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(This is related to this question.)

Define the integral,

$$I_n = \int_0^1\frac{\rm{Li}_n(x)}{1+x}dx$$

with polylogarithm $\rm{Li}_n(x)$. Given the Nielsen generalized polylogarithm $S_{n,p}(z)$,

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

Then it seems,

$$I_1 = -S_{1,1}(-1)-\tfrac12\ln(2)\ln(2)$$

$$I_2 = -5S_{1,2}(-1)+\ln(2)\,\zeta(2)\quad$$

$$\quad\qquad I_3 = -2S_{1,3}(-1)+\ln(2)\,\zeta(3)-\tfrac12\zeta(4)$$

where $S_{1,1}(-1) = -\tfrac12\zeta(2)$, and $S_{1,2}(-1) = \tfrac18\zeta(3)$ and $S_{1,3}(-1)$ has a more complicated closed-form given in the linked post.

Q: What is $I_4$ and $I_5$? In general, can $I_n$ be expressed by the Nielsen generalized polylogarithm?

P.S. Note that $\rm{Li}_n(z), \ln(z), \zeta(z)$ are just special cases of this function.

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  • $\begingroup$ $$I_4=\frac{\pi^4}{90}\ln 2-\frac{\pi^2}{12}\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x}dx$$ Is the last integral known by any chance? $\endgroup$ – LeBlanc May 20 at 19:13
  • $\begingroup$ @Threesidedcoin: Not that I'm aware of. Do you have $I_4$ up to 30 digits? $\endgroup$ – Tito Piezas III May 20 at 19:28
  • $\begingroup$ I barely have 5 digits, but if I can find $\int_0^1 \frac{\operatorname{Li}_2^2(-x)}{x}dx$ then $I_4$ is solved. $\endgroup$ – LeBlanc May 20 at 19:36
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    $\begingroup$ This might help:$$I_4=\int_0^1 \frac{\text{Li}_4(x)}{1+x}dx=\sum_{k\ge 0}(-1)^k\int_0^1x^k \text{Li}_4(x)dx$$ We have integrating by parts: $$I(t)=\int_0^1 x^k\text{Li}_t(x)dt=\frac{\text{Li}_t(1)}{k+1}-\frac{1}{k+1}\int_0^1 x^{k}\text{Li}_{t-1}(x)d=\frac{\zeta(t)-I(t-1)}{k+1}$$ So this will give: $$I_4=\eta(1)\zeta(4)-\eta(2)\zeta(3)+\eta(3)\zeta(2)+\sum_{k\ge 0}\frac{(-1)^k}{(k+1)^3}\int_0^1 x^k\ln(1-x)dx$$ $$=\eta(1)\zeta(4)-\eta(2)\zeta(3)+\eta(3)\zeta(2)-\int_0^1\ln(1-x)\frac{\text{Li}_3(-x)}{x}dx$$ The last integral is similar to this one math.stackexchange.com/a/463200 $\endgroup$ – LeBlanc May 20 at 20:07
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    $\begingroup$ @Zacky we can expand the dilogarithm then apply IBP $$\int_0^1\frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$ $\endgroup$ – Ali Shather Jul 3 at 2:53
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Per @TitoPiezas' request, I will expand on my comment above and essentially close this question as far as the literature we have (unless there are new or unknown developments I am unaware of).

As @Zacky said, a key step in converting this problem to a more conventional one is repeated integration by parts. Using the techniques outlined in the comments above, we get $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=0}^\infty \tfrac{(-1)^{k+n}}{(k+1)^{n-1}}\int_0^1x^k\text{Li}_1(x) dx$$ We now shift the right summation index and use $\text{Li}_1(x) = -\log(1-x)$ to get $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=1}^\infty \tfrac{(-1)^{k+n}}{k^{n-1}}\int_0^1x^{k-1}\log(1-x) dx$$ and recall that$$\int_0^1x^{k-1}\log(1-x) dx= \frac{H_k}{k}$$ So we finally conclude $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+(-1)^nA(n)$$ where $A(n) = \sum_{k=1}^\infty \tfrac{(-1)^k H_k}{k^n}$ is the $n$th "Alternating Euler Sum".

Alternating Euler Sums are well studied; examples just from this website can be found at this basic survey of integral forms here and the most amazing collection of answers ever here.

That latter link, when converted over to the more simple notation outlined here, gives a closed form for $A(2n)$ and shows why $A(2n+1)$ is particularly tricky; using the results on that page we should be able to easily get $A(1)$ and $A(3)$ but once we move higher we get linear dependence chains in the recurrence formulas that seemingly prevent us from expressing our answers in terms of Polylogarithms $\text{Li}_s(z)$ (which includes the Zeta and Dirichlet Eta functions as special values at $z=\pm 1$ respectively) and elementary functions in general (though specific cases, in particular for low $n$, may still work).


Note: I cannot stress enough how amazing that second link is. It essentially settles this question on its own through the amazing answers! Please take the time to go upvote those answers if possible.

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    $\begingroup$ +1 This is great. By a happy coincidence, I know that $$\sum_{k=1}^{\infty}\frac{H_k}{k^a}z^k= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$ with Nielsen polylog $S_{n,p}(z)$. So $I_n$ via the $A(n)$ can indeed be expressed by Nielsen polylogs as I suspected. However, I thought closed-forms of $A(n)$ in terms of ordinary polylogs are unknown for $n>3$, so I need to check that link for $A(5)$ and higher. $\endgroup$ – Tito Piezas III Jun 8 at 4:57
  • $\begingroup$ @TitoPiezasIII You are right; I misremembered the post. We have cleaner recursions for $n= 5, 7$ but they remain at intractable integrals. Unless I find evaluations somewhere I will remove those from my post. With your comment completing the link to Nielsen polylogs, this question seems about as closed as it can be until someone finds a way to evaluate $A(2n+1)$ (if it is even possible) $\endgroup$ – Brevan Ellefsen Jun 8 at 5:10
  • $\begingroup$ No problem. At least this question has closure, and I'm happy that my "guess" for $I_4$ is correct. But $I_5$ already involves the Nielsen polylog $S_{4,2}(-1)$ which has no known representation as ordinary polylogs. $\endgroup$ – Tito Piezas III Jun 8 at 5:53
  • $\begingroup$ Ah, I see. The formula is now correct. $\endgroup$ – Tito Piezas III Jun 8 at 5:54
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Based on Three-sided coin's comments, it seems

$$I_4 = \int_0^1 \frac{\rm{Li}_4(x)}{1+x}dx =\ln(2)\zeta(4)+\tfrac34\zeta(2)\zeta(3)-\tfrac{59}{32}\zeta(5) \approx 0.321352\dots$$

(Note: Confirmed as correct by Brevan's answer.) Connecting it to Three-sided coin's other integrals, then

$$I_4 = \ln(2)\zeta(4)-\tfrac12\zeta(2)\zeta(3)-h_1$$ $$I_4 = \ln(2)\zeta(4)+\tfrac14\zeta(2)\zeta(3)-h_2$$

where,

$$h_1=\int_0^1\frac{\rm{Li}_2(x)\,\color{blue}{\rm{Li}_2(-x)}}{x}dx = -\tfrac54\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$

$$h_2=\int_0^1\frac{\color{blue}{\rm{Li}_3(-x)}\ln(1-x)}{x}dx = -\tfrac12\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$

which implies,

$$h_1-h_2 =\int_0^1\frac{\rm{Li}_2(x)\,\rm{Li}_2(-x)}{x}dx- \int_0^1\frac{\rm{Li}_3(-x)\ln(1-x)}{x}dx =-\tfrac34\zeta(2)\zeta(3)$$


Compare to the similar integrals here that he mentioned,

$$h_3= \int_0^1\frac{\rm{Li}_2(x)\,\color{blue}{\rm{Li}_2(x)}}{x}dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$

$$h_4 = \int_0^1\frac{\color{blue}{\rm{Li}_3(x)}\ln(1-x)}{x}dx =\zeta(2)\zeta(3)-3\zeta(5$$

which has the proven relation,

$$h_3-h_4 = \int_0^1\frac{\rm{Li}_2(x)\,\rm{Li}_2(x)}{x}dx - \int_0^1\frac{\rm{Li}_3(x)\ln(1-x)}{x}dx = \zeta(2)\zeta(3)$$


Update: Per Brevan's answer:

$$I_n = \int_0^1\frac{\rm{Li}_n(x)}{1+x}dx=(-1)^nA(n)+\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)$$ with Dirichlet eta function $\eta(k)$ and $$A(n) = \sum_{k=1}^\infty \frac{H_k}{k^n}(-1)^k$$

is the $n$th "Alternating Euler Sum". However, since,

$$A(n,z) = \sum_{k=1}^\infty \frac{H_k}{k^n}z^k =S_{n-1,2}(z)+\rm{Li}_{n+1}(z)$$

for $-1\leq z\leq 1$, then $I_n$ can be expressed by Nielsen polylogs $S_{n,p}(z)$ as I suspected.

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Lets start with expanding the denominator $$I_n=\int_0^1\frac{\operatorname{Li}_n(x)}{1+x}\ dx=-\sum_{k=1}^\infty(-1)^k\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx$$ and lets have an example of the inside integral so that we have a clear image of generalizing it :$$\int_0^1x^{k-1}\operatorname{Li}_5(x)\ dx\overset{IBP}{=}\frac{\zeta(5)}{n}-\frac{\zeta(4)}{n^2}+\frac{\zeta(3)}{n^3}-\frac{\zeta(2)}{n^4}+\frac{H_n}{n^5}$$ as we can see the pattern clearly, we can write $$\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx=(-1)^{n-1}\frac{H_k}{k^n}-\sum_{i=1}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}$$ which follows: $$\boxed{I_n=\sum_{k=1}^\infty(-1)^{k+n}\frac{H_k}{k^n}+\sum_{k=1}^\infty\sum_{i=1}^{n-1}(-1)^{k+i}\frac{\zeta(n-i+1)}{k^i}}$$

Another generalization is by expanding the polylogarithm: $$\boxed{I_n=\sum_{k=1}^\infty\frac1{k^n}\int_0^1\frac{x^k}{1+x}\ dx=\sum_{k=1}^\infty\frac1{k^n}\left(H_{k/2}-H_k+\ln2\right)}$$

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