Let \begin{equation} f(x)= \frac{1}{\sqrt{(x_{1}^2 x_{2}^2+x_{3}^2)^2+ x_{4}^2}} \quad (x \in \mathbb{R}^4 \backslash \{0 \}). \end{equation} Is $f$ locally integrable?
Thank you very much for your attention.
Is $[(x_{1}^2 x_{2}^2+x_{3}^2)^2+ x_{4}^2]^{-1/2}$ locally integrable?
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real-analysis
integration
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0why shouldn't it – 2017-01-30
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1Please involve some thoughts yourself. When you have come to the level of dealing with locally integrability, I think you could at least show some effort. – 2017-01-30
1 Answers
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Ops, this is a really trivial exercise. I don't know why I could not see that this function is locally integrable at first glance. Maybe I am too tired in the last days. Anyhow, we have \begin{equation} f(x) \leq \frac{1}{\sqrt{x_{3}^4 + x_{4}^2}}, \end{equation} and we have \begin{equation} \int_{-1}^{1} \int_{-1}^{1} \frac{1}{\sqrt{x_{3}^4 + x_{4}^2}} dx_3 dx_4 = \int_{-1}^{1} \log \left( \frac{\sqrt{1+x_{3}^4}+1}{\sqrt{1+x_{3}^4}-1} \right) dx_3 = \int_{-1}^{1} 2 \log \left( \frac{\sqrt{1+x_{3}^4}+1}{x_{3}^2} \right) dx_3 \end{equation} which is clearly integrable since \begin{equation} \int_{-1}^{1} - \log(x^2)dx = -4 \int_{0}^{1} \log(x)dx = 4. \end{equation} Sorry for having posted it!