Let $f$ be a function, $x>0$ and $\theta>0$, and suppose $\int_{0}^{\infty}tf(t)\,dt=1$
How could I show that $\int_0^\infty \frac{x^2}{\theta^2}f(\frac{x}{\theta}) \, d\theta =x$?
I try substituting with $u=\frac{x}{\theta}$ and then $du=-\frac{x}{\theta^2} \, d\theta$, and the integral becomes: $\int_0^\infty \frac{x^2}{\theta^2}f(\frac{x}{\theta}) \, d\theta=\int_\infty^0 -xf(u)\,du=\int_0^\infty xf(u) \, du$... but then I don't seem to be getting anywhere and I don't know how to use that $\int_0^\infty tf(t) \,dt=1$ ... Thanks.
$$ \int_0^\infty \frac{x^2}{\theta^2} f\left(\frac x \theta \right)\,d\theta= \int_\infty^0 x f(u)\,(-du) = \overbrace{\int_0^\infty xf(u)\,du = x \int_0^\infty f(u)\,du} $$
The step under the $\overbrace{\text{overbrace}}$ is valid because $x$ is a constant, i.e. it does not change as $u$ goes from $0$ to $\infty$, or as $\theta$ goes from $0$ to $\infty$.
To finish this off, you would have to know that $\int_0^\infty f(u)\,du=1$.