Direct Comparison Test - Improper Integrals

[SirPlus]

1. Homework Statement [/b]

Use the direct comparison test to show that the following are convergent:


(a)[itex]\int_1^∞[/itex] [itex]\frac{cos x\,dx}{x^2}[/itex]

I don't know how to choose a smaller function that converges similar to the one above. The main problem is i don't know where to start.

A simple one that i could solve is : (b)[itex]\int_0^∞[/itex] [itex] \frac{1\,dx}{e^x + 1}[/itex] where a similar function(yet greater) is --> (c)[itex]\int_0^∞[/itex][itex] \frac{1\,dx}{e^x}[/itex] that converges to 1.

Problem :
If cosines and logs come into the integral - i get confused. What do i do?

 

[dirk_mec1]

Why not choose

[tex] \int_1^{\infty} \frac{-2}{x^2}\ \mbox{d}x [/tex]

 

[vela]

1. Homework Statement [/b]

Use the direct comparison test to show that the following are convergent:


(a)[itex]\int_1^∞[/itex] [itex]\frac{cos x\,dx}{x^2}[/itex]

I don't know how to choose a smaller function that converges similar to the one above. The main problem is i don't know where to start.

You want to find a greater function if you want to prove convergence.

When you have cosines and sines, you typically want to use the fact that cosine and sine are bounded. That is, ##-1 \le \cos x \le 1##.

 

[SirPlus]

So how do i solve (a), where do i begin?

 

[dirk_mec1]

Ah, I should've left out the minus sign!

 

[verty]

I am very confused. These answers seem to be going in the wrong direction. If the integrand is sometimes negative, the partial sums could be negative increasing without bound.

To me, this is straight forward: find integrands that bound the integrand and show that those integrands converge.

 

[lurflurf]

It converges absolutely compare to x^-2
cos(x)/x is a bit harder as it converges conditionally.