Does the change of variable function have to be injective?

Solution 1:

$\phi$ does not have to be monotonic, but $f$ should be continuous (at least then the proof is more easy).

More precisely the assumptions are the following:

$\phi : [a,b] \rightarrow \mathbb{R}$ is continuously differentiable and $f : [c,d] \rightarrow \mathbb{R}$ is continuous with $\phi([a,b]) \subset [c,d]$.

Note that this last condition is automatic if(!) $\phi$ is monotonic and $f : [\phi(a), \phi(b)] \rightarrow \mathbb{R}$.

Now define $F : [c,d] \rightarrow \Bbb{R}, x \mapsto \int_c^x f(t) dt$. By the fundamental theorem of calculus (and because $f$ is continuous), we see that $F$ is differentiable with derivative $F' = f$.

Now $$\frac{d}{dx}F(\phi(x)) = F'(\phi(x)) \cdot \phi'(x) = f(\phi(x)) \cdot \phi'(x).$$

By the fundamental theorem of calculus again, we conclude

$$\int_a^b f(\phi(x)) \cdot \phi'(x) dx = F(\phi(b)) - F(\phi(a)) = \int_{\phi(a)}^{\phi(b)} f(x) dx.$$

Hence we do not need the monotonicity of $\phi$.

Note that we get in particular that

$$\int_a^b f(\phi(x)) \phi'(x) dx = 0$$

if $\phi(a) = \phi(b)$.