Example of f integrable but |f| not integrable

1. Nov 20, 2007

SiddharthM

Define F(x) = integral (from x to 1) f

take limF(x) as x goes to 0.

I would like to give an example of a function on (0,1] so that the above limit exists but if we replace f with |f| the limit does not exist.

So I came up with f= [(2^n)/(n)](-1)^(n+1) for x in (1/(2^n),1/[2^(n-1)]] (which geometrically) at these endpoints gives us the alternating harmonic series, that is the limit above with this function is the limit of the alternating harmonic series.

Is there a simpler example?

2. Nov 20, 2007

Eighty

f(x)=sin(1/x)/x.

3. Nov 20, 2007

Gib Z

$$\int^{\infty}_0 \frac{ \sin x}{x} dx = \frac{\pi}{2}$$

$$\int^{\infty}_0 |\frac{ \sin x}{x}| dx = \infty$$

4. Nov 21, 2007

SiddharthM

i'm looking for examples on the domain [0,1].

5. Nov 21, 2007

morphism

This one works on [0,1] (it's not Lebesgue integrable there IIRC).

6. Nov 21, 2007

Kummer

If f is a bounded function on [0,1] which is integrable then |f| is integrable on [0,1]. It is a simple theorem to prove.

7. Nov 21, 2007

morphism

While this is true, it's not really what the OP asked for. On the other hand, it does tell us that if we are going to find such a function f, it's not going to be (properly) Riemann integrable on [0,1].

8. Nov 22, 2007

SiddharthM

I'm aware of this, but this isn't my question. I'm asking for a function so that f is integrable for every c>0 on [c,1] and with the following properties:

(1) the limit of the integrals as c goes to zero exists.
(2) the same limit as (1) replaced with f with |f| does NOT exist.

Although I did come up with one, I'm asking for more, possibly simpler, ones.

The theorem you mentioned tells us that any such function should NOT be bounded on [0,1].

9. Nov 23, 2007

My solution, perhaps not the best.

This is problem 7 in Rudin, Chapter 6. I used a harmonic series:
$$f(x) = (n+1)(-1)^{n+1},\ x \in (\frac{1}{n+1},\frac{1}{n}],\ n = 1,2,...$$. Then $$\int_{\frac{1}{n+1}}^{\frac{1}{n}} f(x) dx = \frac{(-1)^{n+1}}{n}$$. The integral $$\int_{c}^{1} f(x) dx$$ converges to $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$$ as $$c \rightarrow 0$$. But it's not absolutely convergent.

10. Nov 23, 2007

$$\int_{0}^{1}\frac{sin(\frac{1}{x})}{x}dx$$ does not appear to me to converge. Set $$\frac{sin(\frac{1}{x})}{x} = x\frac{sin(\frac{1}{x})}{x^2}$$, and integrate by parts, with $$G(x)=x, \ f(x) = -sin(u(x))u'(x) = \frac{sin(\frac{1}{x})}{x^2}$$, then $$\int_{e}^{1}x\frac{sin(\frac{1}{x})}{x^2}dx = 1 \cdot cos(\frac{1}{1}) - e \cdot cos(\frac{1}{e}) - \int_{e}^{1}cos(\frac{1}{x})dx$$ does not converge as $$e \rightarrow 0+$$. Using the same integration by parts, you can show $$\int_{0}^{1}\frac{sin(\frac{1}{x})}{\sqrt{x}}dx$$ does converge, but it also converges absolutley, since $$\int_{0}^{1}\frac{1}{\sqrt{x}}dx$$ also converges.

Last edited: Nov 23, 2007
11. Nov 23, 2007

OK I think I'm wrong there... $$\int_{e}^{1}cos(\frac{1}{x})dx$$ must indeed converge as $$e \rightarrow 0$$, if the identity $$\int_{e}^{1}\frac{sin(\frac{1}{x})}{x}dx = \int_{1}^{\frac{1}{e}}\frac{sin(u)}{u}du$$ holds. Then I suppose that's an easier to construct example of an integral that is convergent but not absolutely convergent.

Last edited: Nov 23, 2007
12. Nov 23, 2007

I think I've done enough for now...

What the hell was I thinking?!? Of course the integral of cos(1/x) converges on [0,1] because it's a bounded function that's only discontinuous at one point!

However, the problem with sin(1/x)/x is that doesn't seem like a trivial function to deal with. The limit is in the residue integration section of complex analysis books. But nonetheless I went ahead and calculated that it does not converge absolutely. It was laborous.

The approach I began above was mistaken but in retrospect yields that $$\int_{0}^{1}\frac{sin(\frac{1}{x})}{x}dx = cos(1) - \int_{0}^{1}cos(\frac{1}{x})dx$$. In particular, it converges.

Using the integration of parts for even n, $$\int_{\frac{1}{n\pi}}^{\frac{1}{(n-1)\pi}}|\frac{sin(\frac{1}{x})}{x}|dx \geq \frac{1}{n\pi} + \frac{1}{(n-1)\pi} - \frac{1}{n^2\pi^2}}$$, hence $$\sum_{n}\int_{\frac{1}{n\pi}}^{\frac{1}{(n-1)\pi}}|\frac{sin(\frac{1}{x})}{x}|dx \geq \sum_{even \ n}\frac{1}{n\pi} + \frac{1}{(n-1)\pi} - \frac{1}{n^2\pi^2}} = +\infty$$

Last edited: Nov 23, 2007
13. Nov 23, 2007

SiddharthM

rudinreader: your solution is similar to mine in that it converges to the alternating harmonic series although i think I like yours better! check out my post on uniform convergence it is in regards to problem 4 chapter 7.