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For integrals of the form:


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:


to a singly infinite integral, e.g: $$\intop_0^{\infty} g(f(u))du?$$

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up vote 1 down vote accepted

You can do this in two steps:

  1. First make a change of variable $t\rightarrow \frac{1}{b-a}(t-a)$. This will change the integration to be over the interval $(0,1)$.

  2. Make a secondary change of variable $t\rightarrow\frac{1-t}{t}$. This will change the integration over $(0,1)$ to an integration over $(0,\infty)$.

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