By transforming to polar coordinates, show that

$$\int_{0}^{1} \int_{0}^{x}\frac{1}{(1+x^2)(1+y^2)} \,dy\,dx$$

Is equal to

$$ \int_{0}^{\pi/4}\frac{\log(\sqrt{2}\cos(\theta))}{\cos(2\theta)} d\theta$$

I have tried the standard $x=r\cos(\theta)$ etc. And can easily see that the limits for theta is $0,\pi/4$ with a sketch. However, I'm struggling to compute the integral that comes out, which ranges $r$ from $r=0$ to $r=\frac{1}{\cos{\theta}}$. Also, I can integrate this explicitly using tan substitution but does that help?

  • $\begingroup$ any hints or progress? $\endgroup$ – Dis-integrating Jan 25 at 5:32
  • $\begingroup$ So is it true you have been explicitly asked to first transform the double integral using polar coordinates before evaluating it? $\endgroup$ – omegadot Jan 25 at 6:03
  • $\begingroup$ no, i was asked first to evaluate it the nice way, and secondly to use polar coordinates $\endgroup$ – Dis-integrating Jan 25 at 6:04
  • $\begingroup$ I see. So the "nice" way is definitely how I did it below. As for the transformation to polar coordinates, see @DavidG answer. $\endgroup$ – omegadot Jan 25 at 6:06

You have correctly converted the region of integration and thus we have:

\begin{align} I &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\left(1 + r^2\cos^2(\theta)\right)\left(1 + r^2\sin^2(\theta)\right)} \cdot r\:dr\:d\theta \\ \end{align}

Applying a partial fraction decomposition we arrive at: \begin{align} I &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\left(1 + r^2\cos^2(\theta)\right)\left(1 + r^2\sin^2(\theta)\right)} \cdot r\:dr\:d\theta \\ &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\cos^2(\theta) - \sin^2(\theta)}\left[\frac{\cos^2(\theta)}{1 + r^2\cos^2(\theta)} - \frac{\sin^2(\theta)}{1 + r^2\sin^2(\theta)} \right] r\:dr\:d\theta \\ &= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\int_1^{\sec(\theta)}\left[\frac{r\cos^2(\theta)}{1 + r^2\cos^2(\theta)} - \frac{r\sin^2(\theta)}{1 + r^2\sin^2(\theta)} \right]\:dr\:d\theta \\ &=\int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln\left|1 + r^2\cos^2(\theta) \right| + \ln\left|1 + r^2\sin^2(\theta) \right| \bigg]_1^{\sec(\theta)} \:d\theta \end{align}

You okay from here?

Edit: The rest of the solution:

\begin{align} I &= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln\left|1 + r^2\cos^2(\theta) \right| + \ln\left|1 + r^2\sin^2(\theta) \right| \bigg]_1^{\sec(\theta)} \:d\theta \\ &= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln(2) + \ln\left|\sec^2(\theta) \right| \bigg] \:d\theta \\ &= \int_0^{\frac{\pi}{4}}\frac{1}{\cos(2\theta)}\cdot\frac{1}{2}\ln\left|2\sec^2(\theta)\right|\:d\theta\int_0^{\frac{\pi}{4}}\frac{\ln\left| \sqrt{2}\sec(\theta)\right|}{\cos(2\theta)}\:d\theta \end{align}

  • 1
    $\begingroup$ I suspect that you intended the lower limit of the inside integral to be $\sec(0)$ instead of $0$. $\endgroup$ – John Wayland Bales Jan 25 at 6:36
  • $\begingroup$ @JohnWaylandBales - Yes, thank you for the correction. I will edit now. $\endgroup$ – user150203 Jan 25 at 8:02
  • $\begingroup$ @Dis-integrating - They don't have to be constants to apply a partial fraction decomposition. $\endgroup$ – user150203 Jan 25 at 8:18
  • $\begingroup$ wouldn't it be easier to substitute r^2 = u $\endgroup$ – Dis-integrating Jan 27 at 20:50

Do you really have to transform this double integral first using polar coordinates? It is far easier to just integrate it as is. Here \begin{align} \int_0^1 \int_0^x \frac{dy \, dx}{(1 + x^2)(1 + y^2)} &= \int_0^1 \left [\frac{\tan^{-1} y}{1 + x^2} \right ]_0^x \, dx\\ &= \int_0^1 \frac{\tan^{-1} x}{1 + x^2} \, dx\\ &= \frac{1}{2} \big{[} (\tan^{-1} x)^2 \big{]}_0^1\\ &= \frac{\pi^2}{32} \end{align}

  • $\begingroup$ Indeed. I believe it is meant to be via polar coordinates as the first words of the post "By transforming to polar coordinates, show that" reads very much like a textbook styled problem. Not saying it's a homework problem though. $\endgroup$ – user150203 Jan 25 at 5:58
  • $\begingroup$ I was thinking maybe the OP thought, from the integrands form, conversion to polar coordinates was the way to go thereby overlooking a direct approach. I will ask for clarification. $\endgroup$ – omegadot Jan 25 at 6:01
  • $\begingroup$ i mean i admit its a problem i was assigned but there were two parts. I can't see a way to use this to solve the latter though $\endgroup$ – Dis-integrating Jan 25 at 6:01

Or even easier than omegadot's argument, note this is the $x\ge y$ half of $\left(\int_0^1\frac{dx}{1+x^2}\right)^2$, i.e. $\frac{(\pi/4)^2}{2}=\frac{\pi^2}{32}$.


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