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I would like to have a better form of this series:


where $Q_m(\alpha,\beta)$ is the generalized Marcum Q-function. Does anyone an idea on how to proceed, if possible to make it better (e.g. getting rid of the series)? Thanks!

EDIT: could the result be just proportional to $Q_{\frac{3}{2}}\left(ab,bx\right)$???

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up vote 1 down vote accepted

Well, I guess, in some form of. But at the same time you can get a closed form solution.
For simplicity let's set $\alpha=\left(\frac{ab\sin(c)}{\sqrt{2}}\right)^{2}$ and $\beta=ab\cos(c)$
Taking into account the definition of the generalized Marcum Q-function: $$ \begin{eqnarray} \sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right)&=& \sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,\int_{bx}^{\infty} t \left( \frac{t}{\beta}\right)^{k+\frac{1}{2}} \exp \left( -\frac{t^2 + \beta^2}{2} \right) I_{k+\frac{1}{2}} \left( \beta t \right) \ \mathrm dt\\ &=&\frac{\exp \left(\!-\!\frac{\beta^2}{2}\! \right)}{\sqrt{\beta}}\int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}\sum_{k=0}^{\infty}\frac{\left(\frac{\alpha t}{\beta}\right)^k}{k!}I_{k+\frac{1}{2}}(\beta t) \ \mathrm dt\\ \end{eqnarray} $$ Then one can use Gradshteyn and Ryzhik's Table of Integrals, Series, and Products (Second Part): $$\sum_{k=0}^{\infty}\frac{\xi^k}{k!}I_{k+\nu}(\eta)=\left(\frac{2\xi}{\eta}+1\right)^{-\frac{\nu}{2}}I_{\nu}\left(\sqrt{\eta^2+2\eta\xi}\right), \qquad |2\xi|<|\eta|$$ So $$ \begin{eqnarray} \sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right) =\frac{\exp \left(\!-\!\frac{\beta^2}{2}\! \right)}{\sqrt{\beta}}\left(1+\frac{2\alpha}{\beta^2}\right)^{-\frac{1}{4}} \int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}I_{\frac{1}{2}}(\sqrt{2\alpha+\beta^2}|t|) \ \mathrm dt=\\ =\text{A} \ \left[\sqrt{\pi } \sqrt{2 \alpha +\beta ^2} e^{\alpha +\frac{1}{2} \left(b^2 x^2+\beta ^2\right)+b x \sqrt{2 \alpha +\beta ^2}}\left(\text{erfc}\left(\frac{ \sqrt{2 \alpha +\beta ^2}+b x}{\sqrt{2}}\right)+\text{erfc} \left(\frac{b x-\sqrt{2 \alpha +\beta ^2}}{\sqrt{2}}\right) \right)+\sqrt{2} \left(e^{2 b x \sqrt{2 \alpha +\beta ^2}}-1\right)\right] \end{eqnarray} $$ where $\text{A}=\frac{\exp \left(-\frac{\beta ^2}{2}-\frac{1}{2} b x \left(2 \sqrt{2 \alpha +\beta ^2}+b x\right)\right)}{2 \sqrt{\pi } \sqrt{\left(2 \alpha +\beta ^2\right) \text{sgn}(\beta)}}$ The last equation was obtained via Mathematica (or you can turn to Gradshteyn and Ryzhik ;)). The equation was simplified assuming that $\alpha>0, \ x>0, \ \beta\in\mathbb{R}$. If not the result gets nastier (with lots of square roots). I guess you can turn the last integral into some form of the generalized Marcum Q-function, if you'd like to:

$$\int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}I_{\frac{1}{2}}(\sqrt{2\alpha+\beta^2}|t|) \ \mathrm dt=(2\alpha+\beta^2)^{\frac{1}{4}}e^{\frac{|2\alpha+\beta^2|}{2}}Q_{\frac{3}{2}}\left(\sqrt{2\alpha+\beta^2},bx\right)$$ So $$\sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right)=e^{\alpha}Q_{\frac{3}{2}}\left(\sqrt{2\alpha+\beta^2},bx\right)$$ Noting that $\sqrt{2\alpha+\beta^2}=a \ b $ the result turns out to be even simplier.

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Thanks! there is anyway a mistake in the power of $t$ inside the integral. – JFNJr Jul 7 '13 at 18:15
Correcting the power inside the integral, it turns out to be really what I expected,i.e. proportional to $Q_{\frac{3}{2}}(ab,bx)$. Awesome! Thanks again! – JFNJr Jul 7 '13 at 18:34
Thanks, I fixed it, guess if I'm not loosing anything else, the answer should be correct. :) – Caran-d'Ache Jul 7 '13 at 18:43
Many thanks! almost correct. The $\frac{1}{\sqrt{\beta}}$ cancels out, instead of giving $\frac{1}{\beta}$. Thanks again! – JFNJr Jul 7 '13 at 18:47
Yep! Overlooked it again. :( Too many constants, too poor eyesight. – Caran-d'Ache Jul 7 '13 at 18:50

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