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I am trying to integrate the following sum. I need to get at least first 5 terms (using math or computer). I've tryed wolfram alpha online-did not work. I should find $ \int_0^\infty \sum_{j=0}^\infty c_jx^{2j}e^{-\frac{x^2}{6}} \;dx, $ where $ c_j =\sum_{ m=0}^j\frac{b_m}{6^{j−m}(j−m)!} $ (here $b_m =0$ for 4m>(k−1)n ) and $b_m =(−1)^m n!\sum_{ n=n_1 +\cdots+n_k} \quad\sum_{m=n_2 +2n_3 +\cdots+(k−1)n_k}\frac{ 1}{\prod_{i=1}^{k}[(2i−1)!n_i !]} . $ (I.e. In the formula for $b_m$ the sum on the RHS is taken over all possible collections of subscripts $n_1 ,\ldots,n_k$ such that $n_1 +\cdots+n_k =n$ and $n_2 +2n_3 +3n_4 +\cdots+(k−1)n_k =m$ .)

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    Since the series is uniformly convergent, you can switch the integral and sum to get $\sum_{j=0}^{\infty}c_j\int_0^{\infty}x^{2j}e^{-\frac{x^2}{6}}$ and these terms should be relatively easy to compute (i.e. wolfram alpha can figure out each term).2019-02-19

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