Asymptotic Expansion: First 2 Terms for Integral

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SUMMARY

The discussion focuses on finding the first two terms in the asymptotic expansion of the integral \int^{1}_{0}e^{-x/t}dt as x approaches zero from the right. Participants suggest expanding the integrand using a Taylor series, specifically e^{-x/t}=\sum^{\infty}_{k=0}\frac{(-x/t)^{k}}{k!}. The integral must be rewritten with a limit to address issues that arise as t approaches zero. The correct approach involves evaluating the limit lim_{a\rightarrow0^{+}}\int^{a}_{0}(1-\frac{x}{t}+\frac{x^{2}}{2t^{2}}-\frac{x^{3}}{6t^{3}}+\cdots)dt.

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Homework Statement



Find the first two terms in an asymptotic expansion of the following as x goes to zero from the right (i.e. takes on smaller and smaller positive values).

\int^{1}_{0}e^{-x/t}dt

Homework Equations


The Attempt at a Solution


I'm not exactly sure how to proceed with this. I'm assuming I should expand the integrand as a Taylor series:

e^{-x/t}=\sum^{\infty}_{k=0}\frac{(-x/t)^{k}}{k!}

I'm kinda stuck here, though.
 
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Notice you start having problems as t goes to zero... you should re-write your integral with a limit and then I think it would be best to actually expand the terms of your sum this time.
 
Zondrina said:
Notice you start having problems as t goes to zero... you should re-write your integral with a limit and then I think it would be best to actually expand the terms of your sum this time.

lim_{a\rightarrow0^{+}}\int^{a}_{0}(1-\frac{x}{t}+\frac{x^{2}}{2t^{2}}-\frac{x^{3}}{6t^{3}}+\cdots)dt

Like this? Hmm...
 

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