Post History

It is well worth knowing how to do what I show you how to do below. The books in which you find formulas were not brought down from Heaven by an archangel. They are derived by humans, and you are one of those.

(I am appalled to find that "align" does not work in MathJax here. So you will have to tolerate the formatting below.)

First, notice that (assuming $\nu>1$) as $t^2_{\nu-1}$ gets bigger, so does $\tau_\nu \sim \sqrt{\frac{\nu \, t_{\nu - 1}^2}{\nu - 1 + t_{\nu - 1}^2}},$ and also $\tau_\nu<\nu.$ That can be seen as follows:
$$
\frac{aw}{a-1+w} = a - \frac{a(a-1)}{a-1+w}
$$
and this increases as $w$ increases, but always remains less than $a.$

Hence we have $\tau_\nu\le\text{something}$ if and only if $t_{\nu-1}^2\le \text{something}.$ A bit of algebra tells us what $\text{“something”}$ is: Suppose $0\le u < \nu.$ Then
$$
\tau_\nu \le u
$$

$$
\sqrt{\frac{\nu t^2_{\nu-1}}{\nu-1+t^2_{\nu-1}}} \le u
$$

$$
\text{iff } \nu - \frac{\nu(\nu-1)}{\nu-1+t^2_{\nu-1}} \le u^2
$$

$$
t^2 \le \frac{(\nu-1)u^2}{\nu - u^2}
$$

$$
|t| \le \sqrt{ \frac{(\nu-1)u^2}{\nu - u^2} }.
$$

Therefore
$$
f_{\tau_\nu}(u) = \frac d{du} \Pr(\tau_\nu \le u)
$$

$$
= \frac d{du} \int_{-\sqrt{(\nu-1)u^2/(\nu-u^2)}}^{ +\sqrt{(\nu-1)u^2/(\nu-u^2)} } f(v) ~ dv
$$
where $f$ is the density of Student's t-distribution
$$
= 2 \frac d{du} \int_0^{ +\sqrt{(\nu-1)u^2/(\nu-u^2)} } f(v) ~ dv
$$
because $f$ is an even function and we are integrating over and interval that is symmetric about $0$
$$
= 2f \left( \sqrt{(\nu-1)u^2/(\nu-u^2)} \right) \cdot \frac d{du} \sqrt{(\nu-1)u^2/(\nu-u^2)}.
$$

Evaluate this derivative, and plug the argument to the function $f$ into the density of Student's t-distribution, then do the routine simplifications.