# Proof $\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}$

I'm interested in possible generalizations of the integral $$\int_{-\infty}^\infty\frac{dx}{\left(\cosh x+\frac{1}{2}e^{ix\sqrt{3}}\right)^2}=\frac{4}{3},\tag{1}$$ or equivalently $$\int_0^\infty\frac{dt}{(1+t+t^{\,\alpha})^2}=\frac23, \quad \alpha=\frac{1+i\sqrt3}{2}, \tag{1a}$$ $$\int_0^\infty\frac{e^{x\sqrt{3}}\cos\left(\frac\pi6-x\right)}{\left(2\cos x+e^{x\sqrt{3}}\right)^2}dx=\frac16.\tag{1b}$$ Alternative ways to prove integral $(1)$ are also welcomed because they might show how to obtain generalizations.

Integrals with similar looking integrands has been considered before, however no context was given explaining where did they arize and no closed form solution was given:

integrate $\int \frac{1}{e^{x}+e^{ax}+e^{a^{2}x}} \, dx$

Integrating a Complex Exponential Function

I know only one proof of $(1)$ which is as follows. By contour integration applied to the function $$\frac{1}{z\prod_{k=1}^\infty\left(1+\frac{z^3}{k^3}\right)}$$ one can show that $$\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left|\Gamma(1-\varepsilon n)\right|^2=-\frac13\tag{*}$$ where $\varepsilon=e^{2\pi i/3}$. Then using the integral 3.985.1 from Gradsteyn and Ryzhik

$$\int_0^\infty \frac{\cos{ax} \space dx}{ch^\gamma \beta x} = \frac{2^{\gamma - 2}}{\beta\Gamma(\gamma)} \Gamma\left( \frac{\gamma}{2} + \frac{ai}{2\beta} \right) \Gamma\left( \frac{\gamma}{2} - \frac{ai}{2\beta} \right)$$

one can write this series as an integral $$\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left|\Gamma(1-\varepsilon n)\right|^2=-1+\frac{1}{2}\int_{-\infty}^\infty\frac{dx}{\left(\cosh x+\frac{1}{2}e^{ix\sqrt{3}}\right)^2}.\qquad \qquad \Box$$

This approach also allows to prove that $$\int_\limits{-\infty}^{\infty}\frac{\text{sech}x~e^{\frac{i\sqrt{3}}{2}x}dx}{\sqrt{e^x+e^{-x}+e^{i\sqrt{3}x}}}=\frac{\pi}{3},\tag{2}$$ $$\int_\limits{-\infty}^{\infty}\frac{e^{i \sqrt{3} x} \cosh x}{\left(e^x+e^{-x}+e^{i \sqrt{3} x}\right)^2}dx=\frac{1}{12}.\tag{3}$$ It is interesting that $(3)$ is obtained from the same series evaluation (*) as $(1)$. Note also the equivalent forms $$\int_0^\infty\frac{t^{\alpha-1}dt}{(1+t+t^{\,\alpha})^2}=\frac{1}{3\alpha}, \quad \alpha=\frac{1+i\sqrt3}{2}\tag{3a}$$ $$\int_0^\infty\frac{t^{\alpha}dt}{(1+t+t^{\,\alpha})^2}=\frac{\alpha}{3}\tag{3b}$$

However, I couldn't apply this technique to find any other integrals of similar kind calculable in closed form that have simple form as $(1),(2),(3)$. It is not also clear whether there is a direct computation without summation of any series. Given the simplicity of the closed forms above this might be possible.

Note that it one can write $(1)$ and $(2)$ in terms of real integrands completely avoiding complex numbers, but the current form is more compact.

Q1: Are there any other ways to prove $(1),(2),(3)$?

Q2: What are possible generalizations of these integrals?

• I think people of math over flow can solve these problem. – Takahiro Waki Jan 30 '17 at 14:49
• What sort of generalization interests you? – Argon Jun 29 '17 at 22:01
• @Argon mostly I'm interested in the question is this integral isolated or not? Probably there are not any parametric generalizations, but it is possible that there are some other isolated integrals similar to this. – Nemo Jun 29 '17 at 23:14
• Integrating around a sector of angle $\frac{2 \pi}{3}$, which is what I assume you did, I get $\sum_{k=1}^{\infty} \frac{(-1)^{k} k^{2}}{k!} |\Gamma(ke^{i \pi/3})|= - \frac{1}{3}$. Then using that integral formula, I get $(3)$. But using $(*)$, which is just another way to express the above series, I get $$\int_{-\infty}^{\infty} \left(\frac{2e^{i \sqrt{3} x} \operatorname{sech} x}{\left(2 \cosh x +e^{i \sqrt{3} x}\right)^{2}} + \frac{e^{i \sqrt{3}x}\operatorname{sech}^{2} x }{2\cosh x + e^{i \sqrt{3}x}}\right) \, dx= \frac{2}{3}.$$ I don't quite see how to manipulate this to get $(1)$. – Random Variable Jun 30 '17 at 17:37
• I think this should explain it $\frac{e^{i \sqrt{3}x}\operatorname{sech}^{2} x }{2\cosh x + e^{i \sqrt{3}x}}=\frac{(2\cosh x+e^{i\sqrt{3}x}-2\cosh x)\operatorname{sech}^{2} x }{2\cosh x + e^{i \sqrt{3}x}}$. – Nemo Jun 30 '17 at 18:49

## 1 Answer

The following formula is a generalization of $(1)$ \begin{align} \small\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{a+ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{-ix\sqrt{3}}\right)^2}=1 \end{align} where $|a|$ is sufficiently small. It gives a parametric extension of $(1)$. I think it is hard to expect anything more simple than this.