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I'm interested in the integral $$ I=\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}.\tag{1} $$ So far I have been able to reduce this integral to an integral of an elementary function in the hope that it will be more tractable $$ I=\frac{8\pi}{\sqrt{3}}\int_{-\infty}^\infty\frac{e^{ix\sqrt{3}}\ dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^3},\tag{2} $$ using the approach from this question. In that question it was also proved that $$ \int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3},\tag{3} $$ which gives some indication that the integral in the right hand side of $(2)$ might be calculable.

Also note that the integrand in $(1)$ can be expressed as $$ \Gamma(x+1)\left|\Gamma\left(1+e^{\frac{2\pi i}{3}}x\right)\right|^2. $$

Bending the contour of integration in the integral on the RHS of $(2)$ one obtains an alternative representation $$ I=8\pi\int_0^\infty\frac{e^{x\sqrt{3}}~dx}{\left(2\cos x+e^{x\sqrt{3}}\right)^3}.\tag{4} $$

There are some calculable integrals containing the infinite product $\prod\limits_{k=1}^\infty\left(1+\frac{x^3}{k^3}\right)$, e.g. $$ \int_{0}^\infty\frac{\left(1-e^{\pi\sqrt{3}x}\cos\pi x\right)e^{-\frac{2\pi}{\sqrt{3}}x}\ dx}{x\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}=0. $$

Q: Is it possible to calculate $(1)$ in closed form?

Cave Johnson
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  • I'm not sure to what degree this may be helpful, but have you tried anything along the lines of contour integration? – Simply Beautiful Art Nov 27 '16 at 13:08
  • @SimpleArt yes, as i wrote i used approach from another question. there i write that by contour integration i obtain a series, and then i 'sum' the series in terms of an integral. that's how i obtained (2) – Cave Johnson Nov 27 '16 at 13:19
  • @Nemo Hint: before attempting the integral look at the relation of the infinite product to the sine (or Sinh) functions. The answer to your question is affirmative. – Dr. Wolfgang Hintze Nov 27 '16 at 20:05
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    @Dr. Wolfgang Hintze but $\sin$ has $n^2$ in its infinte product formula, not $n^3$ – tired Nov 27 '16 at 20:12
  • @tired You are right. Sorry, my error. – Dr. Wolfgang Hintze Nov 27 '16 at 20:16
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    @Nemo to be honest i think your questions might be better suited for MathOverflow... – tired Nov 27 '16 at 20:28
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    The numerical value of the integral is 0.9726754529705... – Dr. Wolfgang Hintze Nov 27 '16 at 20:51
  • Have you tried to rexpress your integral as an infinte sum? this should be possible using contour integration since the integrand is invariant under a rotation $x\rightarrow x e^{i 2 \pi /3}$ – tired Nov 27 '16 at 21:44
  • @tired you mean the integral on the right hand side of $(2)$? – Cave Johnson Nov 27 '16 at 21:48
  • no one the left hand side – tired Nov 27 '16 at 21:52
  • @tired Simple Art already asked that question. – Cave Johnson Nov 27 '16 at 21:56
  • ah ok sorry i haven't read the first comments. and this doesn't lead to anything useful? – tired Nov 27 '16 at 21:59
  • @tired it leads to a series like $\sum_{n=1}^\infty\frac{(-1)^n}{(n-1)!}\left|\Gamma(1-\varepsilon n)\right|^2$ where $\varepsilon=e^{2\pi i/3}$. – Cave Johnson Nov 27 '16 at 22:02
  • $$\frac{\pi}{2\sqrt{3}}\sum_{j=1}^{\infty}j\prod_{k\neq j}\frac{1}{1-\left(\frac{j}{k}\right)^3}$$ – Count Iblis Nov 28 '16 at 02:40
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    @Nemo In (2) : shouldn't it read 8 pi instead of 4 pi ? – Dr. Wolfgang Hintze Nov 28 '16 at 13:41
  • @Dr.WolfgangHintze you are right – Cave Johnson Nov 28 '16 at 15:11
  • @Jack D'Aurizio you did not like the criticism and decided to downvote my question? – Cave Johnson Jul 08 '17 at 18:07
  • @JackD'Aurizio did you see a comment above where I gave the same series? And also the linked question where I consider the same type of series? The fact that you gave this series as an answer says that you did not read the question carefully. – Cave Johnson Jul 08 '17 at 18:11
  • @JackD'Aurizio I think I did exactly that but phrased it differently: "using the approach from this question[link]". And that linked question used the same approach as in your answer. – Cave Johnson Jul 08 '17 at 18:18
  • @JackD'Aurizio that's the point. You did not read the question carefully. Moreover, your answer contained an error. – Cave Johnson Jul 08 '17 at 18:24
  • @JackD'Aurizio because there was not any answer to my last 8 questions. The one with the answer has been accepted. – Cave Johnson Jul 08 '17 at 18:29
  • @JackD'Aurizio what same subject? – Cave Johnson Jul 08 '17 at 18:58
  • @JackD'Aurizio well, I have one question for an integral of the product $\prod_{k\geq 1}\left(1+\frac{x^3}{k^3}\right)$ over $\mathbb{R}^+$, and another question over $\mathbb{R}^-$. I have reasons to believe this two cases are quite different. All other 7 questions are not about integrals of this product at all. How are they the same? – Cave Johnson Jul 08 '17 at 19:05
  • @JackD'Aurizio no, 2 questions are about integral of the infinite product, and other two questions are related to the integral of the infinite product, but being related does not mean being the same thing. – Cave Johnson Jul 08 '17 at 19:11
  • Hello @Nemo, do you remember by any chance how to show that the last integral vanishes? I think you could also post an answer here. – Zacky Jan 25 '20 at 10:23

1 Answers1

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define:

$$ Q^n(x) = \prod_{a=1}^n{1-\left(\frac{x}{a}\right)^3} = \left(1-\left(\frac{x}{a}\right)^3\right) \cdot\prod_{b\ne a}{1-\left(\frac{x}{b}\right)^3} = \left(1-\left(\frac{x}{a}\right)^3\right)\cdot Q_a(x) $$

using:

$$\begin{align*} (a) && & 1-\left(\frac{x}{a}\right)^3 = 0 \Leftrightarrow x=az_i, \; (z_i)^3=0.\; i=0,1,2 \\ (b) && & \frac{d}{dx}\left(1-\left(\frac{x}{a}\right)^3\right) = -3\frac{x^2}{a^2}; \; (x=az_i)\; -3\frac{\bar{z}_i}{a} \\ (c) && & (b\ne a) \Rightarrow1-\left(\frac{az_i}{b}\right)^3 = 1-\left(\frac{a}{b}\right)^3 \\ (d) && & \frac{d}{dx}Q^n(x)=\left(1-\left(\frac{x}{a}\right)^3\right)Q'_a(x) - 3\frac{x^2}{a^3}Q_a(x) \\ (a,b,c,d) \Rightarrow (e) && & Q'(az_i)=0-3\frac{\bar{z}_i}{a}Q_a(az_i) = -3\frac{\bar{z}_i}{a} \prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)} = -3\frac{\bar{z}_i}{a}P(a) \end{align*}$$

$$\begin{align*} \frac1{Q^n(x)} && & \stackrel{pfd}{=} \sum_{a=1}^n{\sum_{i=0}^3{\frac1{(x-az_i)\cdot Q'_a(x)}}} \\ && & \stackrel{(e)}{=} \sum_{a=1}^n{\sum_{i=0}^3{\frac1{-3\bar{z_i}(x-az_i)} \cdot \frac{a}{P(a)}}} \\ && & = \sum_{a=1}^n{\frac{a}{P(a)}\sum_{i=0}^3{\frac{z_i}{-3(x-az_i)}}}\\ && & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3-x^3}}\\ \end{align*}$$

$$\begin{align*} \int_0^{\infty}{\frac1{Q_n(-x)}dx} && & = \int_0^{\infty}{ \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3+x^3}}dx} \\ && & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\int_0^{\infty}{\frac{a^2}{a^3+x^3}dx}}\\ && & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}} \\ \end{align*}$$

$$\begin{align*} \lim_{n\to\infty}\int_0^{\infty}{\frac1{Q_n(-x)}dx} && & = \lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}}\\ && & = \frac{2\pi}{3\sqrt{3}}\lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}}\\ \end{align*}$$

Answer:

$$\frac{2\pi}{3\sqrt{3}}\sum_{a=1}^{\infty}{a\prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)^{-1}}}$$

Sam Coutteau
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