I have an example that f is differentiable but not L-integrable. On [0,1] define $f$ s.t. $f(x)=x^2 \sin \frac{1}{x^2}$ $f(0)=0$
And calculate $\int_0^1 |f'|$ I think the value of this integral is infinite, but i cant how to prove it.. Please give me some hints..
The derivative $f'(x)$ is $2x\sin\frac{1}{x^2}-\frac 2x\cos\frac{1}{x^2}$. We can go two ways estimating $\int_0^1 |f'(x)|\,dx$.
First, we can focus on the larger term and say that $\left|\cos\frac{1}{x^2}\right|$ is comparable to a constant $\frac 2\pi$ - larger in some places, smaller in others, but that's the average of $|\cos|$. We estimate that $|\cos$ by a constant on intervals where it's not close to zero, and estimate it by zero when it is close. Done right, those intervals where it's not small cover an appreciable fraction of the interval, and we get that $\int_0^1 |f'(x)|\,dx$ is greater than $\int_0^1 \frac cx\,dx$ for some $c$.
Alternatively, we can use that $f'$ is a derivative to evaluate those integrals. Clearly, $\int_a^b |f'(x)|\,dx \ge |f(b)-f(a)|$. Slice up the interval at maxima and minima of $f$ - well, not exactly the maxima and minima because we don't want to try to solve for the exact zeros of the derivative, but the points where $\sin\frac{1}{x^2}=\pm 1$. This will give us that $\int_a^b |f'(x)|\,dx$ is $\ge$ an infinite sum, and we should be able to show that the sum diverges to $\infty$.
Pick a path and follow it. They'll both work.