1
$\begingroup$

I wonder if there is a way to get resummation of this series? By this way , i am trying to get the integral representation of this series, it could be by Gamma function.

$\sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]$

thank you.

  • 1
    Defacing your questions is quite frowned upon; please don't do this.2013-03-27

2 Answers 2

3

One can start with the Euler asymptotic expansion $ f(z)=\int_0^\infty\frac{e^{-t}}{z+t}\,dt=\sum_{k=0}^\infty(-1)^k\frac{{k!}}{z^{k+1}}. $ Putting $ g(z)=\frac{f(-iz)-f(iz)}{2i}= \sum _{k=0}^{\infty} (-1)^k\frac{(2k)!}{z^{2k+1}} $ we have formally $ g((i+a)^{-1})+g((-i+a)^{-1})= \sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]. $

  • 0
    Then from this$f(z)$function we get the Exponential integral again as @GEdgar 's way. I believe we cannot formulate/represent exponential integral by Gamma function . This is the point these kinda integrals do not have certain answer...2012-04-29
2

It looks like Andrew beat me to it

There is no guarantee that the following makes sense... But...

There is this Borel summation $ \sum_{j = 0}^{\infty} \frac{(-1)^{j} j!}{x^{j + 1 }} = -\operatorname{e} ^{x} \mathrm{Ei} (-x) \tag{1} $ Here, $\mathrm{Ei}$ is the exponential integral. Put $-x$ for $x$, $ \sum_{j = 0}^{\infty} \frac{j!}{x^{j + 1}} = \operatorname{e} ^{-x} \mathrm{Ei} (x) \tag{2} $ Add (1) and (2) $ \sum_{k = 0}^{\infty} \frac{(2 k)!}{x^{2 k + 1}} = -\operatorname{e} ^{x} \mathrm{Ei} (-x) + \operatorname{e} ^{-x} \mathrm{Ei} (x) $ Put $x=iz$, $ \sum_{k = 0}^{\infty} \frac{(-1)^{k} (2 k)!}{z^{(2 k + 1)}} = -i\operatorname{e} ^{i z} \mathrm{Ei} (-iz) + i \operatorname{e} ^{-iz} \mathrm{Ei} (i z) $ Put $z=a+i$ and $z=a-i$ and subtract: $\begin{align} &\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\ &\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) + i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1) \end{align}$

  • 0
    @GEdgar yes, maple and matlab give this solution in terms of Si, Ci also Ei (exponential integral) but it does not make sense. $B$ecause even if we cannot get any certain answer by Si,Ci integrals, we need to at least get approximate answer. That is why by approximation I get this above summation and now I need to re-sum this summation; so that by integral representation I am going to represent it by integral and maybe by way of residue I am going to find its approximate solution;which is what i want, then I am going to take limit as a goes zero to see how it behaves as a goes zero. Many thanks2012-04-27