Search found 308 matches

by Grigorios Kostakos
Tue Jun 13, 2017 6:22 am
Forum: Calculus
Topic: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
Replies: 3
Views: 3788

\(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)

Calculating the series $\sum_{m=1}^{+\infty}\frac{(-1)^{m+1}}{m+n-1}$, I get the half of \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\). Can you suggest an elegant and simple formula for \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\) or \(\psi\big(n+\frac{1}{2}\big)-\psi(n)\), ...
by Grigorios Kostakos
Sun Jun 11, 2017 4:18 pm
Forum: Calculus
Topic: \(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)
Replies: 1
Views: 2975

\(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)

As a continuation of this:

For $0<\alpha<\beta$, evaluate
\[\displaystyle\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\,.\]
by Grigorios Kostakos
Sat Jun 10, 2017 9:22 pm
Forum: Calculus
Topic: \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\)
Replies: 2
Views: 3442

Re: \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\)

Another solution using complex analysis: We consider the complex function $f(z)=\frac{{\rm{Log}}^2{z}}{(z-\alpha)(z-\beta)}\,,\quad z\in\mathbb{C}\setminus\{x+0i\;|\;x\leqslant0\}$, which is meromorphic at \({\mathbb{C}}\setminus\{x+0i\;|\;x\leqslant0\}\) with simple poles $z=\alpha$, $z=\beta$ and...
by Grigorios Kostakos
Wed Jun 07, 2017 11:11 am
Forum: Calculus
Topic: \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\)
Replies: 2
Views: 3442

\(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\)

For $0<\alpha<\beta$, evaluate
\[\displaystyle\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\,.\]
by Grigorios Kostakos
Tue May 30, 2017 11:54 am
Forum: Real Analysis
Topic: Identity function
Replies: 0
Views: 2041

Identity function

Let $h\,:\mathbb{R}\longrightarrow\mathbb{R}$ be a function defined and valued in all real numbers, strictly increasing and continuous. We define the functions $h^{[n]}:\mathbb{R}\longrightarrow\mathbb{R}\,,\; {n}\in\mathbb{Z}$, as $$h^{[n]}(x)=\left\{{\begin{array}{ll} \mathop{(\underbrace{h\circ{h...
by Grigorios Kostakos
Fri May 26, 2017 3:52 am
Forum: Competitions
Topic: Inequality
Replies: 0
Views: 2846

Inequality

Let $a,b,c$, and $d$ be strictly positive real numbers such that the arithmetic mean of the six numbers $ab,\, ac,\, ad,\, bc,\, bd,\, cd$ equals to the square of their geometric mean. Prove that $$a+b+c+d - 2\sqrt{2}\,abcd\leqslant 4 - 2\sqrt{2}\,.$$
When is the equality attained?
by Grigorios Kostakos
Thu May 18, 2017 11:24 am
Forum: Algebraic Structures
Topic: Groups of order $2p$
Replies: 2
Views: 3866

Re: Groups of order $2p$

By Cauchy's theorem for the group $G$ of order $2p$, $p$ prime, we have that there is an element $a\in G$ of order $p$. But then every element $a^{k}\,,\;k\in\{1,2,\ldots,p-1\}$ of the subgroup $\langle{a}\rangle$ has order $p$. By Sylow's 3rd theorem we have that the number of the subgroups of $G$ ...
by Grigorios Kostakos
Thu May 18, 2017 1:51 am
Forum: Algebraic Structures
Topic: Centre of dihedral group
Replies: 1
Views: 2792

Re: Centre of dihedral group

All the distinct elements of the $n$-dihedral group $${\cal{D}}_{2n}=\left\langle{\rho,\,\tau \ | \ \rho^{n}=\tau^2={\rm{id}}, \ \tau\rho\tau=\rho^{n-1}}\right\rangle$$ are of the form $\rho^{k}\tau^{m}\,,\; k\in\{0,1,\ldots,n-1\},\; m\in\{0,1\}$. For $n=1$, we have $Z({\cal{D}}_{2})=\{{\rm{id}},\ta...
by Grigorios Kostakos
Mon May 15, 2017 5:27 pm
Forum: Complex Analysis
Topic: Line integral 01
Replies: 1
Views: 3182

Line integral 01

Evaluate $$\displaystyle\oint_{\gamma}{{\rm{Log}}\,\big(z-\tfrac{1}{2}\big)\,dz}\,,$$ where $\gamma$ has parametric representation $$\gamma(t)=\displaystyle(1-t)\cos({t\pi})+\frac{1}{2}\,\big(-1+(2t-4)\sin(t\pi)\big)\,i\,,\quad t\in\big[-\tfrac{1}{2},\tfrac{1}{2}\big]\,.$$
by Grigorios Kostakos
Wed May 03, 2017 7:50 pm
Forum: Calculus
Topic: \(\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\)
Replies: 5
Views: 6698

Re: \(\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\)

Nice solution mathofusva! Here is a second one: \begin{align*} I&=\displaystyle\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\\ &=\int_{-1}^1 \log(1-x)\,\Big(\log2+\mathop{\sum}\limits_{n=1}^{+\infty}\frac{(-1)^{n-1}}{2^n\,n}\,(x-1)^n\Big)\, dx\\ &=\log2\int_{-1}^1 \log(1-x)\,dx+\int_{-1}^1 \log(...