For integrals of the form:
$$\intop_a^bg(t)dt,$$
we can apply the $tanh$-substitution to transform the integral into a doubly infinite integral, i.e:
$$\intop_a^bg(t)dt = \frac{b-a}{2}\intop_{-\infty}^\infty g \left( \frac{b+a}{2} \frac{b-a}{2}\tanh(u) \right)\mathrm{sech^2(u)} du.$$
My question is - Is there another substitution method we can use to transform:
$$\intop_a^bg(t)dt,$$
to a singly infinite integral, e.g: $$\intop_0^{\infty} g(f(u))du?$$