Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Search found 27 matches

Thu Oct 03, 2019 10:07 am
Forum: LaTeX code testings
Topic: xy.jax
Replies: 4
Views: 1993

### Re: xy.jax

$$\xymatrix @C=0.5pc @R=2.5pc{H_{n-1}(S^{n-1})\ar[dr]_{i_{*}}\ar[rr]^{id} & & H_{n-1}(S^{n-1})\\ & H_{n-1}(D^{n})\ar[ur]_{f_{*}} & }$$
Sat Jun 03, 2017 8:18 pm
Forum: Calculus
Topic: Series & Integral
Replies: 3
Views: 1411

### Re: Series & Integral

Two more solutions for the first series can be found here.
Sat Jun 03, 2017 7:48 pm
Forum: Calculus
Topic: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$
Replies: 2
Views: 1273

A solution given by akotronis : Let $\displaystyle{S_{n}:=\sum_{k=1}^{n}(-1)^k\frac{\ln k}{k}}$. First observe that, by Dirichlet's criterion, the series converges, because $(-1)^k$ has bounded partial sums and $\displaystyle{\frac{\ln k}{k}}$ is eventually decreasing to $0$. Therefore $$\displaysty... Sat May 20, 2017 1:16 pm Forum: Meta Topic: New sub-forum for Measure and Integration theory Replies: 1 Views: 1141 ### Re: New sub-forum for Measure and Integration theory r9m wrote:Would it be a possible to have a sub-forum for 'Measure and Integration Theory' that is covered in most graduate classes? Bien sûr! Voila! Sun Aug 21, 2016 8:28 pm Forum: Calculus Topic: Integral with polylogarithm Replies: 1 Views: 820 ### Re: Integral with polylogarithm A solution by Seraphim We note that$$\zeta(m) - {\rm Li}_m(x)= \sum_{n=1}^{\infty} \frac{1-x^n}{n^m}Therefore \begin{align*} S &= \int_{0}^{1}\frac{\zeta(m)-{\rm Li}_m (x)}{1-x} \log^{m-1} x \, {\rm d}x \\ &= \sum_{n=1}^{\infty} \frac{1}{n^m} \int_{0}^{1}\frac{1-x^n}{1-x} \log^{m-1} x \, {\rm d... Thu Jul 14, 2016 10:44 am Forum: Real Analysis Topic: Absolute convergence criterion Replies: 1 Views: 791 ### Re: Absolute convergence criterion A solution may be found here. Mon Jul 11, 2016 4:02 pm Forum: Real Analysis Topic: $\int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx$ Replies: 1 Views: 689 ### Re: $\int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx$ Replied ex-member by aziiri: Set $e^x-1=t^2$ then : \ Now set $t=\tan y$ to get : \ The latter integral is widely known to be $\frac{-\pi\ln 2}{2}$, then the result is $I=\pi \ln 4$. Mon Jul 11, 2016 10:03 am Forum: Real Analysis Topic: A nice integral involving sum Replies: 6 Views: 1849 ### Re: A nice integral involving sum Replied by ex-member aziiri: Tolaso J Kos wrote:I don't know if induction works, but I believe that your solution can easily be modified for that case also. you just need to substitute $2014$ with $n$ in my solution. Mon Jul 11, 2016 9:59 am Forum: Real Analysis Topic: A nice integral involving sum Replies: 6 Views: 1849 ### Re: A nice integral involving sum Replied by ex-member aziiri: Tolaso J Kos wrote:Azirii, can we generalize that? That is, does the identity\int_{-\pi}^{\pi}\left ( \sum_{k=1}^{n}\sin \left ( kx \right ) \right )^2\,dx=n\pi hold?
Yes, of course.
Mon Jul 11, 2016 9:57 am
Forum: Real Analysis
Topic: A nice integral involving sum
Replies: 6
Views: 1849

### Re: A nice integral involving sum

Replied by ex-member aziiri : A possible first step is to use Lagrange's trigonometric identity: \begin{align*} \mathop{\sum}\limits_{k=1}^n {\sin(kx)}&=\frac{1}{2}\frac{\cos\frac{x}{2}-\cos\bigl({\bigl({n +\frac{1}{2}}\bigr) x}\bigr)}{\sin\frac{x}{2}}\\ &=........................\\ &=\frac{\sin\bi...