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    #11
    Quote Originally Posted by ZaidAlyafey View Post
    5- $ \displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}$
    I have seen a series like this before. The simple (but unsatisfying) explanation is that it must converge by the alternating series test, which may be extended to such unconventionally oscillating terms.

    However, I'm sure there's a more elegant underlying structure if you use some decomposition of Euler's formula. Having suggested it, I will look into it if I have the inclination later.

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    #12
    Quote Originally Posted by ZaidAlyafey View Post
    5- $ \displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}$
    According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

    $$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

    ... we arrive to write...

    $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$

    Kind regards

    $\chi$ $\sigma$

  3. زيد اليافعي
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    #13 Thread Author
    Quote Originally Posted by chisigma View Post
    According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

    $$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

    ... we arrive to write...

    $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$

    Kind regards

    $\chi$ $\sigma$
    Yes, I think also it is solvable by Fourier series .

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    #14
    Quote Originally Posted by chisigma View Post
    According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

    $$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

    ... we arrive to write...

    $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$
    To take that a bit further, the expression in (2) can be simplified as $$\tan^{-1} \Bigl(\frac{\sin x}{1-\cos x}\Bigr) = \tan^{-1} \biggl(\frac{2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}\biggr) = \tan^{-1}\bigl(\cot\tfrac x2\bigr) = \tfrac\pi2 - \tfrac x2.$$ Hence $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = \tfrac12(\pi-x).$$ But that only works provided that $0<x< 2\pi$. At the endpoints of the interval, when $x=0$ or $2\pi$, the sum $\sum _{n=1}^{\infty} \frac{\sin n x}{n}$ is obviously $0$ (since each term vanishes).

    As ZaidAlyafey points out, this sum is a Fourier series, namely for the function $\tfrac12(\pi-x)$ over the interval $[0,2\pi].$

  5. زيد اليافعي
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    #15 Thread Author
    Quote Originally Posted by Opalg View Post
    To take that a bit further, the expression in (2) can be simplified as $$\tan^{-1} \Bigl(\frac{\sin x}{1-\cos x}\Bigr) = \tan^{-1} \biggl(\frac{2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}\biggr) = \tan^{-1}\bigl(\cot\tfrac x2\bigr) = \tfrac\pi2 - \tfrac x2.$$ Hence $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = \tfrac12(\pi-x).$$ But that only works provided that $0<x< 2\pi$. At the endpoints of the interval, when $x=0$ or $2\pi$, the sum $\sum _{n=1}^{\infty} \frac{\sin n x}{n}$ is obviously $0$ (since each term vanishes).

    As ZaidAlyafey points out, this sum is a Fourier series, namely for the function $\tfrac12(\pi-x)$ over the interval $[0,2\pi].$
    Since the series converges for all $x$ , there may be a general solution that works for all $x$ , right ?

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    #16
    Quote Originally Posted by ZaidAlyafey View Post
    Since the series converges for all $x$ , there may be a general solution that works for all $x$ , right ?
    Yes, it is the $2\pi$-periodic function defined on the interval $[0,2\pi)$ by $f(x) = \begin{cases}0&(x=0),\\ \frac12(\pi-x)&(0<x<2\pi). \end{cases}$ It is an example of what is often called a .

  7. زيد اليافعي
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    #17 Thread Author
    6- $ \displaystyle \int^{\infty}_0 \frac{x}{\sqrt{e^x-1}}\, dx$

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    #18
    $\displaystyle \int_{0}^{\infty} \frac{x}{\sqrt{e^{x}-1}} \ dx = \int_{0}^{\infty} \frac{x e^{- \frac{x}{2}}} {\sqrt{1-e^{-x}}} \ dx $

    Let $ \displaystyle u = e^{-\frac{x}{2}}$

    $ \displaystyle = - 4 \int_{0}^{1} \frac{\ln{u}}{\sqrt{1-u^{2}}} \ du $

    Let $v = \arcsin u$

    $ \displaystyle = -4 \int^{\frac{\pi}{2}}_{0} \ln (\sin v) \ dv $


    I'm sure we've all seen evaluations of that last integral. So I'm just going to argue that it converges.


    The only potential issue is at $x=0$.

    But $\displaystyle \ln (\sin x) = \ln \Big( \frac{x \sin x}{x} \Big) = \ln(x) + \ln \Big(\frac{\sin x}{x} \Big) $.

    So near $x=0$, $ \ln(\sin x)$ behaves like $\ln x$, and thus $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx $ converges.

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    #19
    7- $ \displaystyle \int_{0}^{\infty} \frac{\ln (\tan^{2} x)}{1+x^{2}} \ dx$

    8- $ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $
    Last edited by Random Variable; May 29th, 2013 at 14:47.

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    #20
    Quote Originally Posted by TheBigBadBen View Post
    I have seen a series like this before. The simple (but unsatisfying) explanation is that it must converge by the alternating series test, which may be extended to such unconventionally oscillating terms.

    However, I'm sure there's a more elegant underlying structure if you use some decomposition of Euler's formula. Having suggested it, I will look into it if I have the inclination later.
    Why do you consider the Alternating Series test not elegant? I personally find the most simple solution to be the most elegant, because it is the most likely to be understood by others...

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