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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.

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    I don't know what you mean by re-sum. It seems to me that the terms are not going to zero, so the series does not converge.2012-04-26
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    @Gerry: See [resummation](http://en.wikipedia.org/wiki/Resummation), in particular [Borel resummation](http://en.wikipedia.org/wiki/Borel_resummation).2012-04-26
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    @Gerry: I'm guessing he means to ask if there's a way to make the divergent series he has to "make sense", e.g. having $1+2+3+\cdots$ be "equal" to $-\frac1{12}$, for some peculiar definition of "equal"...2012-04-26
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    @J.M.: What are all the scare quotes and the qualification "peculiar" about? Resummation is a perfectly legitimate mathematical tool.2012-04-26
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    @joriki: I know it's legitimate (I've used it a lot), but most people aren't used to it... it's peculiar in that sense.2012-04-26
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    Maybe if you mean what joriki and J. M. think you mean, you could edit that information into your question.2012-04-26
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    @Gerry Motion seconded.2012-04-26
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    Defacing your questions is quite frowned upon; please don't do this.2013-03-27

2 Answers 2

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

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    Nice! (Will upvote when I can do so again). Note that $f(z)$ can be expressed in terms of the exponential integral $\mathrm{Ei}(z)$.2012-04-26
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    @Andrew, how did you or where did you get this euler asymptotic expansion formula? I was not able to get it. In Euler's formula, dont we have the summation ,vice versa,like this but k! is in the denominator and z is in the numerator? Thank you2012-04-27
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    @Bilveys77 It is usually found in textbooks as the first example of asymptotic expansion. For example, in Erdelyi A., "Asymptotic expansions".2012-04-27
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    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
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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: $$\begin{align} &\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\ &\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) + i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1) \end{align}$$

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    GEdgar and Andrew thank you for your helps. As you might seen easily, you just did how I get this summation. It comes from the solution of a^2*y"+y=1/(1+x^2) equation and by using this resummation or by solving this question I want to have at least approximate solution of this equation.2012-04-26
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    @Bilveys77: what will you be using your approximation for? A lot of the modern computing environments are able to evaluate the exponential integral for complex argument. If yours doesn't and you want to implement a numerical method, then that's a different can of worms.2012-04-27
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    @J.M. I am going to find the integral representation of the resummation,maybe gamma or by any other special function. So that I will atleast find the approximate solution of the differential equation. i believe it would be very easy to calculate by Residue...2012-04-27
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    @Bilveys: "find the integral representation of the resummation" - but Andrew already (implicitly) gave you one, using his $f(z)$... if you want to make an approximation from that, you can build a Padé approximant from the series.2012-04-27
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    J.M. and @Andrew , in that equation should g((i+a)^−1) be just g((i+a))? because z=(a+i) and z=(a-i),respectively.2012-04-27
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    For the differential equation $a^2 y"+y=1/(1+x^2)$: Maple finds its solution in terms of Si and Ci functions. Presumably it is what you get using variation of parameters.2012-04-27
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    @GEdgar yes, maple and matlab give this solution in terms of Si, Ci also Ei (exponential integral) but it does not make sense. Because 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