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I've been learning about polar coordinates recently and had a question about the following example problem:

$$\int_{x=0}^{x=1}\int_{y=0}^{y=a}\frac{1}{(1+x^2+y^2)^2}dydx$$

where a is an arbitrary positive constant. I'm required to solve the above problem using polar coordinates. According to the example, the correct answer is supposed to be:

$$\frac{\sqrt2}{4}\arctan\frac{a\sqrt2}{2}+\frac{a}{2\sqrt{a^2+1}}\arctan\frac{1}{\sqrt{a^2+1}}$$

I'm lost as to how the above answer is derived. I tried splitting up the given rectangular region of $0 \leq x \leq 1$ and $0 \leq y \leq a$ and the integrand as follows, in terms of polar coordinates:

$$\int_{\theta=0}^{\theta=\arctan a}\int_{r=0}^{r=\sec \theta}\frac{r}{(1+r^2)^2}drd\theta+\int_{\theta=\arctan a}^{\theta=\pi/2}\int_{r=0}^{r=a\csc \theta}\frac{r}{(1+r^2)^2}drd\theta$$

but the evaluation of the second term above seems to diverge to infinity?

I'd appreciate any help on this. Thanks.

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The inner integral of the second term is, $$ \int_0^{a\csc\theta}\frac{rdr}{(1+r^2)^2} = \frac{1}{2}\int_0^{a^2\csc^2\theta} \frac{du}{(1+u)^2} = \frac{1}{2}\left(1-\frac{1}{1+a^2\csc^2\theta}\right)\\=\frac{1}{2}\left(1-\frac{\sin^2\theta}{a^2+\sin^2\theta}\right)$$

This has no singularities and won't diverge when you integrate it.

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  • $\begingroup$ Thanks, but is there a way to integrate by hand the above in terms of $\theta$ where it goes from $\arctan a$ to $\pi/2$? Or do I have the limits of integration wrong somewhere? I tried plugging the integration into mathematica and get something that's wildly different from the provided answer above, something involving "ConditionalExpression" $\endgroup$ – user3280193 Mar 17 '17 at 2:13
  • $\begingroup$ @user3280193 Yes, it is doable and your limits are right. You can write the integrand as $\frac{1}{2}\frac{a^2\csc^2\theta}{1+a^2\csc^2\theta}$ and then do a subtitution $u=\cot\theta.$ It turns into an arctan-type integral and works out to the second term in your answer above. Right after that first substituation the limits should be $0$ and $1/a$ $\endgroup$ – spaceisdarkgreen Mar 17 '17 at 4:02

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