Constructing f for Int'l Limit Existence, Not |f|

[ehrenfest]

Homework Statement

Suppose f is a real function on (0,1] and f is Riemann-integrable 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.

Construct a function f such that the above limit exists, although it fails to exist with |f| in place of f.

Homework Equations

The Attempt at a Solution

I think f(x) = sin(1/x)/x will work although I am having trouble proving it. It seems intuitively true that the limit exists since the positive and negative humps will cancel and the oscillations in the integral will get get smaller and smaller as c gets smaller and smaller. I am not really sure how to prove that the limit diverges when we take the absolute value of f.

 

[mighty2000]

Any suggestions or corrections would be greatly appreciated.Your proposed function, f(x) = sin(1/x)/x, is a good choice to demonstrate the concept. Here's a proof that the limit exists for f(x) but not for |f(x)|:

First, let's show that the limit exists for f(x). We can rewrite the integral as:

\int_c^1 f(x) dx = \int_c^1 \frac{sin(1/x)}{x} dx

Using u-substitution with u = 1/x, we get:

\int_c^1 \frac{sin(1/x)}{x} dx = \int_{1/c}^1 \frac{sin(u)}{u} du

Now, as c approaches 0, 1/c approaches infinity. This means that the integral will be over a larger and larger interval, but the function we are integrating, sin(u)/u, will become more and more oscillatory. However, since the function is bounded between -1 and 1, the oscillations will not cause the integral to diverge. In other words, as c approaches 0, the integral will be over a larger and larger interval, but the function will be bounded, so the integral will converge.

Next, let's show that the limit does not exist for |f(x)|. Again, using u-substitution, we get:

\int_c^1 \frac{|sin(1/x)|}{x} dx = \int_{1/c}^1 \frac{|sin(u)|}{u} du

Now, as c approaches 0, the integral will be over a larger and larger interval, and the function we are integrating, |sin(u)|/u, will also become more and more oscillatory. However, unlike the previous case, this function is not bounded. In fact, it has a singularity at u = 0, which means that the integral will diverge as c approaches 0.

Therefore, the limit exists for f(x) but not for |f(x)|, as desired.