compute $\int_Q\frac{1}{|x|} \, dx$ on $Q=[0,1]^2$ Let $Q=[0,1]^2$, compute the integral:
$$\int_Q\frac{1}{|x|} \, dx$$
I tried to take $x=(x_1,x_2)$, then the integral is:
$$\int_Q\frac{1}{\sqrt{x_1^2+x_2^2}} \, dx_1 \, dx_2$$
Then I moved to polar coordinates $(x_1,x_2)=(r\cos\theta,r\sin\theta)$, so the integrand is $\frac{1}{\sqrt{r^2}}r=1$.
I could't find the integral's domain. I took $0\leq r\cos\theta\leq1$ and $0\leq r\sin\theta\leq1$ from $(x_1,x_2)\in Q$.
Then I got $0\leq r\leq\sqrt2$, but I didn't mange to find $\theta$'s interval, is it $\displaystyle \Big[0,\frac{\pi}{2}\Big]$?
Final result by WA is: $2\log(1 + \sqrt2)$
 A: $$\begin{eqnarray*}\iint_{(0,1)^2}\frac{da\,db}{\sqrt{a^2+b^2}}&\stackrel{\text{symmetry}}{=}&2\int_{0}^{1}\int_{0}^{a}\frac{1}{\sqrt{a^2+b^2}}\,db\,da\\&\stackrel{b\mapsto a c}{=}&2\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+c^2}}\,dc\,da\\&=&2\,\text{arcsinh}(1)=\color{red}{2\log(1+\sqrt{2}).}\end{eqnarray*}$$
A: $$0 \leq \tan\theta \leq \infty \implies  \theta \in [0,\pi/2]$$
$$0 \leq r\cos\theta \leq 1, 0 \leq r\sin\theta \leq 1 \implies 0 \leq r \leq \min(1/\cos\theta,1/\sin\theta) $$
$$\begin{align}I &= \int_0^{\pi/2}\int_0^{\min(1/\cos\theta,1/\sin\theta)} \,\partial r \,\partial \theta\\ &= \int_0^{\pi/4}\int_0^{1/\cos\theta} \, \partial r \, \partial \theta + \int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} \, \partial r \, \partial \theta \\
&= \int_0^{\pi/4}\frac{1}{\cos\theta} \, \partial  \theta + \int_{\pi/4}^{\pi/2} \frac{1}{\sin\theta} \, \partial \theta \\
&= \log(\tan\theta+\sec\theta)|_0^{\pi/4} -\log(\cot\theta+\csc\theta)|_{\pi/4}^{\pi/2}\\
&= \log(1+\sqrt 2) - 0-0 + \log(1+\sqrt{2}) \\
&= 2\log(1+\sqrt 2)\end{align}$$
Note that I used the fact that $\sin\theta\leq\cos\theta, \forall \theta \in [0,\pi/4]$ and $\sin\theta>\cos\theta, \forall \theta \in (\pi/4,\pi/2]$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\int_{\bracks{0,1}^{2}}{\dd x\,\dd y \over \root{x^{2} + y^{2}}} & =
\int_{\bracks{0,1}^{2}}{\dd x\,\dd y \over r} =
\int_{\bracks{0,1}^{2}}\bracks{\partiald{\pars{x/r}}{x} -
\partiald{\pars{-y/r}}{y}}\,\dd x\,\dd y
\\[5mm] & =
\int_{\bracks{0,1}^{2}}\bracks{%
\nabla\times\pars{-\,{y \over r}\,\hat{x}\ +\ {x \over r}\,\hat{y}}}_{z}
\,\dd x\,\dd y
\\[5mm] & =
\int_{\partial\bracks{0,1}^{2}}{-y\,\dd x + x\,\dd y \over \root{x^{2} + y^{2}}}
\qquad\pars{~Stokes\ Theorem~}
\\[5mm] & =
\int_{0}^{1}{\dd y \over \root{1 + y^{2}}} +
\int_{1}^{0}{-\dd x \over \root{x^{2} + 1}} =
2\int_{0}^{1}{\dd x \over \root{x^{2} + 1}}
\\[5mm] \stackrel{x\ \mapsto\ \tan\pars{x}}{=}\,\,\,& 2\int_{0}^{\pi/4}\sec\pars{x}\,\dd x =
2\bracks{\vphantom{\Large A}\ln\pars{\sec\pars{x} + \tan\pars{x}}}_{\ 0}^{\ \pi/4}
\\[5mm] &=
2\ln\pars{\sec\pars{\pi \over 4} + \tan\pars{\pi \over 4}} =
\bbx{2\ln\pars{\root{2} + 1}} \approx 1.7627
\end{align}
