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I have to evaluate the series:

$$\sum_{n=1}^{\infty}\frac{\sin(\frac{\pi a}{a+b}n)}{n^3}+\frac{\sin(\frac{\pi b}{a+b}n)}{n^3}$$

Where $a$ and $b$ are real numbers.

Since I'm not very good with series I tried brute force by using a complex analysis method which involves multiplying by the cotangent and calculating the residue at $n=0$ (I don't know the name of this method, sorry!), but I didn't get the result my teacher gave me.

I figured this is the Fourier series expansion of some odd function, but I don't know how to guess said function.

So I was wondering if someone could give me some advice.

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  • $\begingroup$ You mean you tried the (Cauchy's) Residue Theorem perhaps? $\endgroup$ – MathematicianByMistake Mar 10 '16 at 14:07
  • $\begingroup$ It's based on that, it's the method mentioned (for example) in this question: math.stackexchange.com/questions/575750/… $\endgroup$ – DR10 Mar 10 '16 at 14:48
  • $\begingroup$ At least we can tell the series is not divergent since $\sin(x)\leq1 \forall x$ $\endgroup$ – Octania Mar 10 '16 at 15:47
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    $\begingroup$ The reason that the proposed methodology in the OP doesn't work here is that we need to find a function $f$ such that $\lim_{N\to \infty}\oint_{C_N}f(z) \cot(z)\,dz=0$, where $C_N$ is the classical square contour centered at the origin with sides of length $2N+1$. But the function $\frac{\sin(\alpha z)}{z^3}$, $\alpha = \pi a/(a+b)$ or $\alpha =\pi b/(a+b)$ is unbounded on that contour. In fact, $\frac{e^{i \alpha z}}{z^3}$ is unbounded in the lower-half plane for $\alpha >0$ on $C_N$. So, it appears that another tact is required. $\endgroup$ – Mark Viola Mar 10 '16 at 20:14
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Let's rewrite the series as

$$ S(a,b)=\Im\sum_{n=1}^{\infty}\left[\frac{(e^{\pi i a/(a+b)})^n}{n^3}+\frac{(e^{\pi i b/(a+b)})^n}{n^3}\right] $$

Now we can apply the definition of the polylogarithmic functions

$$ \text{Li}_m(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^m} $$

we therefore obtain

$$ S(a,b)=\Im\left[\text{Li}_3(e^{\pi i a/(a+b)})+\text{Li}_3(e^{\pi i b/(a+b)})\right] \quad (*) $$

which is i fear the best one can do for arbritary $a,b$. For example if $a,b $ are natural numbers we may rewrite this as a finite sum over Hurwitz Zeta values.

An alternative representation is terms of Clausen functions $\text{Si}_m(z)$

$$ S(a,b)=\text{Si}_3(e^{\pi i a/(a+b)})+\text{Si}_3(e^{\pi i b/(a+b)}) $$

For the special values $a=b$, $a=0$ we might find the particulary nice results

$$ S(a,a)=\frac{\pi^3}{16} \quad \\ S(0,b)=0 \quad $$

All defintions i used and much more can be found here

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    $\begingroup$ And that is all she wrote. +1 ... One could make use of the identity $z-\bar z=2i \text{Im}(z)$ to get rid of the imaginary part operator. But that doesn't do much. And one might also observe that $e^{i\pi b/(b+a) }=-1e^{-i\pi a/(a+b)}$, but that doesn't help much either. $\endgroup$ – Mark Viola Mar 10 '16 at 17:31
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    $\begingroup$ @Dr. MV Good Jesus i was taking an absolute value :( $\endgroup$ – tired Mar 10 '16 at 19:15
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    $\begingroup$ I thought so. It was a valid attempt. I really don't think its expressible in terms of elementary functions for general $a$ and $b$. $\endgroup$ – Mark Viola Mar 10 '16 at 19:20
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    $\begingroup$ @Dr.MV shocking stupidity from my side, i'm very sorry! At least one could rewrite it in terms of on trilog $\endgroup$ – tired Mar 10 '16 at 19:21
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    $\begingroup$ Tired, it wasn't stupidity. I call this error type an oversight or in my case carelessness (or even wrecklessness). And no need at all to apologize. - Mark $\endgroup$ – Mark Viola Mar 10 '16 at 19:23

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