5
$\begingroup$

Evaluate the double integral:

$$\iint_D\arctan e^{xy}\,dy\,dx$$

where $D:\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 4x\}$

For thie integral , since our function does not have elementary anti-derivetives with respect to both variables, I tried to solve this by polar coordinate. But then it was getting worth and I got something very ugly and I cannot simplify, I stuck. Are there any tricks to deal with this?

$\endgroup$
4
  • $\begingroup$ Show some more of your work. Staring this integral down, I agree it probably is hard, but I don't care to work out all the details. Did you ever try integration by parts? $\endgroup$ – abnry Dec 7 '13 at 1:15
  • $\begingroup$ @nayrb : Do you really think integration by parts is what one should start with here? $\endgroup$ – Michael Hardy Dec 7 '13 at 1:18
  • $\begingroup$ My intuition is that we need to integrate something like $\int \arctan x dx$. How do you do this? You use integration by parts. Apparently I didn't think about any symmetries. $\endgroup$ – abnry Dec 7 '13 at 3:47
  • $\begingroup$ @Andrew : polar coordinates was a reasonable idea, since the limits of integration turn out to be simpler, it just doesn't work in this problem. $\endgroup$ – Stefan Smith Dec 8 '13 at 0:56
7
$\begingroup$

Hint:

$$\int_0^4 \int_{-\sqrt{4x-x^2}}^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx=\int_0^4 \int_{-\sqrt{4x-x^2}}^0 \tan^{-1}(\exp(xy))\,dy\,dx+ \cdots $$

$$+\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx=$$

$$\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx +\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(-xy))\,dy\,dx.$$

Use the identity $\tan^{-1}(c)+\tan^{-1}(\frac{1}{c})=\frac{\pi}{2}$ for $c>0$.

By the way, I tried polar coordinates too, and even Maple couldn't do the integral.

The key is to notice that the domain of integration is symmetric with respect to the $x$-axis and find a symmetry trick.

$\endgroup$
2
  • $\begingroup$ @MichaelHardy : Thanks, I just fixed it. $\endgroup$ – Stefan Smith Dec 7 '13 at 1:36
  • $\begingroup$ I think this works. And I believe this problem could be terribly hard without noticing this fact. Thank you very much! $\endgroup$ – Andrew Dec 7 '13 at 1:45

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.