Integral $\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}$

Question

I need your help with this integral: $$\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2\,x^2+2}.$$

I wasn't able to evaluate it in a closed form, although an approximate numerical evaluation suggested its value could be $\frac{\pi}{4}$.

Answer 1

Let us introduce the notation $$\mathcal{I}=\int_0^{\infty}\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}.$$

1. Now observe that \begin{align}\frac{1}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}&= \frac{1}{x^2+1}\cdot\frac{1}{\frac{x^2-1}{x\cos(\pi\ln x)+1}+2}=\\ &=\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+\frac1x}{x+\frac1x+2\cos(\pi\ln x)}. \end{align}

2. Using this formula and making the change of variables $x\leftrightarrow \frac1x$, we can rewrite $\mathcal{I}$ as $$\mathcal{I}=\int_0^{\infty}\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+x}{x+\frac1x+2\cos(\pi\ln x)}dx.$$

3. Summing the last representation with the initial one, we get $$2\mathcal{I}=\int_0^{\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}\quad \Longrightarrow\quad \mathcal{I}=\frac{\pi}{4}.$$