All Questions
4
questions
17
votes
2
answers
894
views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt
=-\frac 1{64}\pi^4
$$
(using a "minimal industry" of relations, ...
- 34.2k
2
votes
1
answer
111
views
Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.
I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
- 125
1
vote
1
answer
121
views
Inverse of Higher logarithms
Th polylogarithm function is defined by $$Li_s(z)=\sum_{k=1}^\infty\frac{z^k}{k^s}.$$ At $s=1$, we have the natural logarithm function. We have the inverse of natural logarithm function as the ...
- 6,245
16
votes
5
answers
1k
views
Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
I proved the following result
$$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$
After ...
- 14.4k