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I am doing evaluation of integral by using Taylor expansion. The integral is: $ {I} = \int_0^\infty {{e^{ - {\mu}x}}g\left( x \right)dx} $ where $ g\left( x \right) = \ln \left( {\sum\limits_{i = 1}^n {{\alpha _i}{e^{ - {\mu _i}x}}} } \right) $ with $\mu,\mu_i, \alpha_i$ are possitive real numbers. Using Taylor Series Expansion, I obtain $ g\left( x \right) = \sum\limits_{l = 0}^R {\frac{{{g^{\left( l \right)}}\left( 0 \right){x^l}}}{{l!}}} + {O_R} $ where $O_R$ is the remainder. Then $I$ can be approximated as $ {I} \approx \int_0^\infty {{e^{ - {\mu }x}}\sum\limits_{l = 0}^R {\frac{{{g^{\left( l \right)}}\left( 0 \right){x^l}}}{{l!}}} dx} = \sum\limits_{l = 0}^R {\frac{{{g^{\left( l \right)}}\left( 0 \right)}}{{\mu^{l + 1}}}} $ The point is I don't know whether these series are convergent. So my proof maybe not valid. If you guys have any ideas, please suggest.

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    So that approximation is valid if $\mu$ is large. What if $\mu$ is small? Are there any method to approximate that integral?2012-12-26

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