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I Asymptotic expansion integral initial step

  1. Oct 9, 2017 #1
    Consider the integral $$ G(x) = \int_0^\infty \frac{e^{-xt}}{1+t}dt$$ which is convergent for x>0.
    For large x, it is dominated by small t so expand:
    $$G(x) = \int_0^\infty e^{-xt}\sum_{m=0}^{\infty}(-t)^mdt$$

    From here my notes say to take out the summation and write:
    $$G(x) = \sum_{m=0}^{\infty}\frac{(-1)^m}{x^{m+1}}\int_0^\infty e^{-t}t^mdt$$.
    I understand that ##\sum_{m=0}^{\infty}(-1)^m## comes from the previous sum on ##(-t)^m## but I'm not sure where the denominator came from. Surely, it's from the exponential but I don't understand how it came in the denominator.

    Any help will be appreciated.
  2. jcsd
  3. Oct 9, 2017 #2


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    [tex]\int_0^\infty e^{-xt} t^mdt=\frac{1}{x^{m+1}}\int_0^\infty e^{-y} y^mdy,\ where\ y=xt[/tex].
    Since y is a dummy, you can call it t.
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