$\displaystyle{\int_{(0,1)\times \mathbb{R}}^{}\frac{|x|}{2\pi(1+x^2y^2)}}d(x,y)$=$\displaystyle{\int 1_{(0,1)}(x)\frac {|x|}{2\pi}d(x)\int \frac{1}{(1+x^2y^2)}}d(y)$. In the work it is given that $\int \frac{1}{1+y^2}=\pi$, so i think im close to being able to calculate the 2nd. integral.
Calculating this double integral.
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integration
lebesgue-integral
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0look up the derivative of $\arctan$ – 2017-01-19
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0furthermore since $0
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0But what is the use of he information I have in the paper? – 2017-01-19
2 Answers
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Through the substitution $y=\frac{u}{x}$, $dy=\frac{du}{x}$, $$\frac{1}{2\pi}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{1+x^2 y^2}\,dy\,dx =\frac{1}{2\pi}\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{1+u^2}\,du\,dx=\frac{1}{2\pi}\int_{0}^{1}\frac{\pi}{2}\,dx=\frac{1}{4}.$$
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Since we can treat $x$ as a constant for the integral over $y$ we use the inverse tangent derivative: $$ \int \frac{1}{1+a^2y^2}dy = \frac{\arctan{ay}}{2\pi a}+ constant$$ which implies that $$\displaystyle{\int_{(0,1)\times \mathbb{R}}^{}\frac{x}{2\pi(1+x^2y^2)}}d(x,y)=\displaystyle{\int_0^1\frac {1}{2\pi}d(x)\int_{-\infty}^\infty \frac{x}{(1+x^2y^2)}}d(y)=\int_0^1 \frac{1}{2\pi} \arctan{xy}|_{-\infty}^{\infty}dx=\frac{1}{2}\int_0^1 dx = \frac{1}{2}.$$