Understanding the substitution theorem of Riemann integration.

Question

Let us say $f$ is an integrable function on $[a,b]$ and we want to evaluate $\int_a^b f(x)dx$ but often the calculation is not easy.So,we have a method of substitution.We substitute $x=\phi(t)$ where $\phi$ is a differentiable function on $[\alpha,\beta]$ such that $\phi(\alpha)=a$ and $\phi(\beta)=b$.Also $\phi'$ is integrable on $[\alpha,\beta]$ and $\phi'(x)\neq 0$ for all $x\in [\alpha,\beta]$.Then we can evaluate the above integral by $\int _a^b f(x)dx=\int_\alpha^\beta f(\phi(t))\phi'(t)dt$.

But I am a little but troubled with so many conditions,I could do the proof but I am having a hard time use the theorem to problems as I often forget the conditions required.So,can someone help me to get a better insight about this theorem of substitution in Riemann integrals?I would also like some counterexamples that show each condition to be essential.

Answer 1

Strong sufficient conditions are that $f$ is continuous and $\phi'$ is integrable. A straightforward proof uses the FTC, and monotonicity of $\phi$ is not needed.

Defining $F(t) = \int_{\phi(\alpha)}^{\phi(t)}f(x) \, dx$, we have $F'(t) = f(\phi(t)) \phi'(t)$ since $f$ is continuous, and

$$\int_a^b f(x) \, dx = \int_{\phi(\alpha)}^{\phi(\beta)}f(x) \, dx = F(\beta)= \int_\alpha^\beta F'(t) \, dt = \int_\alpha^\beta f(\phi(t))\phi'(t) \, dt$$

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On the other hand, we can drop the condition that $f$ is continuous and assume only integrability. To facilitate an easy proof using Riemann sums, we need to assume that $\phi$ is both continuously differentiable and monotone.

Take a partition $\alpha = t_0 < t_1 < \ldots < t_n = \beta$ and form the sum

$$\tag{*}S(P,f\circ\phi \, \phi')= \sum_{j=1}^n f(\phi(\xi_j))\phi'(\xi_j)(t_j - t_{j-1})$$

where we use intermediate points $\xi_j \in [t_{j-1},t_j]$ and which will converge to $\int_\alpha^\beta f(\phi(t)) \phi'(t) \, dt$ as the partition is refined.

If $\phi$ is increasing then a partition $P'$ of $[\phi(\alpha),\phi(\beta)]$ is induced by

$$\phi(\alpha) = \phi(t_0) < \phi(t_1) < \ldots < \phi(t_n) = \phi(\beta),$$

and using the intermediate points $\phi(\xi_j)$, we have a Riemann sum for the integral of $f$ over $[\phi(\alpha),\phi(\beta)]$ of the form

$$S(P',f) = \sum_{j=1}^n f(\phi(\xi_j))(\,\phi(t_j) - \phi(t_{j-1})\,)$$

Note that we need the monotonicity of $\phi$ to ensure that $\phi(\xi_j) \in [\phi(t_{j-1}), \phi(t_j)]$.

Applying the mean value theorem, there exist points $\eta_j \in (t_{j-1},t_j))$ such that

$$\tag{**}S(P',f) = \sum_{j=1}^n f(\phi(\xi_j))\phi'(\eta_j)(t_j - t_{j-1})$$

Notice the similarity between the sums in (*) and (**). Aside from the distinction between $\eta_j$ and $\xi_j$, they are identical. Using the continuity (and, hence, uniform continuity) of $\phi'$ we can show that as the partition is refined and both $\|P\|, \|P'\| \to 0$ we have

$$\lim_{\|P|| \to 0}|S(P,f\circ \phi\,\phi') - S(P',f)| = 0$$

Therefore, $S(P',f)$ converges to both integrals and we have

$$\lim_{\|P'\| \to 0}S(P',f) = \int_{\phi(\alpha)}^{\phi(\beta)} f(x) \, dx = \int_a^b f(\alpha(t)) \alpha'(t) \, dt$$

Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $\phi$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $\phi'$ is assumed.

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The conditions can be weakened further. The result holds if both $f$ and $\phi'$ are integrable, without any assumptions of continuity. This is much more difficult to prove. Here is where you might begin to search for counterexamples.

Answer 2

Think of this as the fundamental theorem applied to a composition. By the chain rule it holds that $(f \circ \phi)'=f'(\phi) \circ \phi'$ so, roughly, $f \circ \phi=\int (f'(\phi) \circ \phi')$. The remainder conditions over the limits of integration are a result of changing variables $\phi (x)= t$.