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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.
 Post subject: Re: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$ 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 $$\displaysty...  Post subject: Re: New sub-forum for Measure and Integration theory Posted: Sat May 20, 2017 1:16 pm Replies: 1 Views: 471  r9m wrote:Would it be a possible to have a sub-forum for 'Measure and Integration Theory' that is covered in most graduate classes?Bien sûr! Voila! Forum: Calculus Topic: Integral with polylogarithm  Post subject: Re: Integral with polylogarithm Posted: Sun Aug 21, 2016 8:28 pm Replies: 1 Views: 458  A solution by Seraphim We note that$$\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  Post subject: Re: Absolute convergence criterion Posted: Thu Jul 14, 2016 10:44 am Replies: 1 Views: 381  A solution may be found here.  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  Post subject: Re: 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  Post subject: Re: 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

 Post subject: Re: 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}}\\ &=........................\\ &=\...
 Post subject: Re: $\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}$ Posted: Sun Jul 10, 2016 4:09 pm

Replies: 1
Views: 340

 Replied by ex-member aziiri : \(1
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