TITLE: definite integral $\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\frac{-1}{2}\ln 3$
QUESTION [5 upvotes]: Problem: Prove that $$\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\frac{-1}{2}\ln 3$$

My attempt:
I am able to prove : $$\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\int_1^\infty \frac {1-x}{(1+x^3)\ln x}dx$$ I don't know if this relation is useful or not. By the way, it seems it's impossible to integrate the function directly so I am looking for an equation driven from integration by parts or other similar tricks but I can't go further.
Any help would be highly appreciated.
P.S: This problem was proposed by "Jalil Ghaderi" from Iran.

REPLY [6 votes]: $$I=\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx$$
Making change $x=e^{-t}$
$$I=-\int_0^\infty \frac {e^{-t}-e^{-2t}}{(1+e^{-3t})t}dt=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}\big)\big(1-e^{-3t}+e^{-6t}-+ ...\big)$$
$$=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}-e^{-4t}+e^{-5t}+e^{-7t}-e^{-8t}+-...\big)$$
$$=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}+e^{-3t}-e^{-4t}+- ...\big)+\int_0^\infty\frac{dt}{t}\big(e^{-3t}-e^{-6t}+e^{-9t}-+ ...\big)$$
$$=-\int_0^\infty\Big(\frac{e^{-t}}{1+e^{-t}}-\frac{e^{-3t}}{1+e^{-3t}}\Big)\frac{dt}{t}=-\int_0^\infty(f(t)-f(3t))\frac{dt}{t}$$
According to Frullani's theorem (https://en.wikipedia.org/wiki/Frullani_integral)
$$I=\big(f(0)-f(\infty)\big)\ln\frac{1}{3}=-\ln3\Big(\frac{1}{2}-0\Big)=-\frac{\ln3}{2}$$

REPLY [3 votes]: Expand $1/(1+x^3)=\sum_{k=0}^\infty(-1)^k x^{3k}$ and use (an instance of Frullani integrals) $$\int_0^1\frac{x^{a-1}-x^{b-1}}{\ln x}\,dx\underset{x=e^{-t}}{=}\int_0^\infty\frac{e^{-bt}-e^{-at}}{t}\,dt=\ln\frac{a}{b},\qquad(a,b>0)$$ which gives, after $\sum_{k=0}^\infty(-1)^k a_k=a_0+\sum_{n=1}^\infty(a_{2n}-a_{2n-1})$, $$\int_0^1\frac{1-x}{1+x^3}\frac{dx}{\ln x}=\sum_{k=0}^\infty(-1)^k\ln\frac{3k+1}{3k+2}=\ln\frac12+\sum_{n=1}^\infty\ln\frac{6n+1}{6n+2}\frac{6n-1}{6n-2}\\=-\ln 2+\sum_{n=1}^\infty\ln\frac{1-(6n)^{-2}}{1-(3n)^{-2}}=-\ln 2+\ln\frac{(6/\pi)\sin(\pi/6)}{(3/\pi)\sin(\pi/3)}=-\frac{\ln 3}{2}$$ using the infinite product for the sine function.