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I would like to know if the following integrals are known, or in case that aren't in the literature we can calculate these in closed-form (in terms of elementary and standard functions). I wondered about it when I tried to get variants of an integral representation from a paper of Zurab (identity $(22)$ from [1], that I refer as Zurab's integral in the title), in next two contexts. If my post is more suitable for Mathematics Stack Exchange or you've feedback (in particular if you know these integrals) please add your feedback in comments.

Question 1. The first, just as comparison to [1], calculate if possible $$\int_0^{\pi}\frac{J_0(x)}{\binom{1}{x/\pi}}dx,\tag{1}$$ where $J_0(x)$ denotes the Bessel function. Many thanks.

For Question 1 I tried to combine an integral representation from the Wikipedia Bessel function and calculations from Wolfram Alpha online calculator, in particular I did an attempt to exploit the section Bessel's integral that refers previous Wikipedia (reference Nico M. Temme, Special Functions: An introduction to the classical functions of mathematical physics (2nd print ed.). New York: Wiley, (1996) pp. 228–231), and the output int x(pi-x)/sin(x) cos(a x) dx that provides me Wolfram Alpha online calculator.

Added see my comments: If isn't possible to get a simple closed-form in terms of particular values of elementary/special functions that I evoke but the integrals (1) can be expressed as series in some simplified form, I should consider it also as an answer.

Question 2. I would like to know as reference request or in other case if it is possible to compute the integrals $$c_{n,m}:=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)dx\tag{2}$$ for integers $m\geq 0$ and integers $n\geq 1$. Many thanks.

Thus for Question 2 if you know a reduction formula or reference for these integrals $(2)$ from the literature helpful here, answer my question as a reference request and I try to search and read from the literature their closed-forms.

If I remember well I could to find just a choice of integration by parts that was suitable in the computations that I did few months ago, but it doesn't solve the problem, since I didn't know how to finish my (tedious) calculation to get in closed-form $(2)$. This is just a curiosity that I wondered few months ago, again from Zurab's integral as starting point in an attempt to combine his integral and a definition of a certain function (and arithmetical functions that I omit here from page 79 of [2]) I wrote $$-\frac{7\zeta(3)}{8\pi^3}=\sum_{n=1}^\infty\sum_{m=0}^\infty\frac{\mu(n)\chi_1(n)}{n^2}\cdot\frac{(-1)^m c_{n,m}}{(2m+1)^2},$$ where $\zeta(s)$ is the Riemann zeta function (and those other arithmetic functions from the mentioned article [2]).

References:

[1] Zurab Silagadze, Sums of Generalized Harmonic Series. For Kids from Five to Fifteen, RESONANCE, September 2015. arXiv:1003.3602

[2] Manuel Benito, Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the "Segundas Jornadas de Teoría de Números" (Madrid, 2007), pp. 63-81. (Semantic Scholar page, pdf)

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  • $\begingroup$ I emphasize that, in case that this question is interesting for MathOverflow, I'm asking as a reference request since I don't know if these definite or indefinite integrals in $(1)$ and $(2)$ are well-known from the literature, and in other case what work can be done to compute the closed-forms of the integrals in Question 1 and Question 2 (hints or some approach) in terms of particular values of elementary functions or standard functions. $\endgroup$ – user142929 Jun 14 at 14:11
  • $\begingroup$ If isn't possible to get simple closed-forms that I evoke but the integrals $(1)$ and/or $(2)$ can be expressed as series in some simplified form, I should consider it also (in the spirit of what work can be done about these questions) as an answer. My motivation is learn from the professors of the site. Many thanks. $\endgroup$ – user142929 Jun 14 at 14:58
  • $\begingroup$ Many thanks for your edit @DavidRoberts $\endgroup$ – user142929 Jul 20 at 6:09
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    $\begingroup$ No problems! I enjoyed reading the Silagadze paper :-) $\endgroup$ – David Roberts Jul 20 at 7:12
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$c_{n,m}=0$ if $n$ is even; for $n$ odd some experimentation indicates it has the form $$c_{2n+1,m}=\frac{1}{\pi^3}\bigl(a_{n,m}-\tfrac{7}{8}(2n+1)(2m+1)\zeta(3)\bigr),\;\;\text{with}\;\;a_{n,m}=a_{m,n}\in\mathbb{Q}\geq 0.$$ I have not found a closed-form expression for the coefficients $a_{n,m}$, they are quite unwieldy.

Some values of $a_{n,m}$ for small $n,m$: $$\{a_{0,0},a_{0,1},a_{0,2},a_{0,3},a_{0,4},a_{0,5}\}=\left\{0,2,\frac{110}{27},\frac{20804}{3375},\frac{3187348}{385875},\frac{323770282}{31255875}\right\},$$ $$\{a_{1,0},a_{1,1},a_{1,2},a_{1,3},a_{1,4},a_{1,5}\}=\left\{2,\frac{3187348}{385875},\frac{88717098057736}{6093243231075},\frac{3060320225351036566502}{146665526640505747875},\frac{2020487623534710670386246274}{74353310943126393099796875},\frac{13319855208745022209825602301502381584}{397806359157656353958636885223796875}\right\}.$$

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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user142929 Jun 14 at 21:46
  • $\begingroup$ Thank you very much again for your propositions in first paragraph for the sequence $c_{n,m}$. $\endgroup$ – user142929 Jun 15 at 19:39
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To evaluate \begin{align} c_{n,m}&=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)\,dx\\ &=-\frac{1}{8}\int_0^{1}\frac{x(1-x)}{\sin^2(\pi x)}\, \sin(\pi(2m+1)nx)\,dx \end{align} we first notice by changing $x\to 1-x$, that the integral vanishes if $n$ is even. In the following $c_{2n+1,m}$ is calculated.

We apply twice a result obtained in this answer: \begin{equation} \int_0^1\frac{x(1-x)}{\sin \pi x}f(x)\,dx=2\sum_{n=0}^\infty \int_0^1x(1-x)f(x)\sin\left( (2n+1)\pi x \right)\,dx \end{equation} With $f(x)=\frac{\sin(\pi(2m+1)(2n+1)x)}{\sin(\pi x)}$, first: \begin{equation} c_{2n+1,m}=-\frac{1}{4}\sum_{p=0}^\infty \int_0^1\frac{x(1-x)}{\sin(\pi x)}\sin(\pi(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\,dx \end{equation} then with$f(x)=\sin(\pi(2m+1)(2n+1)x)$: \begin{equation} c_{2n+1,m}=-\frac{1}{2}\sum_{p,q=0}^\infty \int_0^1x(1-x)\sin(\pi(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\sin\left( (2q+1)\pi x \right)\,dx \end{equation} A decomposition of the product of the three sines is \begin{align} \sin(\pi&(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\sin\left( (2q+1)\pi x \right)\\ =&\frac{1}{4}\left[ \sin\left( (N-2p+2q)\pi x \right)-\sin\left( (N-2p-2q-2)\pi x \right)\right.\\ &\left.-\sin\left( (N+2p+2q+2)\pi x \right)+\sin\left( (N+2p-2q )\pi x\right) \right] \end{align} where $N=(2m+1)(2n+1)$ is an odd integer. Using the identity \begin{equation} \int_0^1 x(1-x)\sin((2s+1)\pi x)\,dx=\frac{4}{\pi^3}\frac{1}{(2s+1)^3} \end{equation} if $s$ is an integer, we obtain \begin{align} c_{2n+1,m}=-\frac{1}{2\pi^3}\sum_{p,q=0}^\infty&\left( \frac{1}{(N-2p+2q)^3}- \frac{1}{(N-2p-2q-2)^3}\right.\\ &\left.- \frac{1}{(N+2p+2q+2)^3}+ \frac{1}{(N+2p-2q)^3} \right) \end{align} \begin{align} =-\frac{1}{2\pi^3}\sum_{p,q=0}^\infty&\left( \frac{1}{(2p+2q+2-N)^3}- \frac{1}{(2p+2q+2+N)^3}\right.\\ &\left.+ \frac{1}{(2p-2q+N)^3} -\frac{1}{(2p-2q-N)^3} \right) \end{align} \begin{align} =-\frac{1}{32\pi^3}\sum_{q=0}^\infty&\left[\psi^{(2)}\left( \frac{N}{2}+1+q \right)-\psi^{(2)}\left(- \frac{N}{2}+1+q \right)\right.\\ &\left.+\psi^{(2)}\left( -\frac{N}{2}-q \right)-\psi^{(2)}\left( \frac{N}{2}-q \right) \right] \end{align} where $\psi^{(2)}$ is a polygamma function. Using recurrence identities and reflection formula, we find \begin{align} c_{2n+1,m}&=-\frac{1}{16\pi^3}\sum_{q=0}^\infty\sum_{k=1}^N\left[\left( \frac{N}{2}+1+q -k\right)^{-3} -\left(\frac{N}{2}-q -k\right)^{-3} \right]\\ &=\frac{1}{32\pi^3}\sum_{k=1}^N\left[ \psi^{(2)}\left(\frac{N}{2}+1-k \right)+\psi^{(2)}\left(- \frac{N}{2}+k \right) \right]\\ &=\frac{1}{16\pi^3}\sum_{k=1}^N \psi^{(2)}\left(\frac{N}{2}+1-k \right)\\ \end{align} With the known value $\psi^{(2)}\left(1/2 \right)=-14\zeta(3)$ and recurrence and reflection identities, \begin{align} c_{2n+1,m}&=\frac{1}{16\pi^3}\sum_{k=1}^N \psi^{(2)}\left(\frac{1}{2}+\frac{N+1}{2}-k \right)\\ &=\frac{1}{16\pi^3}\sum_{r=1}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}+r \right)+\sum_{r=0}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}-r \right)\\ &=\frac{1}{16\pi^3}\left[\psi^{(2)}\left(\frac{1}{2}\right)+2\sum_{r=1}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}+r \right)\right]\\ &=\frac{1}{8\pi^3}\left[-7N\zeta(3)+2\sum_{r=1}^{(N-1)/2}\sum_{s=0}^{r-1}\frac{1}{\left( s+\frac{1}{2} \right)^3}\right] \end{align} which seems to be correct when compared to the values obtained by @CarloBeenakker.

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    $\begingroup$ Many thanks for your answer. I'm going to study the propositions that you're shared. $\endgroup$ – user142929 Jul 21 at 15:02

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