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0 votes
0 answers
50 views

How to integrate $\frac{x^N\log(1+x)}{\sqrt{x^2+x_1^2}\sqrt{x^2+x_2^2}}$?

I am trying to compute the integral $$\int_{x_0}^{1}\frac{x^N\log(1+x)}{\sqrt{x^2+x_1^2}\sqrt{x^2+x_2^2}}\text{d}x$$ where $x_0, x_1$ and $x_2$ are related to some parameters $\kappa_\pm$ by $$x_0=\...
Anders W's user avatar
5 votes
1 answer
429 views

Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$

I encountered a hypergeometric integral while investigating harmonic sums $$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$ Based on my experience I suspect a nice ...
Infiniticism's user avatar
  • 8,654
10 votes
1 answer
578 views

Evaluate $\int_0^1 x^{a-1}(1-x)^{b-1}\operatorname{Li}_3(x) \, dx$

Define $\small f(a,b)=\frac1{B(a,b)}\int_0^1 x^{a-1}(1-x)^{b-1} \text{Li}_3(x) \, dx$$ $$=\frac a{a+b}{}_5F_4(1,1,1,1,a+1;2,2,2,1+a+b;1)$ Where $a>-1$ and $b>0$. $1$. By using contour ...
Kemono Chen's user avatar
  • 8,679
10 votes
1 answer
477 views

A twisted hypergeometric series $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$

Question. I was given that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2=\frac{32}\pi G\ln2+\frac{64}\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)-2\ln^22-\frac53\pi^2$$ ...
Kemono Chen's user avatar
  • 8,679
14 votes
1 answer
480 views

Yet another difficult logarithmic integral

This question is a follow-up to MSE#3142989. Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$) $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \...
Jack D'Aurizio's user avatar
13 votes
2 answers
522 views

On the integral $\int_{0}^{1/2}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz$

This questions is related to my previous one. I am interested in a explicit evaluation in terms of Euler sums for $$ \int_{0}^{\pi/4}\text{Li}_3(\cos^2\theta)\,d\theta = \frac{1}{2}\int_{0}^{1/2}\...
Jack D'Aurizio's user avatar
2 votes
1 answer
186 views

About the integral $\int\arctan\left(\frac{1}{\sinh^2 x}\right)dx$, some idea or feedback

While I was playing with Wolfram Alpha calculator I wondered if it is known a closed-form for $$\int_0^\infty\arctan\left(\frac{1}{\sinh^2 x}\right)dx.\tag{1}$$ Wolfram Alpha provide me the ...
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