Why is it correct to substitute $\tan x$ on $[0,\frac{\pi}{2}]$?

Question

Does the following hold?

$$ \int_0^{\pi/2} f(\tan(x))\cdot \sec^2(x) \,{\rm d}x = \int_0^\infty f(t) \,{\rm d}t $$

If so, why is this substitution allowed, when the proof of the substitution rule, to my understanding, requires $\tan'(x)$ to be continuous on $[0,\pi/2]$ for the chain rule step, when it is only continuous on $(0,2\pi)$ I think?

Also if so, is are there other cases not covered by the standard substitution rule where the result also holds.

Answer 1

Let us recall the following theorem:

Let $g$ be a continuous function on $[a,b]$ which carries its set of point of nondifferentiability into a set of zero measure. Let us set $g^* = g'$ as soon as $g'$ exists and zero elsewhere. Then $$\int_{g(a)}^{g(b)} f = \int_a^b (f \circ g)g^*$$ whenever the integral at right exists.

I suppose that if you ask this question it is because you haven't seen the Lebesgue integral yet (and with it the concept of set of zero measure). In the case of Riemann integral, this theorem is also true if the set of point of nondifferentiability is finite (which is a special case of set of zero measure).