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I am trying to compute the sum of the following series $$\sum _{k=0}^{\infty }\frac{\left(2it (1-H)^{2 (1-H)} \left(\frac{H}{\mu}\right)^{2 H} \right)^k \Gamma \left(\frac{k}{2 (1-H)}+\frac{1}{2 H}\right)}{k!}$$ where $0 < H < 1$, $\mu > 0$ and $t \in \mathbb{C}$, but I am unsure how to proceed forward.

Any suggestions/help would be much appreciated.

Edit I am trying to evaluate a characteristic function for a probability density. Below is the derivation I have so far The Characteristic function is given by \begin{eqnarray}\nonumber \phi_Q\left(t\right) &=& \mathrm {E}\left[\exp \left(itQ\right)\right]\\\nonumber &=& \int_{-\infty}^{\infty} \exp \left(itx\right)f_Q\left(x\right) \mathrm{dx}\\\nonumber &=& c \int_{0}^{\infty} \exp \left(itx\right)x^{\frac{1}{H}-2} \exp \left(-\frac{1}{2} (1-H)^{2 (H-1)} H^{-2 H} \mu ^{2 H} x^{2-2 H}\right) \mathrm{dx}\\\nonumber \end{eqnarray} where $c = \frac{\mu \left(1-H\right)^{\left(2 - \frac{1}{H}\right)}{2^{\left(1 - \frac{1}{2H}\right)}}}{H\Gamma\left(\frac{1}{2H}\right)}$

Upon substituting $y = x^{2-2 H}$, $\mathrm{dx}=\frac{y^{\frac{2H - 1}{2\left(1-H\right)}}}{2\left(1-H\right)}\mathrm{dy}$, $\zeta = \frac{1}{H}$ and $\eta = \frac{H^{\left(2H-1\right)}}{\mu^{2H}\left(1-H\right)^{2\left(H-1\right)}}$ we get \begin{eqnarray}\nonumber \phi_Q\left(t\right) &=& \frac{c}{2\left(1-H\right)}\int_{0}^{\infty}\exp \left(ity^{\frac{1}{2\left(1-H\right)}}\right)y^{\frac{\zeta}{2} - 1}\exp \left(-\frac{\zeta y}{2\eta} \right) \mathrm{dy}\\\nonumber \end{eqnarray} Upon substituting Taylor's series expansion of $\exp \left(ity^{\frac{1}{2\left(1-H\right)}}\right)$ we get \begin{eqnarray}\nonumber \phi_Q\left(t\right) &=& \frac{c}{2\left(1-H\right)}\int_{0}^{\infty}\sum\limits_{k=0}^{\infty}\frac{\left(ity^{\frac{1}{2\left(1-H\right)}}\right)^k}{k!} y^{\frac{\zeta}{2} - 1}\exp \left(-\frac{\zeta y}{2\eta} \right) \mathrm{dy}\\\nonumber &=& \frac{c}{2\left(1-H\right)}\sum\limits_{k=0}^{\infty}\frac{\left(it\right)^k}{k!}\int_{0}^{\infty} y^{{\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2} - 1}\exp \left(-\frac{\left(\frac{k}{1-H} + \zeta\right) y}{2\eta \left(\frac{\frac{k}{1-H} + \zeta}{\zeta}\right)} \right) \mathrm{dy}\\\nonumber &=&\frac{c}{2\left(1-H\right)}\sum\limits_{k=0}^{\infty}\frac{\left(it\right)^k}{k!}2^{\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) }\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) \left(\frac{\zeta}{\eta}\right)^{-\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right)}\\\nonumber & &\int_{0}^{\infty} \frac{\left(\frac{\zeta}{\eta}\right)^{\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right)}}{2^{\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) }\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right)} y^{{\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2} - 1}\exp \left(-\frac{\left(\frac{k}{1-H} + \zeta\right) y}{2\eta \left(\frac{\frac{k}{1-H} + \zeta}{\zeta}\right)} \right) \mathrm{dy}\\\nonumber \end{eqnarray} As the integrand is the pdf of a Gamma Random variable, the integral resolves to 1 and upon using the definition of $c$ we get \begin{eqnarray}\nonumber \phi_Q\left(t\right) &=& \frac{\left(\frac{\zeta}{\eta}\right)^{\frac{\zeta}{2}}}{2^{\frac{\zeta}{2}}\Gamma \left(\frac{\zeta}{2}\right)}\sum\limits_{k=0}^{\infty}\frac{\left(it\right)^k}{k!}2^{\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) }\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) \left(\frac{\zeta}{\eta}\right)^{-\left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right)}\\\nonumber &=& \frac{1}{\Gamma \left(\frac{\zeta}{2}\right)}\sum\limits_{k=0}^{\infty}\frac{\left(it\right)^k}{k!}2^{{\frac{k}{2\left(1-H\right)}} }\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) \left(\frac{\zeta}{\eta}\right)^{\frac{-k}{2\left(1-H\right)}}\\\nonumber &=& \frac{1}{\Gamma \left(\frac{\zeta}{2}\right)}\sum\limits_{k=0}^{\infty}\frac{\left(it\right)^k}{k!}\left(\frac{2\eta}{\zeta}\right)^{{\frac{k}{2\left(1-H\right)}} }\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) \\\nonumber &=& \frac{1}{\Gamma \left(\frac{\zeta}{2}\right)}\left(\frac{2\eta}{\zeta}\right)^{{\frac{1}{2\left(1-H\right)}}}\sum\limits_{k=0}^{\infty}\frac{1}{k!}\left(\frac{2\eta it}{\zeta}\right)^k\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right) \\\nonumber \end{eqnarray} Upon substituting back the values for $\zeta$ and $\eta$ I got the series in my original question. Suggestions to side step this series evaluation are also welcome.

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    $\begingroup$ Is there any motivation or context you can provide for this monster? $\endgroup$
    – David H
    Dec 24, 2014 at 22:21
  • $\begingroup$ @DavidH I have edited my question to provide more context. $\endgroup$ Dec 24, 2014 at 22:58

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In fact, if you know how to express the following series $$ \sum_{k=0}^{\infty}\frac{\Gamma(ak+b)}{k!}x^k \tag1 $$ in terms of a known special function, then it is OK.

Using Mathematica, we obtain that $$ \sum_{k=0}^{\infty}\frac{\Gamma(ak+b)}{k!}x^k = 2^{1-b} e^{x^2/8} \Gamma(2b) D_{-2b}(-2 b,-x/\sqrt{2}) \tag2 $$ where $x \mapsto D_\nu(x)$ is the parabolic function which satisfies the differential equation $$ y''+(\nu+\frac12-\frac{x^2}{4})\:y=0, \quad y(0)=y'(0)=\Gamma(b). \tag3 $$ Finally, using your computation, you may rewrite the charateristic function for your probability density as $$ \phi_Q\left(t\right)=\frac{\Gamma(\zeta) }{\Gamma \left(\frac{\zeta}{2}\right)}\left(\frac{2\eta}{\zeta}\right)^{{\frac{1}{2\left(1-H\right)}}}2^{1-\frac{\zeta}{2}} e^{-\left(\eta t\right)^2/(2\zeta)} D_{-\zeta}\left(-\zeta,-\frac{\sqrt{2}\eta it}{\zeta}\right). $$ Hoping it helps.

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  • $\begingroup$ Hi Oliver, thank you for your answer. I am new to mathematica, so it would be very helpful if u could please share the commands you used to carry out this analysis. Also can you please add a comment about what $D$ and $v$ represent. Thank you very much for your help. $\endgroup$ Dec 25, 2014 at 0:39
  • $\begingroup$ Never mind oliver upon re reading I know what d and v are. $\endgroup$ Dec 25, 2014 at 0:43

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