# Questions tagged [polylogarithm]

For questions about or related to polylogarithm functions.

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### Integrate $\frac1{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)}$ over $[-1,1]^2$?

Problem $$\text{Evaluate} \quad I = \iint\limits_{[-1,1]^2} \frac{du\,dv}{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)}$$ The integral is a special case of one that appeared in this recent post, in which ...
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### Definite integral involving exponential and logarith function

Working with Dilogarimth function, we get the following definite integral $$\int_0^{\infty}\frac{t^2\,\ln^{n}(t)}{(1-e^{x\,t})(1-e^{y\,t})}\,dt$$ with $n=1,2,3,...$ and $x,y>0$. I wonder if is ...
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### How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$? No context, this is just a curiosity o'mine. Yes, I am aware there is no reason to believe a random power ...
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### Behaviour of polylogarithm at $|z|=1$

I have the sum $$\sum_{n=1}^\infty \dfrac{\cos (n \theta)}{n^5} = \dfrac{\text{Li}_5 (e^{i\theta}) + \text{Li}_5 (e^{-i\theta})}{2},$$ where $0\leq\theta < 2 \pi$ is an angle and $\text{Li}_5(z)$ ...
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### Relations between Dilogarithms and Imaginary part of Hurwitz-Zeta function

I'm working through a paper that involves a problem concerning the calculation of the Imaginary part of the derivative of the Hurwitz-Zeta function $\zeta_H(z,a)$ with respect to $z$, evaluated at a ...
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### Connection between the polylogarithm and the Bernoulli polynomials.

I have been studying the polylogarithm function and came across its relation with Bernoulli polynomials, as Wikipedia site asserts: For positive integer polylogarithm orders $s$, the Hurwitz zeta ...
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### I need help evaluating the integral $\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz$

I was playing around with the integral: $$\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz$$ I couldn't find a way of solving it, but I used WolframAlpha to find that the integral evaluated ...
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### Upper bound of a polylogarithm type series

The series is $\sum_{n=0}^{\infty}\left(1+\frac{n}{a}\right)^bx^n,$ where $a$ and $b$ are positive real numbers, $x\in[0,1]$. The sum of the series diverges when $x\to1$. I want to get an upper bound ...
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### Solution of a meme integral: $\int \frac{x \sin(x)}{1+\cos(x)^2}\mathrm{d}x$

Context A few days ago I saw a meme published on a mathematics page in which they joked about the fact that $$\int\frac{x\sin(x)}{1+\cos(x)^2}\mathrm{d}x$$ was very long (and they put a screen shot of ...
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### $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(1+\alpha^N\tan(x)^N)\mathrm{d}x\quad$ where $N\in\mathbb{N}$

$\color{red}{\textrm{Context}}$ I wanted to calculate the following integrals $$\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(1+\tan(x)^N)\mathrm{d}x\qquad\text{for }N\in\mathbb{N}$$ and I used the Feymann ...
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### Prove $\int_{0}^{1}\frac1k K(k)\ln\left[\frac{\left(1+k \right)^3}{1-k} \right]\text{d}k=\frac{\pi^3}{4}$

Is it possible to show $$\int_{0}^{1}\frac{K(k)\ln\left[\tfrac{\left ( 1+k \right)^3}{1-k} \right] }{k} \text{d}k=\frac{\pi^3}{4}\;\;?$$ where $K(k)$ is the complete elliptic integral of the first ...
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### Verification of the generalized polylogarithm formula

Here I posted a generalized formula for the polylogarithm I had discovered. However, for $x=\frac{1}{2}$, $z=\frac{1}{2}$, $p=1$ wolfram alpha yields a result different than what the double integral ...
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### How to show $\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G$

I am trying to prove that $$\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G,$$ where $G$ is the Catalan constant and $\operatorname{Li}_2(x)$ is the dilogarithm ...
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### Efficient calculation for Lerch Transcendent Expression

I've encountered: $$\Phi(z, s, \alpha) = \sum_{k=0}^\infty \frac { z^k} {(k+\alpha)^s}.$$ When trying to compute: $$\frac{1}{x}\sum_{p=0}^m \frac{2}{(2p-1)\ x^{2p-1}}\ s.t. x\in\mathbb{N} =\ ???$$ ...
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### Prove that $-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$
Prove that $$-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$$ where ${\rm Li}_2(x)$ is the Poly Logarithm function and $\zeta(s)$ is the Riemann zeta function Let I=-\int_{0}^{1}...