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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?

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  • $\begingroup$ 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, 2013 at 1:15
  • $\begingroup$ @nayrb : Do you really think integration by parts is what one should start with here? $\endgroup$ Dec 7, 2013 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, 2013 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. $\endgroup$ Dec 8, 2013 at 0:56

1 Answer 1

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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.

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  • $\begingroup$ @MichaelHardy : Thanks, I just fixed it. $\endgroup$ Dec 7, 2013 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, 2013 at 1:45

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