Questions tagged [polygamma]

For questions about, or related to the polygamma function.

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Computing solutions to polygamma function of order 1?

I want to preface this with the fact that I am WAY out of my depth with my mathematical familiarity with these topics. While trying to figure out how to compute the solution to the polygamma function ...
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How do you solve for x in this equation which includes the polygamma function.

How do you solve for x given: $$ 0 = 2x+18 - \int_{0}^{\infty}\left(t^x * e^\left(-t\right) * ln(t)\right) $$ The background is I have this function: $f\left(x\right)\ =x^{2}+\ 19x\ -\ x!$ I took the ...
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Closed form for $\rm{Li }_2\left( -{\frac {i\sqrt {3}}{3}} \right)$

In my personal research with Maple i find this closed form : $$\operatorname{Li }_2\left( -{\frac {i\sqrt {3}}{3}} \right)={\frac {{\pi}^{2}}{24}}+{\frac {\ln \left( 2 \right) \ln \left( 3 \right) }...
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Sum of Fermi-Dirac integrals with opposite chemical potentials: closed form (Le Bellac eq. 1.13)

I am trying to reproduce the result of eq. (1.13) in Le Bellac's Thermal Field Theory book to compute the grand canonical potential of a gas of massless fermions: $$ Ω = - \frac{V T^4}{6 π^2} \int_0^\...
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Evaluate the limit with the Appell F1 function

I encountered the limit with the form $$ \lim_{r\rightarrow 1^+}\frac{1-\frac{1}{r^{d-3}}}{e^{2\kappa r_s(r)}},\quad r_s(r)\equiv r~ F_1\left(-\frac{1}{d-3};-\frac12,1;\frac{d-4}{d-3},-\frac{Q}{r^{d-3}...
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7 votes
2 answers
161 views

Great difficulty in finding the residues of $\frac{\mathrm{Log}\Gamma\left(\frac{z+ai}{2\pi i}\right)}{\cosh(z)+1}$

$\newcommand{\log}{\operatorname{Log}}\newcommand{\res}{\operatorname{Res}}\newcommand{\d}{\mathrm{d}}$Let $\Lambda(z)=\log\Gamma(z)$, $a\gt0$, let $\psi$ denote digamma. It is written here, Page 49, ...
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Simplification of a difficult identity involving the digamma function

EDIT: The question here is not for the reader to laboriously scan all the working, but rather to suggest ways to continue the train of thought; for example, there is perhaps a closed form of the ...
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6 votes
3 answers
326 views

In an attempt to find $I = \int_0^\infty \frac{t}{e^t-1}dt$

I was trying to solve $I = \int_0^\infty \frac{t}{e^t-1}dt$ My approach I took the more general form of integral $f(s) = \int_0^{\infty}\frac{e^{-st}}{e^t-1}dt$ the same way as How to evaluate the ...
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3 votes
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New trigamma identity for $\Psi_1(\frac3{20})+6\,\Psi_1(\frac15)+10\,\Psi_1(\frac25)-\Psi_1(\frac1{20})$

I play with Maple, and I find this relation for the trigamma function: $$\begin{align} \Psi_1\left({\frac{3}{20}}\right)+6\,\Psi_1\left(\frac15\right)+ 10\,\Psi_1\left(\frac25 \right)-\Psi_1\left(\...
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Polygamma sum problem ...

Hello guys i have a problem evaluating the following sum $$\sum_{n=1}^{+\infty}\frac{n(n+1)}{2}\frac{4x(3\pi ^{2}(n+1)^{2}+x^{2})}{(x^{2}-\pi ^{2}(n+1)^{2})^{3}}$$ It is obviously of the polygamma ...
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1 answer
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Polygamma sum problem

I have a problem evaluating the following sum, $$\sum_{n=1}^{+\infty}\frac{4nx(3\pi ^{2}(n+1)^{2}+x^{2})}{(x^{2}-\pi ^{2}(n+1)^{2})^{3}}$$ The sum obviously is of the form of a polygamma function. ...
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Ask for a proof of logarithmically complete monotonicity of a power-exponential function involving the difference of the psi and logarithmic functions

It is common knowledge that the classical Euler gamma function $\Gamma(z)$ can defined by \begin{equation*} \Gamma(z)=\int^\infty_0t^{z-1} e^{-t}\textrm{d}t, \quad \Re(z)>0 \end{equation*} and the ...
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What is the general formula of the sum $\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{m}$ for $m,n\in\mathbb{N}$?

The classical Euler's gamma function $\Gamma(z)$ can be defined by \begin{equation} \Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{\prod_{k=0}^n(z+k)}, \quad z\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}. \end{...
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closed form in terms of a polygamma function? $ f(2,2)=\sum \sum \exp\big(- n^2k^2 \big)? $

Is there a closed form for:$$ f(2,2)=\sum_{n=1}^\infty \sum_{k=1}^\infty \exp\big(- n^2k^2 \big)? $$ Note that I'm defining a function: $$f(x,y)=\sum_{n=1}^\infty \sum_{k=1}^\infty\exp\big(-n^xk^y\...
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1 answer
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A curious summation.

Months ago I was interested in calculating new infinite summations and was able to make a nice summation and also find its closed form as follows : $$\sum \limits_{n=0}^{\infty } \frac{(-1)^{0+1+2+..+....
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8 votes
6 answers
326 views

Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$

I came across the following statements $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$ $$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \...
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Is there a general summation formula for the polygamma function at $z=1/2$? i.e $\psi^{(s)}(\frac{1}{2})$ for all s.

For $s>0$ one has $ \psi^{(s)}(\frac{1}{2}) = s! \cdot \zeta(s+1, \frac{1}{2}) \cdot (-1)^{s+1} $. E.g. $ \psi^{(1)}(\frac{1}{2}) = 3 \cdot \zeta(2) $ $ \psi^{(2)}(\frac{1}{2}) = -14 \cdot \zeta(...
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1 vote
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a limit containing a digamma function

I wonder if the following limit is correct and how to prove it $$ \lim _{x \rightarrow+\infty} \frac{4}{p^{2}} x^{4}\left(\frac{1}{x^{2}}+\frac{2}{x}(\log 2+\Psi(x))+\Psi^{\prime}(x)\right)>0 $$ ...
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1 vote
2 answers
106 views

Simplify this expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Is it possible to simplify this constant expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$? Here $\psi(x)$ is digamma function. Particularly, ...
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4 votes
2 answers
125 views

A convergent series for the Trigamma function $\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $

I just came up with the following convergent series for the Trigamma function defined by $\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $. \begin{align*} \psi_1(n) &=\lim_{m \to \infty} \sum_{j=1}^m \...
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2 votes
1 answer
136 views

Another bizarre sum involving a binomial coefficient and inverse powers of integers.

In the attempt to answer Binomial identity involving Harmonic numbers we stumbled on the following problem. Let $i\ge 0 $ and $k \ge i+2 $ and $p \ge 1$ be integers. Consider a following sum: \begin{...
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Partial sum of polygamma

Do you know what kind of process use mathwolfram in determining \begin{align} &\sum_{m = 1}^{n}\left[\frac{1}{-s + m} + \frac{1}{s + m}\right] \\[3mm] = &\ \Psi_{0}\left(-s + n + 1\right) + \...
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1 answer
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Polygamma function approximation to infinity

Be $\psi_0(s)$ the polygamma function of order zero in $s \in C$. Do you think is correct to write $ \psi_{0}(s) + \psi_{0}(-s) = \sum_{n=1}^{\infty} \frac{1}{-s+n} + \sum_{n=1}^{\infty} \frac{1}{s+n} ...
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1 answer
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Is there a decomposition for the digamma function as a sum of digamma functions?

Let $\psi(x)$ denote the digamma function $$ \psi(x)=\Gamma(x)\frac{\partial}{\partial x} \Gamma(x). $$ Consider $x=x_1 +x_2+\dots +x_m$, where $x_j>0$, for $j=1, \ldots,m$. Is there any formula to ...
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2 votes
1 answer
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What is the root of $\sum _{i=1}^{m-1} i^k=m^k$ given integer k?

Let $m'$ be the root of $\sum _{i=1}^{m-1} i^k=m^k$ given integer $k$ (solving for $m$). What is $m'$?
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1 answer
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Closed formulae of polygamma function of negative order

Wolfram Alpha seems to output nice closed forms for the polygamma function of negative orders with whole and half arguments. I was interested in order $-2$ and I deduced the following $$\psi^{(-2)}\...
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2 votes
1 answer
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Derivatives of Polygamma Functions

I would like to know if there's a quicker way to verify: $$\partial_z^{n-1}\psi(tz)t^n = \partial_z^n\ln[\Gamma(tz)], \,\,\\ n \in \mathbb{N}^+, t \in \mathbb{C}_+\tag{1}\label{1}$$ That's true for $t=...
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1 vote
1 answer
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Upper bound for the nth derivative of $\Gamma(x)^n$

I was trying to find an upper bound for $$ \frac{d^n}{ds^n} \Gamma(s)^{n}|_{s=1}$$ yet, I only get the bound for the nth derivative of gamma, as follow: First, the integral of the nth derivative of ...
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3 votes
2 answers
173 views

Prove $\lim_{n\mapsto 0}[(\psi(n)+\gamma)\psi^{(1)}(n)-\frac12\psi^{(2)}(n)]=2\zeta(3)$

How to prove that $$\lim_{n\mapsto 0}[(\psi(n)+\gamma)\psi^{(1)}(n)-\frac12\psi^{(2)}(n)]=2\zeta(3)\ ?$$ I encountered this limit while I was trying to solve $\int_0^1\frac{\ln x\ln(1-x)}{x(1-x)}dx$ ...
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2 answers
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How can we derive the asymptotic expansion for the second derivative of the gamma-function?

We can expression the first derivative of the gamma function as: $$\Gamma'(s) \sim -\frac{1}{s^2}+\frac{6\gamma^2+\pi^2}{12}+O(s)$$ but what about the second derivative? I do not know how to approach ...
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Log gamma integral with a slightly more complicated polynomial factor

I need some help figuring out how to evaluate this integral. $$\frac{1}{n!}\int_0^1 (x-t)^n \ln\Gamma(t) \, dt$$ A similar integral is known due to the work of Victor Adamchik. But I haven’t been ...
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1 vote
3 answers
130 views

How to Prove : $ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $

How to Prove : $$ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $$ I have tried looking at Series definitions of the Polygamma function from which ...
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Find the function $f(x)=\sum_{n=1}^{\infty}\frac{H_{n-1}(-x)^n}{n!}$

I want to find the function defines by : $$f(x)=\sum_{n=1}^{\infty}\frac{H_{n-1}(-x)^n}{n!}$$ Where $H_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is the Harmonic series. My work We have the ...
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1 vote
3 answers
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How can we evaluate $\sum_{k\geq 0} \frac{1}{(2k+1)^3}$?

I have been looking to evaluate $$\mathcal{A} = \sum_{k=0}^\infty \frac{1}{(2k+1)^3}.$$ We can represent our sum in terms of the Hurwitz zeta function; namely, $$\mathcal{A} = \zeta\left(\frac{1}{2},...
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1 vote
2 answers
53 views

Decomposition of $\psi^{(n)}(1)$ in terms of $\psi^{(n)}(k)$

Accidentally run into this identity: \begin{align} \psi^{(n)}(1) &= 2^{n+1}\, \sum_{k = 2}^\infty (-1)^k\,\psi^{(n)}(k) \tag{1}\label{1} , \end{align} its variation \begin{align} 2^{-n-1} &...
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2 votes
1 answer
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Analytic continuation of $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$?

Is there an analytic continuation of the generalised harmonic number $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$ to the positive reals x, for $k>1$? I can’t find anything useful through Google, just ...
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1 vote
2 answers
118 views

Prove $\lim_{x\to\infty}\sum_{n=1}^x x\log\left(1+\frac1{xn(an+1)}\right)= H_{\frac1a}$

How to prove that $$\large\lim_{x\to\infty}\sum_{n=1}^x x\log\left(1+\frac1{xn(an+ 1)}\right)= H_{\frac1a}, \quad a\in \mathbb{R},\ |a|>1$$ This question is a formulated form of this problem. ...
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7 votes
1 answer
229 views

Evaluate $\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy$

How to prove $$\small\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy= -4 \pi ^2 \left(\frac{\pi }{\sqrt{3}}+\log (2)-\frac{\psi ^{(1)}\left(\frac{1}{6}\right)}{2 \sqrt{3} \pi ...
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0 votes
0 answers
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Sum involving 2nd antiderivative of the digamma function.

Help evaluate the following sum: $$\sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{1}{4n-1}\ln\left(2n-1\right)+\frac{4}{\left(4n-1\right)\left(4n-3\right)}\left(\psi^{(-2)}\left(2n-\frac{1}{2}\right)-\...
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5 votes
3 answers
265 views

Is the closed form for $\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}$ known in the literature?

I managed to find $$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-2^{-m}\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}$$ $$=(1-2^{-m})\left[\left(1+\frac m2\...
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1 vote
3 answers
233 views

$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right)$

Inside the Wolfram Documentation page for the secant function, an identity is given which involves the gamma function, polygamma function, and Catalan's constant. Notes on documentation page: Some ...
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0 votes
1 answer
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Reference request: $q$-gamma, $q$-polygamma, $q$-Pochhammer.

I'm trying to solve a problem related to a few difficult series and am using Mathematica to hammer out the difficult bits. The problem is that QPochhammer,...
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5 votes
3 answers
313 views

Evaluation of $\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx$

I want to show that $$\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx=\frac{\psi^{(1)}(\frac{t}{2\pi})-4\psi^{(1)}(\frac{t}{\pi})}{8\pi}t+\frac14 $$ where $\...
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0 votes
0 answers
59 views

correct property of a Gamma function?

On Wikipedia, https://en.wikipedia.org/wiki/Gamma_function, I read that the following hold true for any positive integer $n$: \begin{aligned}\Gamma \left({\tfrac {1}{2}}+n\right)&={(2n)! \over 4^{...
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4 votes
1 answer
56 views

The limit of the ratio of polygamma functions

I want to calculate this quantity: $$\lim_{x \rightarrow \infty}\frac{\Psi_1 (x)}{\Psi_1 (x + y)}$$ where $$\Psi_1 (x)=\frac{d^2}{dx^2}\log \Gamma (x)=\sum_{k=0}^{\infty}\frac{1}{(x+k)^2}. $$ I ...
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2 votes
2 answers
122 views

Evaluation of a digamma series involving golden-ratio

Let $\varphi =\frac{1}{2} \left(\sqrt{5}+1\right), a=\tan \left(\frac{\sqrt{5} \pi }{2}\right)$, then how can one prove $$\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+\varphi)-\psi ^{(0)}\left(n-\frac{1}{...
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0 votes
1 answer
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Limit involving Polygamma function: $\lim_{x\to \infty} \sum_{k = 0}^\infty \frac{2x^2}{(x + k)^3}$

I don't remember where I got this identity: $$\lim_{x\to \infty} [(-1)^{k + 1} x^k \psi^{(k)}(x)] = (k - 1)!$$ where $\psi^{(k)}$ is the polygamma function. I just need to find the limit above for $k ...
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4 votes
2 answers
455 views

The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $

How to Prove that $$ \sum_{n=0}^{\infty}\frac{(-1)^nH^{(2)}_{n}}{(2n+1)^2} \;\;=\;\;\frac{7 \pi \; \zeta(3)}{4}-\frac{\zeta(2)G}{2}+\frac{45\zeta(4)}{8}-\frac{\Psi^{(3)}\big(\frac{1}{4}\big)}{...
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6 votes
3 answers
635 views

Integrating $\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2} dx$ using real methods

How to evaluate, without contour integration the following integral: $$I=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2}\ dx\ ?$$ @Cody mentioned in this solution that $$I=\frac{\pi^{2}}{6}G+\frac{\pi^{3}}{...
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2 votes
1 answer
117 views

Prove $\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1$

Prove $$\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1\tag{1}$$ where $\psi^{(m)}(x)$ is the polygamma function and $\zeta$ is the Riemann zeta function. This problem ...
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