# Improper Integration

by Redoctober
Tags: improper, integration
 P: 41 Help :@ !! the question states that $$\int^{\infty}_{1}\frac{e^x}{x}.dx$$ Determine whether the integrand is convergent or divergent ?? It tried using the limit comparison test but i fail to select a g(x) to compare it with this i chose $$g(x) = \frac{e^x}{x +1}$$ but this hard to integrate too :S ! I can't find any other function to compare with cause i need to cancel out the e^x this seek for functions to compare the integral seem more like luck factor dependent lol !
 P: 41 Improper Integration I do know that e^x > than x when x>1 thus i know that the integral is divergent . but the question is asking me to prove it via the comparison test $$\lim_{x \to \infty } \frac{f(x)}{g(x)}$$
 P: 41 Its alright i found out the solution It turns out, because f(x) is divergent , i shall choose a smaller function g(x) but diverge too i chose $$g(x) = \frac{e^x}{e^x + 1}$$ $$\int^{\infty}_{1} \frac{e^x}{e^x + 1}.dx$$ the integral is easy to calculate . It will give infinity . Because the g(x) < f(x) and g(x) integral is divergent , thus f(x) integral is divergant too :D !