Check if given function is Lebesgue Integrable

Question

Is $f:\mathbb{R^{2}}\rightarrow [0,\infty)$ with $$f(x,y)=\frac{1}{1+x^{2}+y^{4}} $$ Lebesgue integrable?

I already got the hint to use Fubini-Tonelli and divide the integral of $x$ into two parts, $x^{2}\le 1+y^{4}$ and $x^{2} > 1+y^{4}$. But I didn't really work for me.

Answer 1

Note that Tonelli's theorem applies for measurable functions $\geq 0$, and under these mere conditions it holds that the multiple integral equals any intended repeated integral. Note that $f$ is continuous, so it is Borel measurable, and hence it is measurable. Note that $f \geq 0$, so Tonelli's theorem is applicable. Now fix any $y \in \mathbb{R}$ and find $g(y) := \int_{0}^{+\infty} f(\cdot, y)$ and then find $I := \int_{0}^{+\infty} g(y)d y$. Note that $4I = \int_{\mathbb{R}^{2}}f$, so whether $I <+\infty$ or not answers the question.

Answer 2

Hint: show $$ \iint_{R^2}\frac{1}{1+x^2+y^4}dxdy=4\iint_{(R^+)^2}\frac{1}{1+x^2+y^4}dxdy<\infty$$

Answer 3

Well, $$\int_{0}^{+\infty}\frac{dx}{(1+y^4)+x^2} = \frac{\pi}{2}\cdot\frac{1}{\sqrt{1+y^4}}\tag{1}$$ hence Fubini's theorem applies and $$\iint_{\mathbb{R}^2}\frac{dx\,dy}{1+x^2+y^4} = 2\pi\int_{0}^{+\infty}\frac{dy}{\sqrt{1+y^4}} = \frac{\sqrt{\pi}}{2}\,\Gamma\left(\frac{1}{4}\right)^2\tag{2}$$ by Euler's Beta function.