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Evaluate $\displaystyle \int\limits_0^1 \dfrac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

I was wondering if the above had some kind of a closed form, here some of the special cases have been discussed but this one is really a fascinating one.

I guess there's no general taylor expansion for $\ln^m (1+x)$ and so transforming into a series wouldn't be that easy.

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  • $\begingroup$ Are $m,n \in \mathbb{N}$? $\endgroup$ – mrnovice Feb 25 '17 at 18:07
  • $\begingroup$ Is this $(\ln(x))^m$ or $\ln\circ\ln\circ\dots\circ\ln(x)$? $\endgroup$ – Fimpellizieri Feb 25 '17 at 18:07
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    $\begingroup$ The similar integral $$ \int_{0}^{1}\frac{\log^m(1-x)\log^n(x)}{x}\,dx $$ depends on derivatives of Euler's Beta function, through differentiation under the integral sign. $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 18:13
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    $\begingroup$ Your problem boils down to the previous one by exploiting $\log(1+x)=\log(1-x^2)-\log(1-x)$, the binomial theorem and a suitable change of variable. $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 18:14
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    $\begingroup$ if you don't like beta derivatives, you might find some inspiration here: math.stackexchange.com/questions/2118580/… $\endgroup$ – tired Feb 25 '17 at 18:29
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Stirling numbers of the first kind might be useful here, Consider

$$m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x)$$

$$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{1}{k!} \int^1_0 x^{k-1} \log^n(x)\,dx$$

Now it is easy to see that

$$\int^1_0 x^{k-1} \,dx = \frac{1}{k}$$

By differentiation $n$ times with respect to $k$

$$\int^1_0 x^{k-1} \log^n(x)\,dx = (-1)^n\frac{n!}{k^{n+1}}$$

Substituting back we have

$$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx =(m!)(n!) \sum_{k=m}^\infty (-1)^{k-m+n} \left[k\atop m\right] \frac{1}{k!\, k^{n+1}}$$

Now the Striling numbers could related to Euler sums through equations like

$$\frac{\left[k\atop 3\right]}{k!} =\frac{ (H_{k-1})^2-H^{(2)}_{k-1}}{2k}$$

and

$$\frac{\left[k\atop 4\right]}{k!} =\frac{ (H_{k-1})^3-3H^{(2)}_{k-1}H_{k-1}+2H^{(3)}_{k-1}}{6k}$$

I don't think there exist a simple formula but this procedure should work.


Case $m=2 , n=2$

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx =4 \sum_{k=2}^\infty (-1)^{k} \left[k\atop 2\right] \frac{1}{k!\, k^{3}}$$

Note that

$$\frac{\left[k\atop 2\right]}{k!} = \frac{H_{k-1}}{k}$$

Hence we deduce that

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx =4 \sum_{k=2}^\infty (-1)^{k} \frac{H_{k-1}}{\, k^{4}}$$

Note that

$$\begin{align} \sum_{k=2}^\infty (-1)^{k} \frac{H_{k-1}}{\, k^{4}} &=\sum_{k=2}^\infty (-1)^{k} \frac{H_{k}}{ k^{4}} -\sum_{k=2}^\infty (-1)^{k} \frac{1}{ k^{5}} \\ &=\sum_{k=1}^\infty (-1)^{k} \frac{H_{k}}{ k^{4}} -\sum_{k=1}^\infty (-1)^{k} \frac{1}{ k^{5}}\\ &= \frac{\zeta(2) \zeta(3)}{2} - \frac{ 29\zeta(5)}{32} \end{align}$$

We deduce that

$$\boxed{\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx = 2\zeta(2) \zeta(3)- \frac{ 29}{8}\zeta(5)}$$

This implies we can represent the special case $m=2$

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^n x}{x}\; dx =2 (-1)^n(n!) \left[ \sum_{k=1}^\infty (-1)^{k} \frac{H_k}{ k^{n+2}} + \left(1-2^{-n-2} \right) \zeta(n+3) \right]$$


General formula in terms of nonlinear Euler sums

Define $\{ m\}$ as the $l$ partitions of $m$ where $m = i_1r_1+\cdots i_l r_l$

$$ \frac{1}{(m+1)!} \log^{m+1}(1+x) =\sum_{\{m\}} \sum_{k=1}^\infty \prod^l_{j=1}\frac{(-1)^{i_j+1}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j} \frac{(-x)^k}{k} $$

Substitute back in the integral

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l'}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$

Reference: https://arxiv.org/pdf/math/0607514.pdf

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  • $\begingroup$ I see, (+1) for the efforts. But the last infinite sum is almost impossible to tackle for any general value . So I guess a closed form doesn't exists. $\endgroup$ – Aditya Narayan Sharma Feb 25 '17 at 19:42
  • $\begingroup$ @AdityaNarayanSharma, yeah this seems to generate an arbitrary terms of nonlinear generalized Euler sums. $\endgroup$ – Zaid Alyafeai Feb 25 '17 at 19:45
  • $\begingroup$ @AdityaNarayanSharma, I included some examples $\endgroup$ – Zaid Alyafeai Feb 26 '17 at 0:55
  • $\begingroup$ Great, the answer now looks substantial. You could add the closed form of the very last infinite series, alternating Euler sums $\endgroup$ – Aditya Narayan Sharma Feb 26 '17 at 6:08
  • $\begingroup$ @AdityaNarayanSharma, I guess it can be representeded as a finite sum of the alternating zeta. I remember seeing it somewhere. $\endgroup$ – Zaid Alyafeai Feb 26 '17 at 13:27
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This is not going to be the full answer however it will be sufficient to obtain closed form expressions for some particular values of parameters. Allow me to use my own notation so that I can avoid typos and mistakes. Agan we want to compute the following quantity: \begin{equation} {\mathcal I}^{(p,d)}:= \int\limits_0^1 \frac{[\log(\xi)]^p [\log(1+\xi)]^d}{\xi} d\xi \end{equation} for $p\ge 1$ and $d\ge 1$. We have: \begin{eqnarray} {\mathcal I}^{(p,d)}&=&-\frac{d}{p+1} \int\limits_0^1 \frac{[\log(\xi)]^{p+1} [\log(1+\xi)]^{d-1}}{1+\xi} d\xi\\ &=& -\frac{d}{p+1} \int\limits_1^2 \frac{[\log(\xi)]^{d-1} [\log(\xi-1)]^{p+1}}{\xi} d\xi\\ &=& -\frac{d}{p+1} \sum\limits_{q=0}^{p+1} \binom{p+1}{q} (-1)^{d-1+q} \underbrace{\int\limits_{\frac{1}{2}}^1 \frac{[\log(u)]^{d-1+q}[\log(1-u)]^{p+1-q}}{u} du }_{{\mathcal J}^{(d,p)}_q} \end{eqnarray} In the top line we integrate by parts, in the second line we substituted $1+\xi \rightarrow \xi$ and in the bottom line we wrote $\xi-1 = \xi(1-1/\xi)$ took the logs, squared and and then substituted $1/\xi \rightarrow u$. Now we fix $p$ and $d$ and we evaluate the integrals in the sum top-down, meaning for $q=p+1,p,p-1,p-2,\cdots,0$. It is clear that the complexity increases with decreasing value of $q$. We have: \begin{eqnarray} {\mathcal J}^{(d,p)}_{p+1} &=&- \frac{[\log(\frac{1}{2})]^{d+p+1}}{(d+p+1)}\\ {\mathcal J}^{(d,p)}_{p+0} &=& (-1)^{d+p}(d+p-1)! \cdot \left[ \zeta(1+d+p) - \sum\limits_{l=0}^{d-1+p} \frac{[\log(2)]^l}{l!} Li_{1+d+p-l}(\frac{1}{2}) \right]\\ {\mathcal J}^{(d,p)}_{p-1} &=&2 (-1)^2 \sum\limits_{m\ge 1} \frac{H_{m-1}}{m} \frac{\partial^{d-2+p}}{\partial m^{d-2+p}}\left(\frac{1-2^{-m}}{m}\right)\\ &=&2!(-1)^{d+p}(d+p-2)! \cdot \left[ \zeta(d+p,1) - \sum\limits_{l=0}^{d-2+p} \frac{[\log(2)]^l}{l!} \cdot \zeta_p(2;d+p-l,1)\right]\\ {\mathcal J}^{(d,p)}_{p-2} &=&3 (-1)^3 \sum\limits_{m\ge 1} \left(\frac{[H_{m-1}]^2-H_{m-1}^{(2)}}{m}\right) \frac{\partial^{d-3+p}}{\partial m^{d-3+p}}\left(\frac{1-2^{-m}}{m}\right)\\ &=& 3! (-1)^{d+p} (d-3+p)! \cdot \left[ \zeta(d-1+p,1,1) - \sum\limits_{l=0}^{d-3+p} \frac{[\log(2)]^l}{l!} \zeta_+(2;d+p-1-l,1,1) \right] \end{eqnarray} Here $\zeta_+(t;p_1,p_2,p_3) := \sum\limits_{m_1 > m_2 > m_3 >0} t^{-m_1} \prod\limits_{\xi=1}^3 1/(m_\xi)^{p_\xi}$.

The first identity above is obvious. In the second identity we used integration by parts and in the subsequent lines we just expanded the second log in a series and integrated term by term. It is not hard to see the pattern that emerges. Note that the $\zeta(\cdots)$ quantities are in principle all reducable to single zeta quantities. In order to complete this solution we need to express that $\zeta_+(2;\cdots)$ quantities through single $\zeta$ quantities (whenever it is possible). We will do this in a systematic way by using the integral representations of the later. We will complete this asap.

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Given the integral, $$I(n,p) = \int\limits_0^1 \dfrac{ \ln^{n-1}(x)\ln^p (1+x)}{x}\; dx$$

which is the notation consistent with Nielsen polylogs. Then closed-forms in terms of ordinary polylogarithms are known only for the following cases,

$$I(1,p) \\ I(n,1) \\ I(2k+1,2) \\ I(2,2) \\ I(2,3)$$

and no more. See this more general post on Nielsen polylogs.

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