frac{{\Gamma \left( {\frac{{\left( {m - 2} \right)}}{2}} \right)}}{{\Gamma \left( {\frac{m}{2}} \right)}}\\ &= m{\pi ^{ - \frac{1}{2}}}\left( {\frac{1}{2}\sqrt \pi } \right)\frac{1}{{\left( {\frac{{\left( {m - 2} \right)}}{2}} \right)}}\\ &= \frac{m}{{m - 2}}\end{align}\)
