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

Re: xy.jax

$$\xymatrix @C=0.5pc @R=2.5pc{H_{n-1}(S^{n-1})\ar[dr]_{i_{*}}\ar[rr]^{id} & & H_{n-1}(S^{n-1})\\
& H_{n-1}(D^{n})\ar[ur]_{f_{*}} & }$$
by admin
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.
by admin
Sat Jun 03, 2017 7:48 pm
Forum: Calculus
Topic: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$
Replies: 2
Views: 1273

Re: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$

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...
by admin
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!
by admin
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...
by admin
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.
by admin
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 \(I=\pi \ln 4\).
by admin
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.
by admin
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.
by admin
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...