Consider the integral $$ G(x) = \int_0^\infty \frac{e^{-xt}}{1+t}dt$$ which is convergent for x>0.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# I Asymptotic expansion integral initial step

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