# Calculate $\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}$

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?

• I'm not sure to what degree this may be helpful, but have you tried anything along the lines of contour integration? Nov 27, 2016 at 13:08
• @Dr. Wolfgang Hintze but $\sin$ has $n^2$ in its infinte product formula, not $n^3$ Nov 27, 2016 at 20:12
• @Nemo to be honest i think your questions might be better suited for MathOverflow... Nov 27, 2016 at 20:28
• The numerical value of the integral is 0.9726754529705... Nov 27, 2016 at 20:51
• @Nemo In (2) : shouldn't it read 8 pi instead of 4 pi ? Nov 28, 2016 at 13:41

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*}

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