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 Forum: Calculus   Topic: Series & Integral

 Post subject: Re: Series & Integral
Posted: Sat Jun 03, 2017 8:18 pm 

Replies: 3
Views: 829


Two more solutions for the first series can be found here.

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

Posted: Sat Jun 03, 2017 7:48 pm 

Replies: 2
Views: 741


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...

 Forum: Meta   Topic: New sub-forum for Measure and Integration theory

Posted: Sat May 20, 2017 1:16 pm 

Replies: 1
Views: 471


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!

 Forum: Calculus   Topic: Integral with polylogarithm

Posted: Sun Aug 21, 2016 8:28 pm 

Replies: 1
Views: 458


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 \...

 Forum: Real Analysis   Topic: Absolute convergence criterion

Posted: Thu Jul 14, 2016 10:44 am 

Replies: 1
Views: 381


A solution may be found here.

 Forum: Real Analysis   Topic: \( \int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx \)

Posted: Mon Jul 11, 2016 4:02 pm 

Replies: 1
Views: 369


Replied ex-member by aziiri : Set \(e^x-1=t^2\) then : \[I=\int_0^{\infty} \frac{x}{\sqrt{e^x-1}} \ \mathrm{d}x= \int_0^{\infty} \frac{2\ln(t^2+1)}{t^2+1} \ \mathrm{d}t\] Now set \(t=\tan y\) to get : \[I= -4\int_0^{\frac{\pi}{2}} \ln \cos y\ \mathrm{d}y\] The latter integral is widely known to be ...

 Forum: Real Analysis   Topic: A nice integral involving sum

Posted: Mon Jul 11, 2016 10:03 am 

Replies: 6
Views: 1040


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.

 Forum: Real Analysis   Topic: A nice integral involving sum

Posted: Mon Jul 11, 2016 9:59 am 

Replies: 6
Views: 1040


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.

 Forum: Real Analysis   Topic: A nice integral involving sum

Posted: Mon Jul 11, 2016 9:57 am 

Replies: 6
Views: 1040


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}}\\ &=........................\\ &=\...

 Forum: Real Analysis   Topic: \(\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\)

Posted: Sun Jul 10, 2016 4:09 pm 

Replies: 1
Views: 340


Replied by ex-member aziiri : \(1<a_n<2\) : we have \(1<a_1<2\). Suppose \(1<a_n<2\) then \(1<{\sqrt{2}\,}^{1}<a_{n+1}={\sqrt{2}\,}^{a_n}<{\sqrt{2}\,}^2=2\). \((a_n)_{n\geq 0} \) is increases. On \([1,2)\) define \(f(x)={\sqrt{2}\,}^x-x\) then \(f'(x)= 2^{x/2-1} \ln 2 -1 \leq f'(2)=\ln 2 -1 <0\) th...
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