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.
Approximate an Integral using Taylor Expansion
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calculus
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0http://en.wikipedia.org/wiki/Watson%27s_lemma – 2012-12-26
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0So that approximation is valid if $\mu$ is large. What if $\mu$ is small? Are there any method to approximate that integral? – 2012-12-26