Finding Real Function for Improper Integration

[boombaby]

Homework Statement

Find a real function f on (0,1] and integrable on [c,1] for every c>0, such that \int^{1}_{0} f(x)dx exists but \int^{1}_{0}|f(x)|dx fails to exist

Homework Equations

The Attempt at a Solution

I think such function should behavior like the sequence 1-1/2+1/3-1/4+...+(-1)^n-1 1/n+...so I came up to something like \frac{1}{x^2} sin(\frac{1}{x})
But this function is not integrable on [0,1].
Moreover, I find that I don't know how to deal with the integration with |f(x)| instead. for example, I cannot find a way to integrate \int{\frac{1}{x^2} |sin(\frac{1}{x})|} dx (can it be solved?)
Perhaps the function I need may have a totally different form...
Any hint?

 

[Dick]

You are going to have to show |f(x)| isn't integrable by splitting the integral up into regions where is positive or negative. What's the integral from 1/((n+1)*pi) to 1/(n*pi)? It's exactly like your series.

 

[boombaby]

Umm...I don't see how to use the integral from 1/((n+1)*pi) to 1/(n*pi)..it seems that it is not sufficient to show the integration converges(in any case) nor diverges(in this particular one).
But I have made some progress under your hints, but still have problems unsolved.
for some reason, I tried \int^{1}_{0} \frac{1}{x} sin(\frac{1}{x}) dx instead.
by taking t=1/x,
\int^{1}_{0} \frac{1}{x} sin(\frac{1}{x}) dx= \int^{\infty}_{1} \frac{sin t}{t} dt.
I have proved that \int^{c}_{1} \frac{sin t}{t} dt diverges as c goes to infinity as follows.
\int^{2n\pi}_{n\pi} \frac{|sin t|}{t} dt >= \int^{2n\pi}_{n\pi} \frac{|sin t|}{2n\pi} dt=\frac{1}{2n\pi} \int^{2n\pi}_{n\pi} |sin t|dt>=\frac{n}{2n\pi}\int^{\pi}_{0} sin t dt= \frac{1}{\pi}
But I still have trouble to prove that \int^{\infty}_{1} \frac{sin t}{t} dt exists.
How can I start with it?
Thanks for your patience!

 

[Dick]

Your original idea about the series is good. Why don't you try defining the function piecewise to make it work exactly as you stated it? For example, define f(x)=(n+1)*(-1)^n for x in (1/(n+1),1/n]. What's the contribution from the nth interval?

 

[boombaby]

yea...I agree that your function is more good-looking...
since I see there's a hope to define the sin(1/x)-like function that works, I just stick to it...I don't know why...
BTW, I have proved that \int^{c}_{1} \frac{sin t}{t} dtconverges, as c goes to infinity...though the proof is not a brilliant one in my opinion. I sketch it below (you may check):
Let A_{i}=\int^{\pi (i+1)}_{\pi i} \frac{sin t}{t} dt then \sum A_{i} converges (to B).
For every epsilon>0, there's an N such that n>N implies that
B-\epsilon<\sum^{n}_{1} A_{i}<B+\epsilon.
For any c>(N+1)*pi. there's a n>N such that n*pi<=c<(n+1)*pi. Hence \int^{c}_{\pi} \frac{sin t}{t} dt lies in between \sum^{n}_{1}A_i(call it X) and \sum^{n+1}_{1} A_i (call it X2), whether X>X2 or not is depending on sin c>0 or sin c<0.
if X<X2, then
B-\epsilon&lt;\sum^{n}_{1} A_{i}&lt;\int^{c}_{\pi} \frac{sin t}{t} dt&lt;\sum^{n+1}_{1} A_{i}&lt;B+\epsilon.
if X>X2, blah blah blah..
Hence, blah blah blah...

 

[Dick]

I don't exactly follow the proof. But it has some of the right ingredients in it. You want to show that the integrals of sin(t)/t make a decreasing alternating series. That's enough to show it converges. Then you want to make an estimate to show that the magnitude of the contribution from the nth cycle is greater than k/n for some k.