definite integral $\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\frac{-1}{2}\ln 3$

Question

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 derived 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.

Answer 1

$$I=\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx$$ 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) $$I=\big(f(0)-f(\infty)\big)\ln\frac{1}{3}=-\ln3\Big(\frac{1}{2}-0\Big)=-\frac{\ln3}{2}$$

Answer 2

First I'll use your expression:-

So what we need is $$\frac{1}{2}\int_{0}^{\infty}\frac {1-x}{(1+x^3)\ln x}dx$$

Let $$I(k)=\int_{0}^{\infty}\frac {1-x^{k}}{(1+x^3)\ln x}dx$$

Using Differentiation under the integral sign

So $$I'(k)=\int_{0}^{\infty}\frac {-x^{k}\ln x}{(1+x^3)\ln x}dx=\int_{0}^{\infty}\frac {-x^{k}}{(1+x^3)}dx$$

Now substitute $x=t^{\frac{1}{3}}$.

We get $$\displaystyle I'(k)=\frac{-1}{3}\int_{0}^{\infty}\frac{t^{\frac{k-2}{3}}}{1+t}dt=\frac{-1}{3}\operatorname{B}\left(\frac{k+1}{3},1-\frac{k+1}{3}\right)=\frac{-\pi}{3}\csc\frac{\pi(k+1)}{3}.$$

(I have used Euler's Reflection Formula here)

Now integrating $I'(k)$ from $0$ to $1$ wrt $k$ we get our answer.

$$\int_{0}^{1}\frac{-\pi}{3}\csc\frac{\pi(k+1)}{3}\,dk=-\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\csc t\,dt=-\ln 3.$$

Use antiderivative of $\int \csc x\,dx=-\ln|\csc x+\cot x|+C$.

Now after integration the LHS has become $I(1)-I(0)$, but $I(0)=0$. Hence the our required value is $I(1)=-\ln 3$.

Now look back at the first line. I took $I(k)$ as twice the desired result. So just dividing $I(1)$ by $2$ we get our final answer.

So we get our final answer as $\displaystyle\frac{I(1)}{2}=\frac{-1}{2}\ln 3$

Edit:- For some reason \Beta and \cosec and \cot are not rendering. So I was forced to use Gamma function directly . But I should have used beta function first and then made it into Gamma because the form of the improper integral was of Beta function. Of course $\operatorname{B}(x,1-x)=\Gamma(x,1-x)$

Answer 3

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.