# 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.