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Has anybody seen an integral of the following form or has some intuition on solving the following integral

$$\int\limits_{ - \infty }^{q} {{t_\nu }\left( x \right)\log \left( {\frac{{{x^2}}}{\nu } + 1} \right)} dx $$ where $${t_\nu }\left( x \right) = \frac{{\Gamma \left( {\frac{{\nu + 1}}{2}} \right)}}{{\sqrt {\pi \nu } \cdot \Gamma \left( {\frac{\nu }{2}} \right)}}{\left( {1 + \frac{{{x^2}}}{\nu }} \right)^{ - \frac{{\left( {1 + \nu } \right)}}{2}}}$$

I tried a bunch of substitutions along with integration by parts but none seem to simplify the integral. I tried mathematica too, but it failed. I can evaluate it numerically but was wondering if I missed something obvious

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  • $\begingroup$ $$\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\pi\nu}\cdot\Gamma\left(\frac{\nu}{2}\right)}\int_{-\infty}^{q} \frac{\ln\left({\frac{{{x^2}}}{\nu }+1}\right)}{\left(1+\frac{x^2}{\nu}\right)^{\frac{1+\nu}{2}}}\text{d}x$$ $\endgroup$ – Jan Oct 20 '15 at 16:29
  • $\begingroup$ How does that help? $\endgroup$ – Rohit Arora Oct 20 '15 at 20:10

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