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12

Performing integration by parts by taking $u=\ln(1-x)\ln(1+x)$, then $$\begin{align} \int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx&=\frac{1}{3}\bigg[\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln(1-x)\ln^3x}{1+x}\ dx\bigg]\\ &=\frac{1}{3}\bigg[I-J\bigg]\\ \end{align}$$ Evaluation of $I$ : $$\begin{align} ...


6

Knowing that $$\text{B}\,(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ and $$\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\quad\Longrightarrow\quad\Gamma'(x)=\psi(x)\Gamma(x)$$ where $\Gamma(x)$ is gamma function and $\psi(x)$ is digamma function, then $$\begin{align}\require\cancel\frac{\partial\text{B}}{\partial ...


5

Note the simple fact that $$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1} =\lim_{n\to\infty} \int_{n}^{2n} \frac{2}{2x+1} \ dx= \log(2)$$ Q.E.D.


5

Consider $$F(x) = \dfrac{\exp(ix)}{x}$$ Then $$\dfrac{d^n}{dx^n} F(x) = P_n(1/x) \exp(ix)$$ where $$-t^2 P_n'(t) + i P_n(t) = P_{n+1}(t)$$ EDIT: If we write $P_n(t) = Q_n(t) + i R_n(t)$ where $Q_n$ and $R_n$ have real coefficients, then $$\dfrac{d^n}{dx^n} \text{sinc}(x) = Q_n(1/x) \sin(x) + R_n(1/x) \cos(x)$$ An exponential generating function comes ...


5

Not sure what you did, but all that work seems kinda silly to arrive at your result. $$\sum_{n=1}^\infty\frac{x^n}{(n+2)^2}=\frac{1}{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{(n+2)^2}=\frac{1}{x^2}(-x-\frac{x^2}{4}+\sum_{n=1}^\infty\frac{x^n}{n^2})=\frac{1}{x^2}(-x-\frac{x^2}{4}+\text{Li}_2(x))$$


4

Here's a sketch of an alternative proof for the lower bound. First, note that $$ x\cdot\frac{x^{x-1}}{e^{x-1}} = \frac{(x+1)^x}{e^x} \cdot \frac{e}{\big(1+\frac1x\big)^x} \ge \frac{(x+1)^x}{e^x} $$ Since $\Gamma(x+1)=x\Gamma(x)$, this means that the lower bound for $x$ implies the lower bound for $x+1$; so it suffices to show the inequality for $x\in[2,3]$. ...


4

The problem discussed here can be easily solved if using explicit expressions for $\gamma_1(1/4)$ and $\gamma_1(3/4)$, see the paper by Donal F. Conon "The difference between two Stieltjes constants" and another paper "Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results" which I wrote two years ago ...


4

This equality is equivalent to this statement: There is an integer $n$ strictly between $\log_2(t+1)$ and $\log_2(t+2)$ And this is the same as $t+1< 2^n<t+2$ This happens if and only if $\lceil t+1\rceil $ is a power of $2$ and $t\notin\Bbb Z$.


3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} ...


3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


3

General Leibniz rule $$ \frac{d^n}{dx^n}f(x)g(x) = \sum \left(\matrix{n \\ k}\right)\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^k}{dx^k}g(x) $$ let $$ f(x) = \frac{1}{x}\\ g(x) = \sin(x) =\mathcal{I}\left(\mathrm{e}^{ix}\right) $$ (using @RobertIsrael answer) therefore $$ \frac{d^{n-k}}{dx^{n-k}}\frac{1}{x} = (-1)^{n-k}\frac{(n-k)!}{x^{n-k+1}}\\ ...


3

You can write the numerator as : $$ x^2 +x\log(1-x) -\log(1-x)-x = (1 - \log(1-x))(1-x) +(x-1)(x+1) $$ so that it remains to evaluate the integral: $$ \int_0^1 \frac{1-\log(1-x)-1-x}{x^2} \mathrm dx = - \int_0^1 \frac{\log(1-x)+x}{x^2} \mathrm dx $$


2

Somewhere in your question, you write "Integrate once with respect to $x.$". In the equation following that phrase, you have a $\frac{3}{2}$ on the r.h.s. That's superfluous. The derivatives of the l.h.s. and the r.h.s are equal, and setting $x=0,$ we find that the constant of integration is $0.$


2

To estimate $I_0$, let us set $\displaystyle M(t):=\log\left(\frac{t^2-1}{t^2+1}\right)$, thus $\displaystyle I_0:=\sum_{\ell=2}^{\infty}M\left(\frac{j_{\nu,\ell}}{j_{\nu,1}}\right)$. Now notice that $M$ is non-decreasing for $t\geq 1$ and $\displaystyle\frac{j_{\nu,\ell}}{j_{\nu,1}}\geq 1$ for all $\ell\geq 2$ so that \begin{equation}\displaystyle I_0\leq ...


2

As shown in this answer, $H_{-1/2}=-2\log(2)$. Therefore, $$ \begin{align} \psi(1/2) &=H_{-1/2}-\gamma\\ &=-2\log(2)-\gamma \end{align} $$


2

To show the continuity of $x\mapsto xf(x)$ at $0$ $$|xf(x)|\le |x|\xrightarrow{x\to0}0$$ To show the differentiability of $g:x\mapsto x^2f(x)$ at $0$ $$\left|\frac{g(x)-g(0)}{x-0}\right|=|xf(x)|\le |x|\xrightarrow{x\to0}0$$


2

Consider: $$ f(x)=\log(x)\log(1-x)+\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x).$$ We want to show that $f$ is constant, hence we compute $f'$: $$ f'(x) =\left(\frac{\log(1-x)}{x}-\frac{\log(x)}{1-x}\right)-\frac{\log(1-x)}{x}+\frac{\log x}{1-x}=0. $$ To finish the proof, we just need to compute $f(x)$ in a point, or to compute the limit: $$ \lim_{x\to ...


2

I can give you a closed form for $|\cos(x)| \le r$. Since $r$ will always be less than $\frac \pi 2$, can just say the following: $$ \int_{\arccos(-r)}^{\arccos(r)} \arccos \left(\frac{\cos(x)} r \right)\ dx $$ However, even that's not going to be pretty. First, an identity: $$ \arccos(x) = \frac \pi 2 - \sum_{n=0}^{\infty} \frac{\binom{2n}{n} x^{2n + ...


2

Consider $(3n-2) \times \cdots \times 4 \times 1$ This equals: $3^n \cdot \frac{3n-2}{3} \times \cdots \times \frac{4}{3} \times \frac{1}{3}$ This equals: $3^n \frac{\Gamma(\frac{3n+1}{3})}{\Gamma(\frac{1}{3})}$ because $\Gamma(t+1) = t \Gamma(t)$. To explain this step further, using $\Gamma(t+1) = t\Gamma(t)$: $\Gamma(\frac{4}{3}) = ...


2

Stirling's formula $$\log \Gamma(z) = \bigl(z-\tfrac{1}{2}\bigr) \log z - z + \tfrac{1}{2} \log (2\pi) + O\left(\frac{1}{\operatorname{Re} z}\right)$$ gives us \begin{align} \log \Gamma\bigl(z+\tfrac{1}{2}\bigr) - \log \Gamma(z) &= z\log\bigl(z+\tfrac{1}{2}\bigr) - \bigl(z-\tfrac{1}{2}\bigr)\log z - \tfrac{1}{2} + O\left(\frac{1}{\operatorname{Re} ...


2

Let: $$A_n\triangleq\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}.$$ Then: $$ A_n A_{n+1}=\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}=\frac{n}{2},$$ so $$ A_n\sim\sqrt\frac{n}{2},$$ since $\log\Gamma$ is convex.


1

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1

The best way to tackle such problems is to show that both members satisfy the same ODE with the same initial conditions. So, try to show that the scaled LHS satisfies the Legendre equation.


1

$$z_{n+1}=z_n^2+c$$ In Mandelbrot's sequence, $z_0$ is defined to be $0$. The definition of $z_{n+1}$ requires only $z_n$ to be defined, so a single starting value, $z_0$, is enough to define $z_n$ for all $n$. $$z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$$ In this sequence, to define $z_{n+2}$, you need both $z_{n+1}$ and $z_n$ to be defined. So, to define $z_n$ ...


1

In Progress... First, the sum of squared integrals inside the brackets, which for convenience we shall denote by $f{\left(x;\alpha\right)}$, may be rewritten as a single double integral with a little algebra and trigonometry: $$\begin{align} f{\left(x;\alpha\right)} ...



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