2
$\begingroup$

Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c,1]$ for every $c>0$. Define $\int_0^1 f(x)dx=\lim_{c\to 0} \int_c^1 f(x)dx$ if this limit exists (and is finite).

(a) If $f \in \mathscr{R}$ on $[0,1]$, show that this definition of the integral agrees with the old one.

(b) Construct a function $f$ such that the above limit exists, although it fails to exist with $|f|$ in place of $f$.


This is Problem 7 of Chapter 6 in Principles of Mathematical Analysis by Rudin. For (a), I can prove the equation is correct but I am not sure what does 'definition agrees' mean? For (b), I have no idea.

Thank you in advance.

$\endgroup$
  • 1
    $\begingroup$ For (b) suppose, for $n\in \Bbb N,$ that $\int_{1/(n+1)}^{1/n}f(x)dx=(-1)^n/n .$ Suppose that when $x\in [1/(n+1),1/n]$ then $f(x)\leq 0$ if $n$ is odd, while $f(x)\geq 0$ if $n$ is even. $\endgroup$ – DanielWainfleet Nov 17 '18 at 10:26
  • $\begingroup$ The above answer is more intuitive. $\endgroup$ – Tengerye Nov 20 '18 at 9:02
4
$\begingroup$

$b$) Using the well known integral $$ \int_{1}^\infty \frac{\sin x}{x}\mathrm dx $$ which converges conditionally, we reflect reflect everything to near $0$ by sending $$ x\to 1/x $$ and find $$ \int_{1}^\infty\frac{\sin x}{x}\mathrm dx= \int_0^1\frac{\sin(1/y)}{y}\mathrm dy $$

For part $a$, you must prove using the definition of the Riemann integral that for a function which is integrable on $[0,1]$, the number $$ \int_0^1 f(x)\mathrm dx $$ is the same as the number $$ \lim_{c\to 0^+}\int_c^1f(x)\mathrm dx $$

$\endgroup$
  • $\begingroup$ I think, it should be $\int_0^1\frac{\sin(1/y)}{1/y}\mathrm dy$ in third display. Awesome answer by the way. I had seen a discrete construction before for this, and now know a well-known function can also serve the purpose. $\endgroup$ – Silent Dec 12 '18 at 2:29
  • $\begingroup$ @Silent thank you. Did you forget the derivative term in the substitution? $\endgroup$ – qbert Dec 12 '18 at 16:47
  • 1
    $\begingroup$ Thank you very much, for pointing that out. It is integration by substitution! $\endgroup$ – Silent Dec 13 '18 at 3:10

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.