<s>[INST] Prove an improper double integral is convergent I need to prove the following integral is convergent and find an upper bound
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{1+x^2+y^4} dx dy$$
I've tried integrating $\frac{1}{1+x^2+y^2} \lt \frac{1}{1+x^2+y^4}$ but it doesn't converge
 [/INST] Continuing from zhw.'s answer,
$$ \int_{0}^{+\infty}\frac{\sqrt{r}}{2(1+r^2)}\,dr = \int_{0}^{+\infty}\frac{u^2\,du}{1+u^4}=\frac{\pi}{2\sqrt{2}} $$
by the residue theorem, while:
$$ \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin\theta}}=\int_{0}^{1}\frac{du}{\sqrt{u(1-u^2)}}=\frac{1}{2\sqrt{2\pi}}\,\Gamma\left(\frac{1}{4}\right)^2$$
from the Euler beta function and the $\Gamma$ reflection formula. Hence we have:

$$ \iint_{(0,+\infty)^2}\frac{dx\,dy}{1+x^2+y^4}=\color{red}{2\sqrt{\pi}\cdot\Gamma\left(\frac{5}{4}\right)^2}=2.9123736927\ldots$$

A simple upper bound may be derived from:
$$ \int_{0}^{+\infty}\frac{u^2}{1+u^4}\,du \leq \int_{0}^{1}u^2\,du+\int_{1}^{+\infty}\frac{du}{u^2}=\frac{4}{3},$$
$$ \int_{0}^{1}\frac{2\,du}{\sqrt{1-u^4}}\leq\int_{0}^{1}\frac{2\,du}{\sqrt{1-u^2}}=\pi. $$

Another simple proof comes from:
$$ \int_{0}^{+\infty}\frac{dx}{1+y^4+x^2}=\frac{\pi}{2}\cdot\frac{1}{\sqrt{1+y^4}}$$
and obviously:
$$ \int_{0}^{+\infty}\frac{dy}{\sqrt{1+y^4}}\leq \int_{0}^{1}dy+\int_{1}^{+\infty}\frac{dy}{y^2}$$
giving $\color{red}{I\leq \pi}$. That can be improved through Cauchy-Schwarz inequality:
$$ \int_{0}^{+\infty}\frac{du}{\sqrt{1+u^4}}\leq \sqrt{\left(\int_{0}^{+\infty}\frac{1+u^2}{1+u^4}\,du\right)\cdot\left(\int_{0}^{+\infty}\frac{du}{1+u^2}\right)}$$
leads to:

$$ \color{red}{I} \leq \frac{\pi^2}{2\sqrt{2\sqrt{2}}}=2.93425\ldots\color{red}{< 3}.$$

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