3

$$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$

$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$

Using binomial series here

$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\over8}+{x^3\over 16}+\cdots=\sum_{k=0}^{\infty}{x^k\over2^{k+1}}$$

$$I=\sum_{k=0}^{\infty}{1\over 2^{k+1}}\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x$$

$$\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x=-{1\over (1+k)^2}$$

$$I=-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}$$

So this is where I got to, unable to determine this infinite sum. Do anyone know how to prove that,

$$-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}={\ln^2(2)-\zeta(2) \over 2}$$

Ѕᴀᴀᴅ
  • 34,263
gymbvghjkgkjkhgfkl
  • 8,090
  • My boss have a lot of these strange integral and I try solve them my self. Where he got them I never asked. He is owner of choice tutor.co.uk maths specialist. – gymbvghjkgkjkhgfkl Jun 01 '16 at 18:07

3 Answers3

5

$$ \int_{0}^{1}\frac{\log x}{2-x}\,dx = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}\left(\frac{x}{2}\right)^n\log(x)\,dx =-\sum_{n\geq 1}\frac{1}{2^n n^2}=-\text{Li}_2\left(\frac{1}{2}\right)$$ that can be computed through the dilogarithm reflection formula: $$ \text{Li}_2(x)+\text{Li}_2(1-x) = \zeta(2)-\log(x)\log(1-x) $$ by substituting $x=\frac{1}{2}$.

Community
  • 1
Jack D'Aurizio
  • 353,855
  • @user1952009: oh, yes, that gives $-\text{Li}_2\left(\frac{1}{2}\right)$ in few steps. – Jack D'Aurizio May 31 '16 at 19:35
  • wolfram tells me $\int_0^1 \ln(1-x) x^n = \frac{H_{n+1}}{n+1}$ hence $\int_0^1 \frac{\ln x}{2-x} dx = \int_0^1 \frac{\ln (1-x)}{1+x} dx = \sum_{n=0}^\infty (-1)^{n}\int_0^1 \ln (1-x)x^n dx =\sum_{n=1}^\infty (-1)^{n+1} \frac{H_{n}}{n} $ but I can't finish with elementary arguments.. – reuns May 31 '16 at 19:54
  • @user1952009: well, that is an interesting path: try to compute the Taylor series of $\log^2(1-x)$, by exploiting the fact that the derivative is $\frac{1}{1-x}$ times $\ldots$ – Jack D'Aurizio May 31 '16 at 20:00
  • too late I don't find the trick for proving wolfram's formula.. but your trick works wonderfully – reuns May 31 '16 at 21:18
  • next step : generalize it to $\sum_{n=1}^\infty \frac{n^{-s}}{2^n}$ for any $s$ and prove the Riemann hypothesis.. :) – reuns May 31 '16 at 21:24
  • 1
    @user1952009: if that works, you owe me half a million dollars :D – Jack D'Aurizio May 31 '16 at 21:33
  • Why the downvote? – Jack D'Aurizio May 31 '16 at 21:57
2

start with $$\int_0^1 \ln(1-x) x^{n-1}dx = -\frac{H_{n}}{n}$$ change of variable and expand in power series $$\int_0^1 \frac{\ln(x)}{2-x}dx = \int_0^1 \frac{\ln(1-x)}{1+x} = \sum_{n=0}^\infty (-1)^n \int_0^1\ln(1-x) x^n dx = \sum_{n=1}^\infty (-1)^n \frac{H_{n}}{n}$$ and use a trick found by Jack D'Aurizio $$\frac{\ln^2(1-x)}{2} = \int_0^x \frac{-\ln(1-t) }{1-t} dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \int_0^x\frac{t^{n+k}}{k}dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \frac{x^{n+k+1}}{k(n+k+1)}$$ $$ = \sum_{m= 2}^\infty \frac{x^{m}}{m} \sum_{k=1}^{m-1} \frac{1}{k} = \sum_{m =2}^\infty \frac{x^{m}}{m} H_{m-1}$$
hence $$\frac{\ln^2(2)}{2} =\frac{\ln^2(1-(-1))}{2} = \sum_{m =2}^\infty \frac{(-1)^{m}}{m} H_{m-1} = -\sum_{m=1}^\infty \frac{(-1)^m}{m^2} +\sum_{m=1}^\infty (-1)^{m}\frac{H_m}{m} $$ $$= \eta(2) +\int_0^1 \frac{\ln(x)}{2-x}dx$$

i.e. with $\eta(2) = (1-2^{1-2})\zeta(2)$ $$\int_0^1 \frac{\ln(x)}{2-x}dx = \frac{\ln^2(2)}{2} - \frac{\zeta(2)}{2}$$

proof of the first formula :

$\displaystyle \frac{1-x^n}{1-x} = \sum_{m=0}^{n-1} x^m\quad$ hence $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \sum_{m=0}^{n-1} \frac{1}{m+1} = H_n\quad$ and integrating by parts $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \lim_{a \to 1^-} (x^n-1)\ln(1-x)\mid_0^a $ $- \int_0^1 n x^{n-1} \ln(1-x)dx = -n\int_0^a x^{n-1} \ln(1-x)dx$

reuns
  • 77,999
  • About the first line: compute $I(\alpha)=\int_{0}^{1}x^n(1-x)^\alpha,dx$, then consider $I'(\alpha)$. – Jack D'Aurizio May 31 '16 at 21:49
  • @JackD'Aurizio : I found a proof from $H_n = \int_0^1 \frac{1-x^n}{1-x}dx$ and integrating by parts – reuns May 31 '16 at 21:55
  • @JackD'Aurizio : the idea of showing $\ln(1-x)$ by differentiating $(1-x)^\alpha$ at $\alpha=0$ is indeed quite nice – reuns May 31 '16 at 21:57
  • (+1) @user1952009, I just came back from work didn't have time. I forgot to log off, so it seem i am still on the site. – gymbvghjkgkjkhgfkl Jun 01 '16 at 18:00
  • @Chinacat : (see the comments that was collaborative work with Jack). and my question was where does it come from ? an exercice (seems too hard, or there must be a much simpler way) ? or mathematica solved it for you ? – reuns Jun 01 '16 at 18:01
2

On the path of user1952009,

Perform the change of variable $y=1-x$,

$\displaystyle I=\int_0^1 \dfrac{\ln(1-x)}{1+x}dx$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\begin{align} I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1+x}\right)}{1+x}dx\\ &=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1+x)}{1+x}dx\\ &=\ln 2\Big[\ln(1+x)\Big]_0^1+\int_0^1 \dfrac{\ln x}{1+x}dx-\Big[\dfrac{1}{2}(\ln(1+x))^2\Big]_0^1\\ &=\dfrac{1}{2}(\ln 2)^2+\int_0^1 \dfrac{\ln x}{1+x}dx \end{align}$

$\begin{align}\int_0^1 \dfrac{\ln x}{1+x}dx&=\int_0^1 \ln x\left(\sum_{n=0}^{\infty} (-1)^n x^n\right)dx\\ &=\sum_{n=0}^{\infty} \left(\int_0^1 (-1)^n x^n\ln x dx\right)\\ &=-\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(1+n)^2}\\ &=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}\\ &=\sum_{n=1}^{\infty} \dfrac{1}{(2n)^2}-\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}\\ &=\dfrac{1}{4}\zeta(2)-\left(\zeta(2)-\dfrac{1}{4}\zeta(2)\right)\\ &=-\dfrac{1}{2}\zeta(2) \end{align}$

therefore,

$\boxed{I=\dfrac{1}{2}(\ln 2)^2-\dfrac{1}{2}\zeta(2)}$

FDP
  • 13,647