# Show that f(x)= $x^{\frac{-1}{2}}\sin(\frac{1}{x})$ with $f(0)=0$ is intregable on $[0,1]$

Show that

$$f(x) = \begin{cases} x^{\frac{-1}{2}}\sin(\frac{1}{x}) , & 0

is integrable on $$[0,1]$$. Intregable means that $$\displaystyle\int_{[0,1]} |f(x)|\,\mathrm dx < \infty$$.

My attempt:

First, I checked to see if $$f(x)$$ had a bounded derivative on $$[0,1]$$. If it did, then it would be a Lipschitz function, hence absolutely continuous and hence Integrable. Unfortunately, the derivative was not bounded.

Okay, so then I noticed that if $$0 < a \leq 1$$ then $$f(x)$$ is a well defined continuous function and so Riemann integrable on $$[0,1]$$, and hence Lebesgue integrable on $$[0,1]$$.

Next, I need to figure out how to prove that $$f$$ is continuous. I wouldn't mind some help with this part. Then I have that $$\lim_{a \rightarrow 0^+} f(a)=f(0)$$.

So, $$f(1)-f(0) = \lim_{a \rightarrow 0^+} f(1)-f(a) = \lim_{a \rightarrow 0^+} \int_{[a,1]} f(x)\,\mathrm dx = \lim_{a \rightarrow 0^+} \int_a^1 f(x)\,\mathrm dx.$$

Ugh idk honestly I'm lost.. Would appreciate help. I'm pretty slow so lots of details appreciated! Thanks!

You can just do it like $$\displaystyle\int_{0}^{1}|f(x)|\,\mathrm dx\leq\int_{0}^{1}x^{-1/2}\,\mathrm dx=2<\infty$$.
• $|\sin u|\leq 1$ no matter what. Dec 3, 2019 at 2:33
• I missed the absolute sign to the $f$. It suffices to check for the absolute integrand, and bound it by less than $\infty$, that's it. Dec 3, 2019 at 2:35