Double Integral help: $ \int_{x=0.5}^{x=1.5}\int_{-0.5}^{0.5}\frac{dx\;dy}{\sqrt{x^2+y^2}} $ So it's been a while since I've had to do any remotely difficult integration, and one of my profs kind of sprung this on us.
$$
\int_{x=0.5}^{x=1.5}\int_{-0.5}^{0.5}\frac{dx\;dy}{\sqrt{x^2+y^2}}
$$ 
If I convert to polar coordinates and integrate, I get r times theta or:
$$
\sqrt{x^2+y^2}\cdot \tan^{-1}(\frac{y}{x})
$$
In rectangular coordinates. However, whenever I tired using the limits given, I arrive at an answer of 0.0933, when the expected value is close to 1. Does anybody know what could have gone awry? I think the integration method is correct, but I'm not sure.
 A: Computing the integrals numerically I got $\approx 1.038049736$
Polar coordinates are more complicated for a rectangular area (the length must depend of the angle) so let's try more directly :
$$\int_{\frac 12}^{\frac 32}  dx\int_{-\frac 12}^{\frac 12} \frac{dy}{\sqrt{x^2+y^2}}
=2\int_{\frac 12}^{\frac 32} \left[\log\left(y+\sqrt{x^2+y^2}\right)\right]_0^{\frac 12} dx$$
(or $\int \operatorname{argsinh}(\frac yx) dx$ if you prefer)
$$
=2\int_{\frac 12}^{\frac 32} \log\left(\frac 12+\sqrt{x^2+\frac 14}\right)- \log(x)\ dx
$$
$$=2\left[x\left(\log\left(\frac 12+\sqrt{x^2+\frac 14}\right)-\log(x)\right)+\frac 
{\operatorname{argsinh}(2x)}2\right]_{\frac 12}^{\frac 32}$$
(with some help from WolframAlpha)
A: Hint: Can you get
$$\frac{1}{2}I=\int_0^{\theta_1}\int_{\frac{0.5}{\cos\theta}}^{\frac{1.5}{\cos\theta}}dr d\theta+\int_{\theta_1}^{\pi/4}\int_{\frac{0.5}{\cos\theta}}^{\frac{0.5}{\sin\theta}}drd\theta
$$
where $\tan \theta_1=\frac{1}{3}$?
A: According to Maple, the answer is $$\frac12\,\ln  \left( \sqrt {2}-1 \right) -\frac32\,\ln  \left( 1+\sqrt {2}
 \right) -\frac32\,\ln  \left( -1+\sqrt {10} \right) -\ln  \left( 
-3+\sqrt {10} \right) +\frac32\,\ln  \left( 1+\sqrt {10}
 \right)$$
This agrees with Raymond's answer.
