Closed form for a difficult integral

Question

Can any one help to find the closed form for this difficult integral, it seems very tough to me.

$$\large{\int}_0^\infty\small \dfrac{\log^2\left|\cot(\dfrac{ax}{2}-\dfrac{\pi}{4})\right|}{x^4+4}dx$$

Answer 1

Long story short, I was able to obtain the following representation:

\begin{align*} &\int_{0}^{\infty} \frac{\log^2\left| \cot\left(\frac{\alpha x}{2} - \frac{\pi}{4}\right)\right|}{x^4 + 4} \, dx \\ &\hspace{3em} = \frac{1}{8}\sum_{\omega \ : \ \omega^4 = -4} \omega \left[ \operatorname{Li}_3\left(\frac{\cos(\alpha \omega)}{\sin(\alpha\omega)+1}\right) + \operatorname{Li}_3\left(\frac{\cos(\alpha \omega)}{\sin(\alpha\omega)-1}\right) \right]. \tag{1} \end{align*}

Here the sum in the right-hand side runs over all $\omega$ such that $\omega^4 = -4$, i.e., $\omega = \pm 1 \pm i$. Also $\operatorname{Li}_3$ is the trilogarithm function. For instance, for $\alpha = 2$ we have

But to be honest, I have no idea whether this simplifies further or not.

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The idea is as follows: Let $I$ denote the integral.

1. It is not hard to check that $x \mapsto \log^2\left| \cot\left(\frac{\alpha x}{2} - \frac{\pi}{4}\right)\right|$ is an even $\frac{\pi}{\alpha}$-periodic function. So

   \begin{align*} I &= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\log^2\left| \cot\left(\frac{\alpha x}{2} - \frac{\pi}{4}\right)\right|}{x^4 + 4} \, dx \\ &= \frac{1}{2}\int_{-\frac{\pi}{2\alpha}}^{\frac{\pi}{2\alpha}} \left( \sum_{k=-\infty}^{\infty} \frac{1}{(x - \frac{k\pi}{\alpha})^4 + 4} \right) \log^2\left| \cot\left(\frac{\alpha x}{2} - \frac{\pi}{4}\right)\right| \, dx \\ &= \int_{0}^{\frac{\pi}{2\alpha}} \left( \sum_{k=-\infty}^{\infty} \frac{1}{(x - \frac{k\pi}{\alpha})^4 + 4} \right) \log^2\left| \cot\left(\frac{\alpha x}{2} - \frac{\pi}{4}\right)\right| \, dx. \tag{2} \end{align*}

2. We have the following partial fraction decomposition

   $$ \frac{1}{x^4 + 4} = -\frac{1}{16} \sum_{\omega \ : \ \omega^4 = -4} \frac{\omega}{x - \omega}. $$

   Using this decomposition and the partial fraction decomposition of cotangent, we find that

   $$ \sum_{k=-\infty}^{\infty} \frac{1}{(x - \frac{k\pi}{\alpha})^4 + 4} = -\frac{\alpha}{16} \sum_{\omega \ : \ \omega^4 = -4} \omega \cot(\alpha x - \alpha \omega). \tag{3} $$

3. Apply the substitution $u = \log \cot \left( \frac{\pi}{4} - \frac{\alpha x}{2} \right)$, assuming that $\alpha > 0$. Then it is easy to check that

   $$ \tan(\alpha x) = \sinh u, \qquad dx = \frac{du}{\alpha \cosh u}.$$

   Putting this and $\text{(2)-(3)}$ altogether, we obtain

   $$ I = -\frac{1}{16} \int_{0}^{\infty} u^2 \left( \sum_{\omega \ : \ \omega^4 = -4} \frac{\omega \cosh u}{\sinh u - \tan(\alpha \omega)} \right) \, du. \tag{4} $$

4. With $A_{\pm}(\omega) = \frac{\cos(\alpha \omega)}{\sin(\alpha \omega) \pm 1}$, we can check that

   \begin{align*} -\sum_{\omega \ : \ \omega^4 = -4} \frac{\omega \cosh u}{\sinh u - \tan(\alpha \omega)} &= \sum_{\omega \ : \ \omega^4 = -4} \omega \left( \frac{1}{1 - A_+(\omega) e^{-u}} + \frac{1}{1 - A_-(\omega) e^{-u}} \right) \\ &= \sum_{n=1}^{\infty} \left( \sum_{\omega \ : \ \omega^4 = -4} \omega( A_+(\omega)^n + A_-(\omega)^n) \right) e^{-nu}. \end{align*}

   Plugging this back to $\text{(4)}$ and simplifying using the gamma integral yields the desired identity $\text{(1)}$.