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A double arctan integral

Let \Delta:\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 4x\}. Evaluate the integral:

    \[\mathcal{J} = \iint \limits_{\Delta} \arctan e^{xy}\,\mathrm{d}y\,\mathrm{d}x\]


The key observation to nail the integral is that the domain of integration is symmetric with respect to the x axis.

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Hence , a symmetry might as well work. So,

    \begin{align*} \mathcal{J} &= \iint \limits_{\Delta} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &=\int_{0}^{4} \int_{-\sqrt{4x-x^2}}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &=\int_{0}^{4} \int_{-\sqrt{4x-x^2}}^{0} \arctan e^{xy} \, \mathrm{d}(y, x) + \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &= \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) + \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{-xy} \, \mathrm{d}(y, x) \\ &= \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \frac{\pi}{2} \, \mathrm{d}(y, x) \\ &= \frac{\pi}{2} \int_{0}^{4} \sqrt{4x-x^2} \, \mathrm{d}x \\ &= \frac{\pi}{2} \cdot 2 \pi \\ &= \pi^2 \end{align*}


    \[\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2} \quad \text{forall} \;\; x>0\]

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