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Do you have any suggestions to solve the following integral:

$\int\limits_0^\infty {{e^{ - a{x^2}}}{Q_1}\left( {bx,cx} \right)dx}$

Thank you very much.

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  • $\begingroup$ it's an elliptic integral, no hope I think for an expression in elementary functions for general $a,b,c$ $\endgroup$ Commented Sep 24, 2014 at 12:07

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$\begin{array}{l} I = \int\limits_0^\infty {{e^{ - a{x^2}}}Q\left( {bx,cx} \right)dx} \\ = \sum\limits_{k = 0}^\infty {{{\left( {\frac{b}{c}} \right)}^k}\int\limits_0^\infty {{e^{ - \left( {a + \frac{{{b^2} + {c^2}}}{2}} \right){x^2}}}{I_k}\left( {bc{x^2}} \right)dx} } \\ = \sum\limits_{k = 0}^\infty {\sum\limits_{l = 0}^\infty {\frac{{{b^{2k + 2l}}{c^{2l}}}}{{{2^{k + 2l}}l!\Gamma \left( {l + k + 1} \right)}}\int\limits_0^\infty {{e^{ - \left( {a + \frac{{{b^2} + {c^2}}}{2}} \right){x^2}}}{x^{2k + 4l}}dx} } } \\ = \sum\limits_{k = 0}^\infty {\sum\limits_{l = 0}^\infty {\frac{{{b^{2k + 2l}}{c^{2l}}\Gamma \left( {0.5 + k + 2l} \right)}}{{{2^{k + 2l + 1}}l!\Gamma \left( {l + k + 1} \right){{\left( {a + \frac{{{b^2} + {c^2}}}{2}} \right)}^{0.5 + k + 2l}}}}} } \end{array}$

where I used [1, 9.6.10 and 9.6.12], [2, 3.326.2], and [3, 4.35].

[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 1970.

[2] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products - Seventh Edition, 2007.

[3] M. K. Simon, and M. -S. Alouini, Digital Communications over Fading Channels - Second Edition, 2005.

To evaluate the above infinite sum, we can put the upper limits (for k and l) sufficient large numbers. ^^

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Making a change in the sum $l+k=j$ we immediately evaluate this integral in terms of the Appell hypergeometric function, if the aim was to classify it via something known: $$ I=\frac{1}{2}\sqrt{\frac{\pi}{d}} F_2(1/2,1,1;1,1;x,y), $$ $$ d=a+\frac{b^2+c^2}{2}, x=\frac{b^2}{2d},y=\frac{c^2}{2d}. $$ May be this function may be also reduced to some known and more simple, I do not know.

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