I am trying to derive the mean of right truncated $t$-distribution but cannot solve the integration further. I tried to find the truncated mean as follow:
The pdf of right truncated distribution is;
$\frac{{\Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{\sqrt {\pi n} \Gamma \left( {\frac{v}{2}} \right)\left( {1 - \Phi \left( z \right)} \right)}}{\left( {1 + \frac{{{t^2}}}{n}} \right)^{ - \left( {\frac{{n + 1}}{2}} \right)}}$
Where $\Phi \left( z \right)$ is cdf of $t$-distribution
The Mean of distribution is written as;
$E\left( t \right) = \frac{{\Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{\sqrt {\pi n} \Gamma \left( {\frac{v}{2}} \right)\left( {1 - \Phi \left( z \right)} \right)}}\int\limits_z^\infty t {\left( {1 + \frac{{{t^2}}}{n}} \right)^{ - \left( {\frac{{n + 1}}{2}} \right)}}$
Substituting
$k = \frac{{{t^2}}}{n};$
$\sqrt {kv} = t;$
$dt = \frac{n}{{2\sqrt {kn} }}dk.$
Putting we get;
$E\left( t \right) = \frac{{\Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{\sqrt {\pi n} \Gamma \left( {\frac{v}{2}} \right)\left( {1 - \Phi \left( z \right)} \right)}}\int\limits_{\frac{{{z^2}}}{n}}^\infty {\sqrt {kv} } {\left( {1 + k} \right)^{ - \left( {\frac{{n + 1}}{2}} \right)}}\frac{n}{{2\sqrt {kv} }}dk$
$E\left( t \right) = \frac{{\sqrt n \Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{2\sqrt \pi \Gamma \left( {\frac{v}{2}} \right)\left( {1 - \Phi \left( z \right)} \right)}}\int\limits_{\frac{{{z^2}}}{n}}^\infty {{{\left( {1 + k} \right)}^{ - \left( {\frac{{n + 1}}{2}} \right)}}} dk$
Can someone help me how to solve it further?