# Improper Integration

by Redoctober
Tags: improper, integration
 P: 40 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 !
 Sci Advisor P: 1,670 Try to use that e^x >= x when x >= 1.
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Well, isn't e^x greater than 1 for ALL positive values for x above x=1?
P: 40

## 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

$$\lim_{x \to \infty } \frac{f(x)}{g(x)}$$
 PF Patron P: 1,930 Compare it to g(x)=1, then.
 P: 40 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 !

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