# Principles of math analysis by Rudin, Chapter 6 Problem 7

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.

• 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. – DanielWainfleet Nov 17 '18 at 10:26
• The above answer is more intuitive. – Tengerye Nov 20 '18 at 9:02

## 1 Answer

$$b$$) 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 $$\lim_{c\to 0^+}\int_c^1f(x)\mathrm dx$$

• 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. – Silent Dec 12 '18 at 2:29
• @Silent thank you. Did you forget the derivative term in the substitution? – qbert Dec 12 '18 at 16:47
• Thank you very much, for pointing that out. It is integration by substitution! – Silent Dec 13 '18 at 3:10