6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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}

$\endgroup$
  • 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
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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. $\frac{(\pi/4)^2}{2}=\frac{\pi^2}{32}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.