# function is continuous and y→f(x,y) is integrable

$$f:\mathbb{R^{2}} \to \mathbb{R}:\mathcal{f(x,y)} = \left\{ \begin{matrix} \mathcal{x^{2}\sin{\frac{1}{x}}\exp(-x^{2}|y|)} & x\neq0 \\ \mbox{0} & x=0 \end{matrix}\right.$$

I want to prove that this function is continuous and $$y\to f(x,y)$$ is integrable.

To prove that $$f$$ is continuous , I think it's enough to prove that $$\lim_{x\to 0} f=0$$. So $$\lim_{x\to 0} f = \lim_{x \to 0}x^{2}\sin{\frac{1}{x}}\exp(-x^{2}|y|)$$. Because $$\lim_{x \to 0}x^{2}\exp(-x^{2}|y|)=0$$ and $$-1 \leq \sin(\frac{1}{x}) \leq 1$$ follows that $$\lim_{x\to 0} f=0$$.

Now to prove that $$y \to f(x,y)$$ is integrable for every $$x \in \mathbb{R}$$ I think it's enough to prove that the integrand is smaller than$$\infty$$. Can someone confirm this?

EDIT: new idea. I can say that $$|f(x)| \leq x^{2}\exp(-x^{2}|y|)$$ because of the same reason I explained with the sinus above this EDIT.

EDIT2: I don't know. Can someone explain how I can prove that the function $$y\to f(x,y)$$ is integrable.

OFFICIAL EDIT: Is it possible that I can try to solve the integral , because $$f$$ is always a positiv , so measurable, function. And that I can switch the limit and integral because $$f$$ is continuous? So the integral would be $$0$$ which is $$<\infty$$?

As you need to prove $$y\to f(x,y)$$ is integrable, here $$x$$ is effectively a constant for you. Thus, we effectively need to prove the integrability of $$g(y) = \begin{cases} Ae^{-B|y|} & x \neq 0 \\ 0 & x = 0 \\ \end{cases}$$ where $$A=x^2\sin\frac{1}{x}$$ and $$B=x^2$$ are constants.
Now, as $$g(y)$$ is continuous for each $$x$$ (easy to see), it is integrable.