## $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, \mathrm{d}x$

Calculus (Integrals, Series)
Grigorios Kostakos
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### $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, \mathrm{d}x$

Evaluate
$\displaystyle\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, {\rm d}x\,.$
Grigorios Kostakos

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mathofusva
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### Re: $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, {\rm d}x$

Let the integral be $I$. We show that $I = -\frac{1}{2}\zeta(4) = - \frac{\pi^4}{180}$.

To see this, recall the generating function for harmonic numbers:
$$\sum_{k=1}^\infty\,H_kx^k = -\,\frac{1}{1-x}\ln(1-x).$$
Integrating from $0$ to $x$ yields
$$\sum_{k=1}^\infty\,\frac{H_k}{k+1}\,x^{k+1} = \frac{1}{2}\,\ln^2(1-x).$$
Thus,
$$I = 2\int_0^1\,\sum_{k=1}^\infty\,\frac{H_k}{k+1}\,x^{k}\ln x\,dx$$
$$= 2\sum_{k=1}^\infty\,\frac{H_k}{k+1}\,\int_0^1\,x^{k}\ln x\,dx = - 2\sum_{k=1}^\infty\,\frac{H_k}{(k+1)^3}.$$
Here the exchange the summation and integration is justified by the monotone convergence theorem. Appealing to the well-known Euler sum:
$$\sum_{k=1}^\infty\,\frac{H_k}{(k+1)^3} = \frac{1}{4}\,\zeta(4)$$
(for example, see http://mathworld.wolfram.com/EulerSum.html), we find that $I = -\frac{1}{2}\zeta(4) = - \frac{\pi^4}{180}$ as claimed.

Remark. This problem is related to the Monthly problem 11921.
Tolaso J Kos
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### Re: $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, {\rm d}x$

Here is a generalization.

Let $\mathbb{N} \ni k \geq 2$. Then:

\begin{align*}
\int_{0}^{1}\frac{\log x \log^k (1-x)}{x} \, {\rm d}x &= \int_{0}^{1} \frac{\log(1-x) \log^k x}{1-x} \, {\rm d}x \\
&= -\int_{0}^{1} \log^k x \sum_{n=1}^{\infty} \mathcal{H}_n x^n \, {\rm d}x\\ &= - \sum_{n=1}^{\infty} \mathcal{H}_n \int_{0}^{1} x^n \log^k x \, {\rm d}x\\
&= - (-1)^k k! \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{(n+1)^{k+1}}\\
&= (-1)^{k-1} k! \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n+1}- \frac{1}{n+1}}{(n+1)^{k+1}}\\
&= (-1)^{k-1} k! \left [ \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n+1}}{(n+1)^{k+1}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{k+2}} \right ]\\ &=(-1)^{k-1} k! \left [ \sum_{n=2}^{\infty} \frac{\mathcal{H}_n}{n^{k+1}} - \zeta(k+2) +1 \right ] \\
&=(-1)^{k-1} k! \left [ \sum_{k=1}^{\infty} \frac{\mathcal{H}_n}{n^{k+1}} - \zeta(k+2) \right ] \\
&= (-1)^{k-1} k! \left [ \left ( 1+ \frac{k+1}{2} \right ) \zeta(k+2) - \frac{1}{2} \sum_{j=1}^{k-1} \zeta(j+1) \zeta(k+1-j) - \zeta(k+2)\right ] \end{align*}

since $\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n^q} = \left ( 1+ \frac{q}{2} \right ) \zeta(q+1) - \frac{1}{2}\sum_{j=1}^{q-2} \zeta(j+1) \zeta(q-j)$. Now for $k=2$ we have that:

$$\int_{0}^{1} \frac{\log x \log^2 (1-x)}{x} \, {\rm d}x = - \frac{\pi^4}{180}$$

Playing around with $k$ we can produce some interesting results:

$$\begin{array}{||c|c|c|c|c||} \hline \text{Integral}&k &\text{Value} \\ \hline \int_0^1 \frac{\log x \log^3 (1-x)}{x} \, {\rm d}x&3& 12 \zeta(5) - \pi^2 \zeta(3) \\\\ \int_0^1 \frac{\log x \log^4 (1-x)}{x} \, {\rm d}x&4& 12 \zeta^2(3) - \frac{2\pi^6}{105} \\\\ \int_0^1 \frac{\log x \log^5 (1-x)}{x}\, {\rm d}x&5 & 300 \zeta(7) - \frac{4 \pi^4 \zeta(3)}{3} - 20 \pi^2 \zeta(5) \\ \hline \end{array}$$

@mathofusva: What does problem 11921 of Monthly actually state? Does it state that:

$$\log^2 2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{(n+1)2^{n+1}} + \log 2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{(n+1)^2 2^n} + \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{(n+1)^3 2^n} = \frac{\zeta(4) + \log^4 2}{4}$$
Imagination is much more important than knowledge.
Tolaso J Kos
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### Re: $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, {\rm d}x$

Now let us continue with an extension of the problem. Well, let us consider the double parametric family of integrals, that is:

$$\mathcal{J}(m, k)= \int_0^1 \frac{\log^m (1+x) \log x \log^k (1-x)}{x} \, {\rm d}x$$

Well , here is a list of results that I leave them as an exercise.

$$\begin{array}{||c|c|c|c|c||} \hline \text{Integral}&\mathcal{J}(m, k)& m& k &\text{Value} \\ \hline \int_{0}^{1} \frac{\log x \log(1-x)}{x} \, {\rm d}x&\mathcal{J}(0, 1)&0 &1& \zeta(3)\\\\ \int_0^1 \frac{\log x \log^2 (1-x)}{x} \, {\rm d}x&\mathcal{J}(0, 2)&0 &2 & - \frac{\pi^4}{180} \\\\ \int_0^1 \frac{\log (1+x) \log x \log(1-x)}{x}\, {\rm d}x&\mathcal{J}(1, 1) &1 &1 & -\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} + 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \\\\ \int_{0}^{1} \frac{\log x \log(1-x) \log^2 (1+x)}{x} \, {\rm d}x&\mathcal{J}(2,1)&2 &1& \frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5) \\\\ \hline \end{array}$$

It is quite interesting for someone to prove that

$$\int_{0}^{1} \frac{\log x \log(1-x) \log^2 (1+x)}{x} \, {\rm d}x= \frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$

P.S 1: We have seen $\displaystyle \int_0^1 \frac{\log x \log (1-x)}{x} \, {\rm d}x$ here.

P.S 2: A similar problem by Ovidiu Furdui is the following:

$$\int_{0}^{1}\ln^k (1-x)\ln x\, {\rm d}x =(-1)^{k+1}k! \left ( k+1 - \zeta(2)-\zeta(3)- \cdots -\zeta(k+1) \right )$$

where $\mathbb{Z} \ni k \geq 1$ and $\zeta$ is the Riemann zeta function.
Imagination is much more important than knowledge.
mathofusva
Posts: 33
Joined: Tue May 10, 2016 3:56 pm

### Re: $\int_{0}^{1} \frac{\log{x}\,\log^2(1-x)}{x}\, {\rm d}x$

Another approach for extension:

Recall the Euler integral
$$\int_0^1\,(1-t)^pt^{q-1}\,dt = \frac{\Gamma(1+p)\Gamma(1+q)}{q\Gamma(1+p+q)}.$$
Differentiating with respect to $p$ and $q$ yields
$$\int_0^1\,\ln^n(1-t)\ln^mt\frac{dt}{t} = D^n_pD^m_q\left[\frac{\Gamma(1+p)\Gamma(1+q)}{q\Gamma(1+p+q)}\right]|_{p=q=0}.$$
Let
$$y = \ln\Gamma(1 +p) +\ln\Gamma(1+q) - \ln\Gamma(1 + + q).$$
In view of
$$\psi^{(n)}(z) = (-1)^{n+1}n!\zeta(n+1, z),$$
we find that
$$D^n_pD^m_q y = (-1)^{n+m-1}(n+m-1)!\zeta(n+m,1+p+q).$$
Hence
$$D^n_pD^m_q y|_{p=q=0} = (-1)^{n+m-1}(n+m-1)!\zeta(n+m),$$
and
$$y = \sum_{n=1}^\infty\sum_{m=1}^\infty\,\frac{D^n_pD^m_q y|_{p=q=0}}{n!m!}p^nq^m.$$
Thus
$$\int_0^1\,\ln^n(1-t)\ln^mt\frac{dt}{t} = D^n_pD^m_q\left[\frac{1}{q}\exp\left(-\sum_{k=1}^\infty\sum_{l=1}^\infty\,A_{k,l}p^kq^l\right)\right]|_{p=q=0},$$
where
$$A_{k,l} = (-1)^{k+l}\frac{(k+l-1)!}{k!\,l!}\zeta(k+l).$$

In particular, if $n = 2$, we obtain
$$\int_0^1\,\ln^mt\,\ln^2(1-t)\frac{dt}{t} = (-1)^m\,m!\left[(m+2)\zeta(m+3) - \sum_{i=1}^m\,\zeta(1+i)\zeta(m+2-i)\right].$$

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