Example of f integrable but |f| not integrable

[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?

 

[Eighty]

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

 

[Gib Z]

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

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

 

[SiddharthM]

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

 

[morphism]

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

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

 

[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.

 

[morphism]

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.

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].

 

[SiddharthM]

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.

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].

 

[rudinreader]

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.

 

[rudinreader]

About sin(1/x)/x

$$\int_{0}^{1}\frac{sin(\frac{1}{x})}{x}dx$$ 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.

 

[rudinreader]

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.

 

[rudinreader]

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 $$\sum_{n}\int_{\frac{1}{n\pi}}^{\frac{1}{(n-1)\pi}}|\frac{sin(\frac{1}{x})}{x}|dx \geq \sum_{even \ n}\frac{1}{n\pi} + \frac{1}{(n-1)\pi} - \frac{1}{n^2\pi^2}} = +\infty$$

 

[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.