# I Asymptotic expansion integral initial step

1. Oct 9, 2017

### spacetimedude

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. Oct 9, 2017

### mathman

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