What better to start the year than a dazzling integral?

$$\int_{0}^{\infty}\left[1+\left(\frac{2013}{x+2013}+\cdots +\frac{2}{x+2}+\frac{1}{x+1}-x\right)^{2014}\,\right]^{-1}\,dx$$

Happy New Year to the mathematical community!

(I am not too familiar with the posting policies on this site, hopefully this is not a major breach of rules)


closed as off-topic by Carl Mummert, Charles, JonMark Perry, user228113, Tim Raczkowski Dec 19 '15 at 1:53

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  • $\begingroup$ Happy new year! The integral would be more interesting if there weren't any $2014$ in it, though. $\endgroup$ – Ian Mateus Jan 1 '14 at 4:26
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    $\begingroup$ @IanMateus Ah but it would not converge if there weren't any $2014$ in it! $\endgroup$ – user118524 Jan 1 '14 at 4:56
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    $\begingroup$ Are you sure that this is recreational mathematics ? $\endgroup$ – Claude Leibovici Jan 1 '14 at 5:43
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    $\begingroup$ @ClaudeLeibovici Well, in the description for said tag it reads "[...] or mathematics done just for fun", so I figured this fell under that category. $\endgroup$ – user118524 Jan 1 '14 at 5:58
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    $\begingroup$ @user118524. I was joking ! Your post is the first I read in 2014. I shall follow your post ! Happy New Year !! $\endgroup$ – Claude Leibovici Jan 1 '14 at 6:00

The integral from $-\infty$ to $\infty$ is $$\frac{2\pi}{2014}\csc\left[\frac{\pi}{2014}\right]$$ See M.L. Glasser, A remarkable property of definite integrals, Math. Comp. 40, 261 (1981).

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    $\begingroup$ larry, unfortunately this is behind a paywall. Could you please sketch the main points of the argument so we can get some hints on how to do it? $\endgroup$ – Ian Mateus Jan 6 '14 at 17:01
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    $\begingroup$ A free copy of the reference is available from AMS here. $\endgroup$ – user119961 Jan 8 '14 at 21:07

Enfim, depois de passar alguns dias a fio tentando resolver esse desafio, acredito que a resposta numérica final seja $0{,}5 \cdot e \approx 1{,}35914$, onde $e \approx 2{,}71828$ é o número de Euler.

Anyway, after spending a few days on end trying to solve this challenge, I believe that the final numerical answer is $0{,}5 \cdot e \approx 1{,}35914$, where $e \approx 2{,}71828$ is the Euler number.

This is a numerical approximation, not the exact result. The function being integrated relies only on the sum $n / (x + n)$, where $n$ is an integer and belongs to the interval $[1, 2013]$. The sum has a single root in the interval $[0, \infty)$. That is, $x ≈ 938{,}17268...$ the value of the sum is zero. Well, then the function being integrated is equal to $1$. Therefore, we can easily see that the function being integrated in the interval $[0; 938{,}17268...)$ is growing. And in the interval $(938{,}17268; \infty)$ is decreasing.

Using the method of trapezoids in the vicinity of $x \approx 938{,}17268$ we can determine an approximation for the numerical integration. As the function values ​​are dwindling, there is no significant change in the first 5 digits houses. So I suggested that $0{,}5 \cdot e \approx 1{,}35914$ is an approximation to the result. Furthermore, it is possible to prove this result using Maple with interactive integration algorithm.

It is important to remember that the function is not continuous the full extent of the real numbers. How can there be integral from $-\infty$ to $\infty$, mentioned in the first reply?


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