# Show that ∫dx from 0 to 1 ∫(x^2-y^2)/(x^2+y^2)dy from 0 to 1=∫dy from 0 to 1 ∫(x^2-y^2)/(x^2+y^2)dx from 0 to 1?

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Steve M Share
Apr 13, 2018

We seek to show that:

${\int}_{0}^{1} {\int}_{0}^{1} \setminus \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \setminus \mathrm{dy} \setminus \mathrm{dx} = {\int}_{0}^{1} {\int}_{0}^{1} \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \setminus \mathrm{dx} \setminus \mathrm{dy}$

i.e., that in this particular case, the order of integration is unimportant (In general, this is not the case)

Consider the LHS integral:

${I}_{L} = {\int}_{0}^{1} {\int}_{0}^{1} \setminus \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \setminus \mathrm{dy} \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} {\left[2 x \arctan \left(\frac{y}{x}\right) - y\right]}_{y = 0}^{1} \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} \left\{\left(2 x \arctan \left(\frac{1}{x}\right) - 1\right) - \left(2 x \arctan \left(0\right) - 0\right)\right\} \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} 2 x \arctan \left(\frac{1}{x}\right) - 1 \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus = {\left[{x}^{2} \text{arccot} x - \arctan x\right]}_{x = 0}^{1}$

$\setminus \setminus \setminus \setminus = \left(\text{arccot} 1 - \arctan 1\right) - \left(0 - \arctan 0\right)$

$\setminus \setminus \setminus \setminus = \frac{\pi}{4} - \frac{\pi}{4}$

$\setminus \setminus \setminus \setminus = 0$

And the RHS:

${I}_{R} = {\int}_{0}^{1} {\int}_{0}^{1} \setminus \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \setminus \mathrm{dx} \setminus \mathrm{dy}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} {\left[x - 2 y \arctan \left(\frac{x}{y}\right)\right]}_{x = 0}^{1} \setminus \mathrm{dy}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} \left\{\left(1 - 2 y \arctan \left(\frac{1}{y}\right)\right) - \left(0 - 2 y \arctan 0\right)\right\} \setminus \mathrm{dy}$

$\setminus \setminus \setminus \setminus = {\int}_{0}^{1} 1 - 2 y \arctan \left(\frac{1}{y}\right) \setminus \mathrm{dy}$

$\setminus \setminus \setminus \setminus = {\left[\arctan x - {x}^{2} \text{arccot} x\right]}_{x = 0}^{1}$

$\setminus \setminus \setminus \setminus = \left(\arctan 1 - \text{arccot} 1\right) - \left(\arctan 0 - 0\right)$

$\setminus \setminus \setminus \setminus = \frac{\pi}{4} - \frac{\pi}{4}$

$\setminus \setminus \setminus \setminus = 0$

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