How to evalute $\int_{0}^{1}\frac{x\log x}{\log(1-x)}dx$

Solution 1:

Partial solution

\begin{align} I&=\int_0^1\frac{1-x}{\ln x}\ln(1-x)\ dx=\int_0^1\left(-\int_0^1x^y\ dy\right)\ln(1-x)\ dx\\ &=\int_0^1\left(-\int_0^1x^y\ln(1-x)\ dx\right)\ dy=\int_0^1\left(\sum_{n=1}^\infty\frac1n\int_0^1x^{n+y}\ dx\right)\ dy\\ &=\int_0^1\left(\sum_{n=1}^\infty\frac{1}{n(n+y+1)}\right)\ dy=\sum_{n=1}^\infty\frac1n\int_0^1\frac{dy}{n+y+1}\\ &=\sum_{n=1}^\infty\frac{\ln(n+2)-\ln(n+1)}{n} \end{align}

Solution 2:

Not a closed form, but quite a few curious series.

$$I=\int_{0}^{1}\frac{(1-x)\log(1-x)}{\log x}dx=-\int_0^1 \int_0^1\frac{x(1-x)dx dt}{\log x (1- x t)}$$

Now make a substitution $x=e^{-u}$:

$$I=\int_0^1 \int_0^\infty \frac{e^{-2u}(1-e^{-u})du dt}{u (1- t e^{-u} )}$$

Now let's expand the bracket in the numerator as a series:

$$I=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k!} \int_0^1 \int_0^\infty \frac{e^{-2u}u^{k-1} du dt}{ 1- t e^{-u} }$$

$$\int_0^\infty \frac{e^{-2u}u^{k-1} du}{ 1- t e^{-u} }=(k-1)! \Phi(t,k,2) $$

Where $\Phi$ is so called Lerch transcendent.

Which gives us:

$$I=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \int_0^1 \Phi(t,k,2) dt$$

The following is tricky and I don't have proof for this so far (I derived it using Mathematica):

$$\int_0^1 \Phi(t,k,2) dt=k-\sum_{l=2}^k \zeta(l)$$

Which gives us:

$$I=\sum_{k=1}^\infty (-1)^{k+1} \left(1- \frac{1}{k} \sum_{l=2}^k \zeta(l) \right) \tag{1}$$

Where for $k=1$ we take the sum inside to be $0$, as is the usual convention when the upper limit is smaller than the lower limit.

It converges quite well (not as fast as some other series). For example we have:

$$\sum_{k=1}^{60} (-1)^{k+1} \left(1- \frac{1}{k} \sum_{l=2}^k \zeta(l) \right)=0.86062019285313836404 \ldots$$

Where all the digits are correct.

Separating even and odd terms we have:

$$I=\sum_{n=0}^\infty \left(\frac{1}{2n+2} \sum_{l=2}^{2n+2} \zeta(l)-\frac{1}{2n+1} \sum_{l=2}^{2n+1} \zeta(l) \right)$$

$$I=\sum_{n=0}^\infty \left(\frac{\zeta(2n+2)}{2n+2}- \sum_{l=2}^{2n+1} \left( \frac{\zeta(l)}{2n+1}-\frac{\zeta(l)}{2n+2} \right) \right)$$

$$I=\frac{1}{2} \sum_{n=0}^\infty \frac{1}{n+1} \left(\zeta(2n+2)- \frac{1}{2n+1} \sum_{l=2}^{2n+1} \zeta(l) \right) \tag{2}$$

Where (2) now converges twice as fast as (1).

Let's rewrite the sum inside in a way that makes it converge for $n \to \infty$:

$$-\frac{1}{2n+1} \sum_{l=2}^{2n+1} \zeta(l)=-\frac{2n}{2n+1} -\frac{1}{2n+1} \sum_{l=2}^{2n+1} (\zeta(l)-1)$$

Let's use the definition of the zeta function:

$$\sum_{l=2}^{2n+1} (\zeta(l)-1)=\sum_{q=2}^\infty \sum_{l=2}^{2n+1} \frac{1}{q^l}$$

But this is geometric sum:

$$\sum_{l=2}^{2n+1} \frac{1}{q^l}= \frac{1}{q^2} \frac{1-\frac{1}{q^{2n}}}{1-\frac{1}{q}}=\frac{1}{q^{2n+1}} \frac{q^{2n}-1}{q-1}=\frac{1}{q(q-1)}-\frac{1}{q^{2n+1}(q-1)}$$

We have:

$$\sum_{q=2}^\infty \frac{1}{q(q-1)}=1$$

Which means that:

$$\sum_{l=2}^{2n+1} (\zeta(l)-1)=1-\sum_{q=2}^\infty \frac{1}{q^{2n+1}(q-1)}$$

Going back to our series:

$$I=\frac{1}{2} \sum_{n=0}^\infty \frac{1}{n+1} \left(\zeta(2n+2)-\frac{2n}{2n+1}- \frac{1}{2n+1}+ \frac{1}{2n+1} \sum_{q=2}^\infty \frac{1}{q^{2n+1}(q-1})\right) $$

$$I=\frac{1}{2} \sum_{n=0}^\infty \frac{1}{n+1} \left(\zeta(2n+2)-1+ \frac{1}{2n+1} \sum_{q=2}^\infty \frac{1}{q^{2n+1}(q-1)}\right) $$

We can sum the last part w.r.t. $n$:

$$\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1) q^{2n+1}}= 2 \operatorname{arctanh} \frac{1}{q} +q \log \frac{q^2-1}{q^2}$$

We can now separate the series into three parts, each of them converges on its own:

$$I_1=\frac{1}{2} \sum_{n=0}^\infty \frac{\zeta(2n+2)-1}{n+1}= \frac{\log 2}{2}$$

$$I_2= \sum_{q=2}^\infty \frac{1}{q-1} \operatorname{arctanh} \frac{1}{q} $$

$$I_3= \sum_{q=2}^\infty \frac{q}{2(q-1)}\log \left(1- \frac{1}{q^2} \right)$$

$$I=\frac{\log 2}{2}+I_2+I_3$$

I apologize, there was a typo in the final expressions, I accidentally wrote $\arctan$ instead of $\operatorname{arctanh}$. Now it's fixed and the resulting expressions give the correct value.

Solution 3:

I am not sure that I could find a closed formula for the result.

Beside numerical integration, I should use the classical series expansion of $\log(1-x)$ and use the long division to get $$\frac 1 {\log(1-x)}=-\frac{1}{x}+\frac{1}{2}+\frac{x}{12}+\frac{x^2}{24}+\frac{19 x^3}{720}+\frac{3 x^4}{160}+\frac{863 x^5}{60480}+\frac{275 x^6}{24192}+O\left(x^7\right)$$ making $$\frac {x \log(x)} {\log(1-x)}=\log(x) \left(-1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{24}+\frac{19 x^4}{720}+\frac{3 x^5}{160}+\frac{863 x^6}{60480}+\frac{275 x^7}{24192}+O\left(x^8\right)\right)$$ and now we face the problem of $$I_n=\int x^n \log(x) \,dx=\frac{x^{n+1} ((n+1) \log (x)-1)}{(n+1)^2}$$ that is to say $$J_n=\int_0^1 x^n \log(x) \,dx=-\frac{1}{(n+1)^2}$$ Using this truncated series, we should end with $\frac{2721985571}{3161088000}\approx 0.861091$ while the numerical integration would give $0.860620$.

Solution 4:

Different approach

let $I$ denotes our integral $\int_0^1\frac{(1-x)\ln(1-x)}{\ln x}\ dx$ and let $I_n=\int_0^1\frac{(1-x^n)\ln(1-x)}{\ln x}\ dx,\quad I_0=0$ and $I_1=I$ $$I^{\large'}_n=-\int_0^1x^{n}\ln(1-x)\ dx=\sum_{k=1}^\infty \frac{1}{k}\int_0^1 x^{n+k} \ dx=\sum_{k=1}^\infty\frac{1}{k(n+k+1)}$$ Then $$I=I_1=\int_0^1I^{\large'}_n\ dn=\sum_{k=1}^\infty\frac1k\int_0^1\frac{dn}{n+k+1}=\sum_{k=1}^\infty\frac{{\ln(k+2)-\ln(k+1)}}{k}$$