2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ You don't need the condition $\phi'(x)\ne 0$. $\endgroup$ Jul 5 '20 at 14:39
  • $\begingroup$ @hamam_Abdallah why not? $\endgroup$ Jul 5 '20 at 14:40
  • $\begingroup$ You just need the continuity of $ f $ at $[a,b]$, of $ \phi'$ at $[\alpha,\beta]$ and $\phi([\alpha,\beta]\subset [a,b]$. $\endgroup$ Jul 5 '20 at 14:42
  • $\begingroup$ @hamam_Abdallah This would be stronger conditon.... $\endgroup$ Jul 5 '20 at 14:43
  • 1
    $\begingroup$ To prove thr substitution formula, we use differentiation, and for this we need continuity of $ f$. $\endgroup$ Jul 5 '20 at 14:46
4
$\begingroup$

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

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


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.


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.

$\endgroup$
0
$\begingroup$

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 $\phi (x)= t$.

$\endgroup$
9
  • $\begingroup$ No your answer is not correct.$f$ may be a function which is Riemann integrable but yet it may not have any antiderivatice $\int f$ ,so it is not always application of fundamental theorem...it is not that simple....For example See Volterra function....a function which is differentiable but its derivative is not Riemann integrable. $\endgroup$ Jul 5 '20 at 14:37
  • $\begingroup$ Your method is useful if $f$ is continuous or if $f$ is an integrable function and $f$ has an antiderivative on $[a,b]$. $\endgroup$ Jul 5 '20 at 14:38
  • $\begingroup$ He asked for a way to grab the idea, and I think this approach works for that purpose. If he wanted to add more patological behaviour he probably would have turned to a book and not a forum.Also, $f$ integrable is among his hypothesis. So your comment is out of place. $\endgroup$
    – astro
    Jul 5 '20 at 14:39
  • $\begingroup$ I am studying real analysis.To me,grabbing the idea means knowing everything about a theorem....each condition....I did not ask for an intuitive way...of just thinking.... $\endgroup$ Jul 5 '20 at 14:41
  • $\begingroup$ Then you shoouldn't have asked for a better insight but instead for a thoroug explanation. $\endgroup$
    – astro
    Jul 5 '20 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.