23rd January

Consider this following integral $\color{#E81990}I(\beta)= \int_0^∞ \frac{x^{\beta}}{1+x^6}dx$ Take $x^6= t$ the integral becomes $\frac{1}{6} \int_0^∞\frac{t^{\frac{\beta+1}{6}-1}}{(1+t)^{\frac{\beta+1}{6}+\frac{5-\beta}{6}}}dt=\frac{1}{6}\Beta{(\frac{\beta+1}{6},\frac{5-\beta}{6})}$ $\color{#E81990}\Gamma{(a)}\Gamma{(1-a)}=π\cosec(πa)$ $= \frac{1}{6} \frac{\Gamma{(\frac{\beta+1}{6})}\Gamma{(\frac{5-\beta}{6})}}{\Gamma{(1)}}= \frac{π}{6} \cosec( \frac{\beta+1}{6}π)$ Now $\frac{\partial{I(\beta)}}{\partial\beta}=I'(\beta)= -\frac{π^2}{36}\cosec(\frac{\beta+1}{6}π)\cot(\frac{\beta+1}{6}π)= \int_0^∞ \frac{x^{\beta}\log(x)}{1+x^6}dx$ Now $\beta=0$ and answer will be $\boxed{\frac{-π^2}{6\sqrt{3}}= \int_0^∞ \frac{\log(x)}{1+x^6}dx}$ Answer is $\color{#20A900}\boxed{a-2b=0}$

I've found that for general $\int_0^∞ \frac{\log(x)}{1+x^n}dx= -\frac{π^2}{n^2}\cosec(\frac{π}{n})\cot(\frac{π}{n})$

Relevant wiki

- Beta function

- Gamma function

Let $\omega = e^{\pi i / 6}.$ Let $I$ be the integral we want. Then take the pie-piece contour $C$ from $r$ to $R,$ then $R$ to $R\omega^2$ in a circular arc, then $R\omega^2$ to $r\omega^2,$ then $r\omega^2$ to $r$ in a circular arc. As $R\to\infty,$ the integral over the outer arc goes to $0,$ and as $r\to 0,$ the integral over the inner arc goes to $0$ (it's approximated by a constant times $r \ln(r),$ which does go to $0$ by L'Hopital). The integral over $C$ equals $2\pi i$ times the residue at the one pole inside, namely $z=\omega.$ Putting this all together gives $\begin{aligned} \int_0^\infty \frac{\ln(x)}{1+x^6} \, dx - \int_0^\infty \frac{\ln(x) + i \pi/3}{1+x^6} \omega^2 \, dx &= 2\pi i \, {\rm Res}_{z=\omega} \frac{\ln(z)}{1+z^6} \\ I(1-\omega^2) - \int_0^\infty \frac{i \pi/3}{1+x^6} \omega^2 \, dx &= 2\pi i \frac{\pi i/6}{6\omega^5} \\ I(\omega^{10} - 1) - \int_0^\infty \frac{i \pi/3}{1+x^6} \, dx &= -\frac{\pi^2 \omega^5}{18} \\ \end{aligned}$ and now we can take real parts, to get $\begin{aligned} -I/2 &= -\frac{\pi^2 (-\sqrt{3}/2)}{18} \\ I &= -\frac{\pi^2 \sqrt{3}}{18} \\ I & = -\frac{\pi^2}{6\sqrt{3}}, \end{aligned}$ so the answer is $6-2\cdot 3 = \fbox{0}.$

In general, replacing $6$ by $n,$ we get $\begin{aligned} I(\cos(2 \pi/n) - 1) &= \frac{2\pi^2}{n^2} \cos(\pi/n) \\ I &= -\frac{\pi^2}{n^2} \frac{2\cos(\pi/n)}{1-\cos(2\pi/n)} \\ I &= -\frac{\pi^2}{n^2} \frac{\cos(\pi/n)}{1-\cos^2(\pi/n)} \\ I &= -\frac{\pi^2}{n^2} \frac{\cos(\pi/n)}{\sin^2(\pi/n)} \\ I &= -\frac{\pi^2}{n^2} \cot(\pi/n) \csc(\pi/n) \end{aligned}$ as in the other solution.