# Help with logarithmic definite integral: $\int_0^1\frac{1}{x}\ln{(x)}\ln^3{(1-x)}\, dx$

I'm look for a closed form evaluation of the following improper definite integral involving logarithms:

\begin{align} I:&=\int_{0}^{1}\frac{1}{x}\ln{(x)}\ln^3{(1-x)}\,\mathrm{d}x\\ &\approx 0.579307. \end{align}

My attempt:

My first idea was to transform the integral using the substitution $u=-\ln{(x)}$, but this didn't really result in anything recognizably easier than the original integral:

\begin{align} I&=\int_{0}^{1}\frac{1}{x}\ln{(x)}\ln^3{(1-x)}\,\mathrm{d}x\\ &=\int_{0}^{\infty}(-u)\ln^3{(1-e^{-u})}\,\mathrm{d}u\\ &=-\int_{0}^{\infty}u\ln^3{(1-e^{-u})}\,\mathrm{d}u. \end{align}

The second thing I tried was a substitution I've known to be successful with similar problems:

\begin{align} u&=-\ln{(1-x)},\\ -u&=\ln{(1-x)},\\ e^{-u}&=1-x,\\ x&=1-e^{-u},\\ \mathrm{d}x&=e^{-u}\,\mathrm{d}u. \end{align}

Applying the substitution, the integral becomes:

\begin{align} I&=\int_{0}^{1}\frac{1}{x}\ln{(x)}\ln^3{(1-x)}\,\mathrm{d}x\\ &=\int_{0}^{\infty}\frac{(-u)^3}{1-e^{-u}}\ln{(1-e^{-u})}\,e^{-u}\,\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{u^3\,e^{-u}}{1-e^{-u}}\ln{(1-e^{-u})}\,\mathrm{d}u\\ &=\int_{0}^{\infty}\frac{u^3}{1-e^{u}}\ln{(1-e^{-u})}\,\mathrm{d}u. \end{align}

This looks like it has a shot at being evaluated in terms of series expansions, but I'm stuck at this point.

Suggestions welcome.

Update:

Using the series expansions $\ln{(1-e^{-x})}=-\sum_{n=1}^{\infty}\frac{1}{n}e^{-nx}$ and $\sum_{k=1}^{\infty}e^{-kx}=\frac{1}{e^{x}-1}$, the integral becomes:

\begin{align} I&=\int_{0}^{\infty}\frac{x^3}{1-e^{x}}\ln{(1-e^{-x})}\,\mathrm{d}x\\ &=\int_{0}^{\infty}\frac{x^3}{e^{x}-1}\sum_{n=1}^{\infty}\frac{1}{n}e^{-nx}\,\mathrm{d}x\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}\frac{x^3\,e^{-nx}}{e^{x}-1}\,\mathrm{d}x\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}x^3\,e^{-nx}\sum_{k=1}^{\infty}e^{-kx}\,\mathrm{d}x\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{\infty}\int_{0}^{\infty}x^3\,e^{-(k+n)x}\,\mathrm{d}x\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{\infty}\frac{6}{(k+n)^4}\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\psi^{(3)}(n+1), \end{align}

where $\psi^{(n)}(x)$ is the $n$-th derivative of the digamma function. Unfortunately, I don't know how to evaluate this last series. But this feels like progress.

• Check this technique. Jun 30, 2014 at 5:46

We could use the series expansion $$\frac{\ln(1-x)}{1-x}=-\sum_{n=1}^\infty H_nx^n,$$ where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the harmonic number. Then we get $$I=\int_0^1\frac{\ln(1-x)\ln^3x}{1-x}dx=-\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3 x\,dx=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}.$$ The last series could be evaluated using Euler's formula $$\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ (see formula (20) here: http://mathworld.wolfram.com/HarmonicNumber.html): $$I=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}=6\sum_{n=1}^\infty \left(\frac{H_{n+1}}{(n+1)^4}-\frac{1}{(n+1)^5}\right)=6(2\zeta(5)-\zeta(2)\zeta(3)).$$

• (+1) I really like this clever use of series. Also, the named series involving harmonic numbers is new to me and looks incredibly useful, so thank you for introducing me to them. Jun 28, 2014 at 1:39

Introduce a two-parameter deformation $$\mathcal{I}(a,b)=\int_0^1 x^{a-1}\left[(1-x)^b-1\right] dx=\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}-\frac1a.$$ The integral we are looking for is obtained as \begin{align*} I&=\frac{\partial^3}{\partial b^3}\left[\frac{\partial\mathcal{I}(a,b)}{\partial a}\biggl|_{a=0}\right]_{b=0}=\\ &=\frac{\partial^3}{\partial b^3}\left[\frac{\gamma^2}{2}+\frac{\pi^2}{12}+\gamma\,\psi(1+b)+\frac{\psi^2(1+b)-\psi'(1+b)}{2}\right]_{b=0}=\\ &=12\zeta(5)-6\zeta(2)\zeta(3). \end{align*} At the last step, one needs to use that $\psi^{(n)}(1)=(-1)^{n+1}n!\,\zeta(n+1)$.

• This looks promising. Give me some time to go through the details for myself and I'll probably accept it. Jun 28, 2014 at 0:22
• I ended up accepting CuriousGuest's answer instead because a) it's easier and b) it seems to be more easily generalizable to a class of problems I'm working on. That said, I'd like to reiterate that I appreciate your response. Jun 28, 2014 at 1:35
• +1. That was the approach I had in mind. Given the integrand structure it was quite natural. Nice answer. Jun 28, 2014 at 1:57

If you're interested in evaluating that $\psi'''(n+1)$ series you obtained in your original evaluation, a fun way to go about it is to use contours. A method popularized by Flajolet and Salvy in their famous paper on the topic.

$$\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

Begin with the kernel $$f(z)=\pi \cot(\pi z)\psi'''(-z)$$

This has poles at the positive integers, n; negative integers, -n, and 0.

The series at the positive integers is:

$$\frac{6}{(z-n)^{5}}-\frac{2\pi^{2}}{(z-n)^{3}}-\frac{\psi'''(n+1)}{z-n}+\cdot\cdot\cdot$$

Thus, the sum of the residues is:

$$Res(f,n)=\frac{6}{4!}\frac{d^{4}}{dz^{4}}\left[\frac{1}{(z-n)^{5}}\cdot \frac{1}{n}\right]-\frac{2\pi^{2}}{2!}\frac{d^{2}}{dz^{2}}\left[\frac{1}{(z-n)^{3}}\cdot \frac{1}{n}\right]-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

$$=6\sum_{n=1}^{\infty}\frac{1}{n^{5}}-2\pi^{2}\sum_{n=1}^{\infty}\frac{1}{n^{3}}-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

$$=6\zeta(5)-2\pi^{2}\zeta(3)-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

the series at the negative integers has a simple pole.

$$Res(f, -n)=\sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}$$$The residue at z=0 can be found by the Laurent expansion and noting the coefficient of the 1/z term: $$Res(f,0)=24\zeta(5)$$ Now, put it altogether and set equal to 0 due to there being no poles in the contour: $$6\zeta(5)-2\pi^{2}\zeta(3)+24\zeta(5)+\sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}=0$$ solve for the series in question. Note, since$\frac{\psi'''(n)-\psi'''(n+1)}{n}=\frac{6}{n^{5}}$, this means that$\displaystyle \sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}$is equal to the series in question plus$6\zeta(5)$. In other words by calling our series S:$\displaystyle \sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}=S+6\zeta(5)\$

$$-2S+24\zeta(5)-2\pi^{2}\zeta(3)=0$$

$$S=12\zeta(5)-\pi^{2}\zeta(3)$$