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Dec
6
comment Closed form for Euler sum with $H_{2n}$?.
I emended the heading to be more specific about what I was asking.
Dec
6
revised Closed form for Euler sum with $H_{2n}$?.
edited title
Dec
6
comment Closed form for Euler sum with $H_{2n}$?.
Yes, of course. Can anyone find a closed form for said Euler sum?.
Dec
6
asked Closed form for Euler sum with $H_{2n}$?.
Dec
3
comment another product of log integral
Hey nospoon, here is another integral I found if you're interested. I think it can be done the way you done the last one. But, I found: $$\int_{0}^{1}\log(1-x^{3})\log(1+x^{3})dx=1/4\left(-72+2 \gamma^{2}+\pi^{2}+4 \gamma \psi(1/6)+2\psi(1/6)-2\psi_{1}(7/6)\right)-\frac{1}{2}\left(3\sqrt{3}\pi +\frac{\pi^{2}}{6}+9\log(3)+12\log(2)-36\right)-\sum_{n=1}^{\infty}\frac{H_{2n}}‌​{(n(6n+1))}$$. I got stuck on that last sum....so far.
Dec
3
comment another product of log integral
Wow, thanks nospoon. Very nice work. There are lots of polylog identities. Maybe reference to Lewin's book we can find a way to simplify them. I managed to discover a solution, but it is not as efficient as yours. As mentioned above, I used a bunch of Euler sums.
Dec
3
accepted another product of log integral
Nov
29
revised another product of log integral
added another sum.
Nov
29
answered another product of log integral
Nov
26
comment another product of log integral
Well, fellas, I did indeed make some progress. Albeit tedious, but doable. Just picking at it as I get time. Here is one I found: $$\sum_{n=1}^{\infty}\frac{H_{2n}}{2n+1}\cos\left(2\pi n/3\right)=-1/32\log^{2}(3)+1/48\log(3)(5\pi\sqrt(3)+54)-\frac{\pi\sqrt{3}}{2}\l‌​og(2)-\frac{31\pi^{2}}{288}+\frac{3\pi\sqrt{3}}{8}-3+\Re\left[e^{2\pi i/3}\left(3Li_{2}\left(\frac{1-e^{\pi i/3}}{2}\right)-\frac{\pi^{2}}{4}+3/2\log^{2}(2)\right)\right]$$
Nov
24
comment another product of log integral
Thanks for the fast responses. Appreciate it:). I would enjoy seeing your work when and if you find time, nospoon. I tried your idea, tired. I have to admit, I don't see where it is that 'straightforward' but it does indeed work, as you said. I have not put all of the pieces completely together yet, but I have it written in terms of various Euler sums I can handle. I would like to see how you handles it if you want to outline your work. We get two for the price of one this way because $\Re\int_{0}^{1}\log(1+x)\log(1-xe^{2\pi i/3})dx=\Re\int_{0}^{1}\log(1+x)\log(1-xe^{4\pi i/3})dx$.
Nov
23
asked another product of log integral
Nov
4
comment Recurrence for $\int \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x$
See here: math.stackexchange.com/questions/146741/…
Oct
30
awarded  Critic
Oct
28
answered How to find the value of $\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$
Oct
24
comment integral with square of log of a quadratic $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$
Very nice, r9m. Turning it into a log trig integral is the ticket. :)
Oct
24
accepted integral with square of log of a quadratic $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$
Oct
17
comment integral with square of log of a quadratic $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$
Nice swp. I derived it a different way by using $$Li_{2}(e^{\pm \pi i/3})$$, but your method is nice and straightforward.
Oct
17
comment integral with square of log of a quadratic $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$
Thanks robjohn. Very insightful. I would not have thought of using that integral second from the top. Very clever, as usual. May I ask, is there some way we can now relate this to $$2/3\zeta(3)-4/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$?.
Oct
16
revised integral with square of log of a quadratic $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$
added 711 characters in body