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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. –  Gerry Myerson Apr 26 '12 at 12:32
    
@Gerry: See resummation, in particular Borel resummation. –  joriki Apr 26 '12 at 12:34
    
@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"... –  J. M. Apr 26 '12 at 12:36
    
@J.M.: What are all the scare quotes and the qualification "peculiar" about? Resummation is a perfectly legitimate mathematical tool. –  joriki Apr 26 '12 at 12:41
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Defacing your questions is quite frowned upon; please don't do this. –  Arthur Fischer Mar 27 '13 at 10:35

2 Answers 2

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 $\mathrm{Ei}(z)$. –  J. M. Apr 26 '12 at 14:41
    
@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 you –  Bilgis77 Apr 27 '12 at 14:11
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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". –  Andrew Apr 27 '12 at 17:34
    
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... –  Bilgis77 Apr 29 '12 at 19:35

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. –  Bilgis77 Apr 26 '12 at 15:11
    
@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. –  J. M. Apr 27 '12 at 0:56
    
@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... –  Bilgis77 Apr 27 '12 at 13:04
    
@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. –  J. M. Apr 27 '12 at 13:16
    
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. –  Bilgis77 Apr 27 '12 at 13:37

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