Show that if $F \sim F(m, n)$, then $\dfrac{1}{F} \sim F(n, m)$. Here $F(m,n)$ is Fisher distribution, $F$ is random variable.
$$f(F) =\frac{1}{B(\frac m2, \frac n2)}\left(\frac{m}{n}\right)^{m/2}x^{m/2-1}\left(1+\frac mn x\right)^{-(m+n)/2}$$
Let $Y=1/F; F(Y \le y)=1-F(F\le1/y)$
$$F(F\le1/y)=\frac{1}{B\left(\frac m2, \frac n2\right)}\left(\frac{m}{n}\right)^{m/2}\int_0^{1/y}x^{m/2-1}\left(1+\frac mn x\right)^{-(m+n)/2}dx\\=\frac{1}{B(\frac m2,\frac n2)}\left(\frac{m}{n}\right)^{m/2}\int_0^{1/y}x^{m/2-1}\left(1+\frac mn x\right)^{-[(m+n+2)/2]-1}dx$$
What can I do next?
Let $U=1/F\implies F=1/U\implies|J|=1/u^2$ $$\implies f(U)=f_F\left(\frac1U\right)\frac1{u^2}=\left(\frac{1}{B\left(\frac m2,\frac n2\right)}\left(\frac{m}{n}\right)^{m/2}\left(\frac1u\right)^{m/2-1}\left(1+\frac mn \frac1u\right)^{-(m+n)/2}\right)\frac1{u^2}$$