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16 votes
2 answers
889 views

Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $

Evaluate : $$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$ Introduction : I have a friend on another math platform who proposed a summation ...
MathGod's user avatar
  • 5,558
3 votes
1 answer
513 views

Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \...
user 1591719's user avatar
  • 44.4k
19 votes
3 answers
709 views

Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$

It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about the cubic version, that is $$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$ What are ...
user 1591719's user avatar
  • 44.4k
6 votes
2 answers
201 views

Double integral with a product of dilog $\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x)\ dx \ dy$

One of the integrals I came across these days (during my studies) is $$\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x) \ dx \ dy$$ that can be turned into a series, or can be approached by ...
user 1591719's user avatar
  • 44.4k
4 votes
1 answer
330 views

A possible dilogarithm identity?

I'm curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before? $$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{...
user 1591719's user avatar
  • 44.4k
29 votes
7 answers
1k views

Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$

I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral \begin{equation} \int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(...
Anastasiya-Romanova 秀's user avatar
4 votes
2 answers
296 views

Integral Involving Dilogarithms

I came across the identity $$\int^x_0\frac{\ln(p+qt)}{r+st}{\rm d}t=\frac{1}{2s}\left[\ln^2{\left(\frac{q}{s}(r+sx)\right)}-\ln^2{\left(\frac{qr}{s}\right)}+2\mathrm{Li}_2\left(\frac{qr-ps}{q(r+sx)}\...
SuperAbound's user avatar
  • 5,604
30 votes
3 answers
921 views

A closed form for a lot of integrals on the logarithm

One problem that has been bugging me all this summer is as follows: a) Calculate $$I_3=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \ln{(1-x)} \ln{(1-xy)} \ln{(1-xyz)} \,\mathrm{d}x\, \mathrm{d}y\, \mathrm{...
Shivam Patel's user avatar
  • 4,044
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 ...
Turbo's user avatar
  • 6,245
22 votes
4 answers
1k views

Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?

Ramanujan gave the following identities for the Dilogarithm function: $$ \begin{align*} \operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{...
Shobhit Bhatnagar's user avatar
2 votes
1 answer
314 views

A new proof a non-linear Euler sum

According to Nielsen we have the following : If $$f(x)= \sum_{n\geq 0}a_n x^n $$ Then we have the following $$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\...
Zaid Alyafeai's user avatar
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 ...
Zaid Alyafeai's user avatar
20 votes
4 answers
1k views

Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure.
user91500's user avatar
  • 5,626
3 votes
1 answer
205 views

Generating function for $\sum_{k\geq 1} H^{(k)}_n x^ k $

Is there a generating function for $$\tag{1}\sum_{k\geq 1} H^{(k)}_n x^ k $$ I know that $$\tag{2}\sum_{n\geq 1} H^{(k)}_n x^n= \frac{\operatorname{Li}_k(x)}{1-x} $$ But notice in (1) the fixed $...
Zaid Alyafeai's user avatar
8 votes
2 answers
391 views

Closed form of $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\varphi-1)$

I am trying to calculate the dilogarithm of the golden ratio and its conjugate $\Phi = \varphi-1$. Eg the solutions of the equation $u^2 - u = 1$. From Wikipdia one has the following \begin{align*}...
N3buchadnezzar's user avatar

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