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Double integral involving the signum function

Real & Complex Analysis, Calculus & Multivariate Calculus, Functional Analysis,
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akotronis
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Double integral involving the signum function

#1

Post by akotronis » Thu Jul 07, 2016 12:51 pm

Show that

\(\displaystyle \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\rm sign}(x)\,{\rm sign}(y)e^{-\frac{x^2+y^2}{2}}\sin (xy)\,dx\,dy=2\sqrt{2}\log (1+\sqrt{2})\).
ZardoZ
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Re: Double integral involving the signum function

#2

Post by ZardoZ » Thu Jul 07, 2016 12:52 pm

A small comment, with a very brief examination we can easily deduce that $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm{sign}}(x)\,{\rm{sign}}(y)e^{-\frac{x^2+y^2}{2}}\sin(xy) \;dx\;dy=4\int_{0}^{\infty}\int_0^\infty e^{-\frac{x^2+y^2}{2}}\sin(xy)\;dx\;dy\,.$$
ZardoZ
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Re: Double integral involving the signum function

#3

Post by ZardoZ » Thu Jul 07, 2016 12:54 pm

Computation of \(\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}\mathbb{e}^{-\frac{x^2+y^2}{2}}\sin(xy)\;\mathbb{d}x\;\mathbb{d}y\).

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\[\begin{eqnarray*}\int_{0}^{\infty} \int_{0}^{\infty}\mathbb{e}^{-\frac{x^2+y^2}{2}}\sin(xy)\;\mathbb{d}x\;\mathbb{d}y &=& \int_{0}^{\infty}\int_{0}^{\infty}\mathbb{e}^{-\frac{x^2+y^2}{2}}\sum_{k=0}^{\infty}\frac{(-1)^{k}(x\cdot y)^{2k+1}}{(2k+1)!}\;\mathbb{d}x\;\mathbb{d}y\\ &=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{e}^{-\frac{x^2+y^2}{2}}\left(x\cdot y\right)^{2k+1}\;\mathbb{d}x\;\mathbb{d}y\\ &=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\int_{0}^{\infty}\mathbb{e}^{-\frac{x^2}{2}}x^{2k+1}\;\mathbb{d}x\right)\left(\int_{0}^{\infty}\mathbb{e}^{-\frac{y^2}{2}}y^{2k+1}\;\mathbb{d}y\right)\\&=&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\int_{0}^{\infty}\mathbb{e}^{-\frac{x^2}{2}}x^{2k+1}\;\mathbb{d}x\right)^{2}\\&=&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\int_{0}^{\infty}\frac{\mathbb{e}^{-x}\left(\sqrt{2x}\right)^{2k+1}}{\sqrt{2x}}\;\mathbb{d}x\right)^{2}\\&=&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\int_{0}^{\infty}\mathbb{e}^{-x}\left(2x\right)^{k}\;\mathbb{d}x\right)^{2}\\&=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}2^{2k}\left(\int_{0}^{\infty}\mathbb{e}^{-x}x^{k}\;\mathbb{d}x\right)^{2}\\&=&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}2^{2k}\Gamma(k+1)^{2}\\&=& \sum_{k=0}^{\infty}\frac{(-1)^{k}2^{2k}(k!)^2}{(2k+1)!}\\&=&\frac{\textrm{arcsinh}(1)}{\sqrt{2}} \end{eqnarray*}\]

so \(\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}\textrm{sign}(x)\,\textrm{sign}(y)\mathbb{e}^{-\frac{x^2+y^2}{2}}\sin(xy)\;\mathbb{d}x\;\mathbb{d}y=4\frac{\textrm{arcsinh(1)}}{\sqrt{2}}=2\sqrt{2}\textrm{arcsinh}(1)=2\sqrt{2}\log(1+\sqrt{2})\).
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