Solve Improper Integration: Integrand Convergence

[Redoctober]

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 !

 

[disregardthat]

Try to use that e^x >= x when x >= 1.

 

[arildno]

Well, isn't e^x greater than 1 for ALL positive values for x above x=1?

 

[Redoctober]

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

$$\lim_{x \to \infty } \frac{f(x)}{g(x)}$$

 

[Char. Limit]

Compare it to g(x)=1, then.

 

[Redoctober]

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 !