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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?

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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.

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

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