Transformation outside injectivity 'domain' in Int. by substitution ok? And outside the integration interval?

In the theorem to prove the finding of primitization by substitution we restrict ourselves to bijective functions $x=\phi(t)$.

However, when proving integration by substitution, there's no mention of a bijection. Why?

Here's the text of the theorem on this wikipedia's page for Integration by Substitution.

Let $I ⊆ ℝ$ be an interval and $\phi : [a,b] → I$ be a differentiable function with integrable derivative. Suppose that $f: I → ℝ$ is a continuous function. Then $\int_{\phi(a)}^{\phi(b)} f(x)\,dx = > \int_a^b f(\phi(t))\phi'(t)\, dt$

As an example take $f(x)=\sqrt{1-x^2}$. If I want to integrate this in an interval, I'll use a primitive. However, when I "primitivated" $f$, I restricted myself to an interval where the function $x=\sin(t)$ had an inverse... Can I use that primitive, or should I integrate interval by interval?

And outside the integration interval, as long as we're in the domain where the transformation is continuously differentiable?

There's no need to invoke bijectivity, in the proof we just need $\phi$ to be continuously diff, so that $\phi '$ be continuous and $f(\phi(t))\phi'(t)$ be a continuous and hence integrable function.

Instead of the example above, I'll use the following integral: $\int^2_1 \frac{1}{\sqrt{e^x-1}}$, using the substitution $x=\log(t^2+1)$. This is not a bijective transformation in $\mathbb{R}$.

$x \in [1,2]$ is equivalent to $t \in [-\sqrt{e^2-1},-\sqrt{e-1}] \cup [\sqrt{e-1},\sqrt{e^2-1}]$. In each interval, the substitution is injective.

Doing the transformation we get $\int^b_a \frac{1}{\sqrt{t^2}}\frac{2t}{t^2-1}$. The expression for $\sqrt{t^2}$ will depend on $t$ belongs.

For $t \in [\sqrt{e-1},\sqrt{e^2-1}]$, we have $\sqrt{t^2}=t$, and thus $\int^b_a \frac{1}{\sqrt{t^2}}\frac{2t}{t^2-1}=\int^{\sqrt{e^2-1}}_{\sqrt{e-1}} \frac{2}{t^2-1}=2(\arctan(\sqrt{e^2-1})-\arctan(\sqrt{e-1}))$

What would happen if we had chosen the other interval? Nothing, i.e., we get the same integral: for $t \in [-\sqrt{e^2-1},-\sqrt{e-1}]$, we have $\sqrt{t^2}=-t$, and thus $\int^b_a \frac{1}{\sqrt{t^2}}\frac{2t}{t^2-1}=\int^{-\sqrt{e^2-1}}_{-\sqrt{e-1}} \frac{-2}{t^2-1}$. Notice that $b=-\sqrt{e^2-1}$, since we must have $\phi(b)=2$, and analogously for $a$.

$\int^{-\sqrt{e^2-1}}_{-\sqrt{e-1}} \frac{-2}{t^2-1}=\int^{-\sqrt{e-1}}_{-\sqrt{e^2-1}} \frac{2}{t^2-1}=2(\arctan(\sqrt{e^2-1})-\arctan(\sqrt{e-1}))$

And what if we wanted to choose $x$ over both intervals? We could, and moreover, we do not need to worry if the transformation is valid over the whole interval, as long as we have $\phi(a)=1$ and $\phi(b)=2$ and $\phi$ continuously differentiable (which it is on the whole line). Here's how.

Let's integrate for $t \in [-\sqrt{e-1},\sqrt{e^2-1}]$. We have $\sqrt{t^2}=-t$ for $t \in [-\sqrt{e-1},0]$, and $\sqrt{t^2}=t$ for $t \in [0,\sqrt{e^2-1}]$. Hence $\int^b_a \frac{1}{\sqrt{t^2}}\frac{2t}{t^2-1}=\int^{\sqrt{e^2-1}}_{-\sqrt{e-1}} \frac{1}{\sqrt{t^2}} \frac{2}{t^2-1}=\int^{\sqrt{e^2-1}}_{0} \frac{2}{t^2-1}+\int^{\sqrt{0}}_{-\sqrt{e-1}} \frac{-2}{t^2-1}$. Notice that limits $a$ and $b$ are respected. If we continue our calculations, we reach the same result.

• If someone could confirm whether my answer is correct, I would be very appreciative. – An old man in the sea. Sep 2 '16 at 23:13