- #1
spacetimedude
- 88
- 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.
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.