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I want to give you this exercise to see if there are simpler solutions to the one that I've chosen! I found it a very interesting exercise.

Question

Find the closed form for all N of following function $$f(x)=\int\cdots\int\delta\left(\sum_{n=1}^{N}x_{n}-x\right)\prod_{n=1}^{N}\frac{1}{1+x_{n}^2}dx_1\cdots dx_N$$ where $\delta(x)$ is the Dirac delta function and all the integrals are to be taken on the entire real line.

My answer

I found very useful for this type of problems to take some N and see the pattern:

  • N=1 $$f(x) = \int_{\mathbb{R}}\delta(x_1-x)\frac{1}{1+x_1^2}dx_1=\frac{1}{1+x^2}$$ easy. Next comes the fun part.

  • N=2 $$f(x) = \iint_{\mathbb{R^2}}\delta(x_1+x_2-x)\frac{1}{1+x_1^2}\frac{1}{1+x_2^2}dx_1dx_2 = \\ = \iint_{\mathbb{R^2}}\delta(x_1-(x-x_2))\frac{1}{1+x_1^2}\frac{1}{1+x_2^2}dx_1dx_2 = \\ = \int_{\mathbb{R}}\frac{1}{1+(x-x_2)^2}\frac{1}{1+x_2^2}dx_2$$ this is just the convolution of two same functions $g(x)=\frac{1}{1+x^2}$. Using the theorem for convolution of the Fourier transform I get $$f(x) = \mathcal{F}^{-1}\left\{\mathcal{F}^{2}\left\{\frac{1}{1+x^2}\right\}\right\} = \mathcal{F}^{-1}\left\{\pi e^{-2k}\right\} =\frac{2\pi}{4+x^2}$$ Here I used the fact that $$\mathcal{F}\left\{\frac{1}{a^2+x^2}\right\} = \frac{\pi}{a}e^{-a|k|}$$

  • N=3 $$f(x)=\iiint_{\mathbb{R}^3}\delta(x_1+x_2+x_3-x)\frac{1}{1+x_1^2}\frac{1}{1+x_2^2}\frac{1}{1+x_3^2}dx_1dx_2dx_3 = \\ = \iiint_{\mathbb{R}^3}\delta(x_1-(x-x_2-x_3))\frac{1}{1+x_1^2}\frac{1}{1+x_2^2}\frac{1}{1+x_3^2}dx_1dx_2dx_3 = \\ = \iint_{\mathbb{R}^2}\frac{1}{1+(x-x_2-x_3)^2}\frac{1}{1+x_2^2}\frac{1}{1+x_3^2}dx_2dx_3$$ which is now a double convolution, one equal to before $$\left(\frac{1}{1+(x-x_3)^2}*\frac{1}{1+x^2}\right) = \frac{2\pi}{4+(x_3-x)^2}$$ so then $$\int_{\mathbb{R}} \frac{2\pi}{4+(x-x_3)^2}\frac{1}{1+x_3^2}dx_3$$ which is yet another convolution, leading to $$f(x) = \frac{3\pi^2}{9+x^2}$$

So my final answer for a generic N would be: $$f(x) = \frac{N\pi^{N-1}}{N^2+x^2}$$

Final thoughts

Now, is there a simpler way to get this result? Probably there are some constant problems but it's not the point, point is that the function is of the form $$f(x)=\frac{c(N)}{N^2+x^2}$$

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You can compute the Fourier transform of $f$: $$\widehat{f}(\xi)=\int f(x) e^{ix \xi}dx=\int \int\cdots\int\delta\left(\sum_{n=1}^{N}x_{n}-x\right)\prod_{n=1}^{N}\frac{1}{1+x_{n}^2} e^{i x \xi}dx_1\cdots dx_N dx$$ so: \begin{align}\widehat{f}(\xi)&= \int\cdots \int\left( \int\delta\left(\sum_{n=1}^{N}x_{n}-x\right) e^{i x \xi} dx \right)\prod_{n=1}^{N}\frac{1}{1+x_{n}^2} dx_1\cdots dx_N\\ &= \int\cdots \int\left( \prod_{n=1}^{N} e^{i x_n \xi} \right)\prod_{n=1}^{N}\frac{1}{1+x_{n}^2} dx_1\cdots dx_N\\ &=\left( \int e^{i x \xi} \frac{1}{1+x^2}\right)^N=\pi^N e^{-N |x|} \end{align} and using the inverse Fourier transform one can obtain your result.

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  • $\begingroup$ This is clever! Converting the delta with that product is ingenious. Thank you so much! $\endgroup$ Commented Jul 12, 2018 at 16:17

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