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6 votes
1 answer
285 views

Calculate $\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$

this integral got posted on a mathematics group by a friend $$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$$ I tried seeing what I'd get from ...
logandetner's user avatar
0 votes
1 answer
148 views

Evaluate: ${{\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx}}$

Evaluate: $${{I=\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx.}}$$ The answer is given below: $$ I=-\frac{7}{12}\pi^4\ln^2(2)-\...
Setness Ramesory's user avatar
1 vote
1 answer
164 views

Integral involving product of dilogarithm and an exponential

I am interested in the integral \begin{equation} \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast) \end{equation} where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to ...
o0BlueBeast0o's user avatar
10 votes
1 answer
790 views

A generalized "Rare" integral involving $\operatorname{Li}_3$

In my previous post, it can be shown that $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\...
Setness Ramesory's user avatar
17 votes
1 answer
1k views

A rare integral involving $\operatorname{Li}_2$

A rare but interesting integral problem: $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\...
Setness Ramesory's user avatar
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, ...
dan_fulea's user avatar
  • 34.2k
23 votes
3 answers
2k views

Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$

The following problem is proposed by a friend: $$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$ $$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
Ali Shadhar's user avatar
  • 25.8k
8 votes
1 answer
497 views

How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.

How can i evaluate $$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$ $$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }...
user avatar
1 vote
1 answer
111 views

How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?

where $H_n$ denotes the harmonic number. I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even $$-\sum_{n\geq 1}H_{2n}^2\...
zalm's user avatar
  • 125
2 votes
1 answer
86 views

Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?

Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
zalm's user avatar
  • 125
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 ...
zalm's user avatar
  • 125
1 vote
2 answers
150 views

How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$

I am having a difficult time evaulating $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$ I have tried the following relation: $$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$ ...
user avatar
1 vote
1 answer
256 views

Looking for closed form for $\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$

I was evaluating and integral involving iterated logarithms when the following integral appeared: $$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \...
user avatar
8 votes
2 answers
270 views

Is there a closed form for $\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x$?

Do we know if there is a closed form for $$ I :=\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x\mathrm{?} $$ Wolfram alpha gives an approximation of $2.66989$ which may be equivalent to: $$10\...
user avatar
8 votes
0 answers
404 views

Powerful Integral $\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt$

This integral can be found in Cornel's book, (Almost) Impossible Integral, Sums and Series page $97$ where he showed that $$\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt=\frac14\left(\frac12\ln^2(1+x^2)-2\...
Ali Shadhar's user avatar
  • 25.8k

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