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3 votes
3 answers
386 views

$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum

Working on the same lines as This/This and This I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$: $$\operatorname{Li}_{2} \left(\frac{1}{e^{\...
Srini's user avatar
  • 862
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
3 votes
0 answers
186 views

how to find closed form for $\int_0^1 \frac{x}{x^2+1} \left(\ln(1-x) \right)^{n-1}dx$?

here in my answer I got real part for polylogarithm function at $1+i$ for natural $n$ $$ \Re\left(\text{Li}_n(1+i)\right)=\left(\frac{-1}{4}\right)^{n+1}A_n-B_n $$ where $$ B_n=\sum_{k=0}^{\lfloor\...
Faoler's user avatar
  • 1,647
11 votes
0 answers
255 views

Solve the integral $\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx$

I tried to solve this integral and got it, I showed firstly $$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$ and for other integral $$\int_0^...
Faoler's user avatar
  • 1,647
4 votes
1 answer
257 views

Find closed-form of: $\int_{0}^{1}\frac{x\log^{3}{(x+1)}}{x^2+1}dx$

I found this integral: $$\int_{0}^{1}\frac{x\log^{3}{(x+1)}}{x^2+1}dx$$ And it seems look like this problem but i don't know how to process with this one. First, i tried to use series of $\frac{x}{x^...
OnTheWay's user avatar
  • 2,702
12 votes
3 answers
460 views

How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$

I am trying evaluating this $$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$ For $k=1$, there has $$J(1)=\frac{\pi^4}{96}.$$ Maybe $J(k)$ ...
Setness Ramesory's user avatar
4 votes
3 answers
211 views

Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form?

I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$ $$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\...
T-Ray's user avatar
  • 179
1 vote
1 answer
83 views

Integral of a modified softplus function

In a manuscript I am currently reading, the authors propose a modified softplus function $$g(a)=\frac{\log\left(2^a +1 \right)}{\log(2)}$$ for some $a \in \mathbb{R}$. The authors then claim that if $...
J.Galt's user avatar
  • 961
4 votes
0 answers
341 views

How to evaluate $\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$

I am trying to evaluate $$\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$$ But I am not sure what to do since integration by parts is not possible here. I tried using a ...
user avatar
5 votes
1 answer
223 views

How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$

I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$ But I've not been successful in doing so, what I tried is $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\...
user avatar
0 votes
0 answers
50 views

Further question on Logarithm product integral

How to perform $\int_0^1 \frac{\left(a_0\log(u)+a_1\log(1-u)+a_{2}\log(1-xu)\right)^9}{u-1} du $? Method tried: Intgration-by-parts Series expansion change of variable $\log(u)=x$ But I still can't ...
YU MU's user avatar
  • 99
1 vote
0 answers
149 views

Evaluating the indefinite integral $\int x^k \log (1-x) \log (x) \log (x+1) \, dx$

Recently I have calculated the long resisting indefinite integral $\int \frac{1}{x} \log (1-x) \log (x) \log (x+1) \, dx$ (https://math.stackexchange.com/a/3535943/198592). Similar cases, but for ...
Dr. Wolfgang Hintze's user avatar
2 votes
2 answers
407 views

Integral related to the softplus function

Let $$ f(x) = \log(1+e^{2x+1}) - 2\log(1 + e^{2x}) + \log(1 + e^{2x-1}). $$ According to Wolfram Alpha, $$ \int_{-\infty}^\infty f(x)\,dx = \frac 12.\tag{$*$} $$ $f(x)$ is a "bump function" built out ...
fmg's user avatar
  • 350
1 vote
0 answers
74 views

Integrate $\int_{-\infty}^\infty [4(\log r_1 - \log r_2) - 2(x_1^2/r_1^2 - x_2^2/r_2^2)]^2 dx$

As the title suggests, I am having trouble evaluating the following definite integral: $$\int_{-\infty}^\infty \left[4\left(\log r_1 - \log r_2\right) - 2\left(\frac{x_1^2}{r_1^2} - \frac{x_2^2}{r_2^...
Son Pham-Ba's user avatar
10 votes
4 answers
477 views

Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$

I am trying to evaluate the following integral $$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$ with $0< k < 1$. My attempt By performing ...
Sanmar's user avatar
  • 168

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