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

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 $<\infty$?

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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.

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  • $\begingroup$ yes, i totally forget that there is |y| so that for -inf as +inf the exp will be zero. $\endgroup$
    – questmath
    Jan 5 at 15:58

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