Mastering Integrals: Tips and Tricks for Convergence and Divergence

[JustaNickname]

Been a long time I had my integral class so I forgot almost everything I knew... I need to integrate to see if the serie converge (limn→∞ an = 0). Thus, there is a theorem of the integral, if you evaluate the limit of the integral of a serie when it tends to the infinite minus when x=1 you can determine if the serie is convergent or divergent...

I can`t find a way to start the problems:

1. limt→∞ ∫ [1, t] sin(1/x) dx
2. limt→∞ ∫ [1, t] xe^(-2x) dx
3. limt→∞ ∫ [1, t] 1/(1+x^(1/2)) dx

You can also dirrectly evaluate the limit when the serie tend to infinite but I haven't found any way to match with the answers... (1. Divergent 2. Convergent 3. Divergent)

The #1 when x tend to infinite, 1/x tend to zero thus sin(1/x) tend to zero also... The serie should be converging if so ??
The #2 should give me an integral like this one:

1/(-2)^2 (-2x-1)e^(-2x) + C
-1/4 * (2x+1)e^(-2x) + C

But you can see right away if you evaluate the integral that it tends to infinite which is not converging

For the two others I have no idea! Any help is welcome thanks!

 

[HallsofIvy]

Been a long time I had my integral class so I forgot almost everything I knew... I need to integrate to see if the serie converge (limn→∞ an = 0). Thus, there is a theorem of the integral, if you evaluate the limit of the integral of a serie when it tends to the infinite minus when x=1 you can determine if the serie is convergent or divergent...

I can`t find a way to start the problems:

1. limt→∞ ∫ [1, t] sin(1/x) dx
2. limt→∞ ∫ [1, t] xe^(-2x) dx
3. limt→∞ ∫ [1, t] 1/(1+x^(1/2)) dx

You can also dirrectly evaluate the limit when the serie tend to infinite but I haven't found any way to match with the answers... (1. Divergent 2. Convergent 3. Divergent)

The #1 when x tend to infinite, 1/x tend to zero thus sin(1/x) tend to zero also... The serie should be converging if so ??

x is not going to infinity. It is the upper limit on the integral that is going to infinity. x ranges from 1 to infinity in that integral.

The #2 should give me an integral like this one:
1/(-2)^2 (-2x-1)e^(-2x) + C
-1/4 * (2x+1)e^(-2x) + C

But you can see right away if you evaluate the integral that it tends to infinite which is not converging y

For the two others I have no idea! Any help is welcome thanks!

What you get for the integral is correct but, as t goes to infinity, it does NOT go to infinity. Why would you think it does?

For (3) let $u= 1+ x^{1/2}$

 

[JustaNickname]

Just making sure I understand you correctly (I am natively speaking french).

The serie Ʃ∞n=1 ne^(-2n) = e^-2 + 2e^(-4) + ... + ne^(-2n)

I can clearly see here the serie is decreasing and convergent because it gets closer to a number since an > an+1.

I would like to prove it with the integral theorem but I can't find a way to make the limt→∞ ∫ [1, t] xe^(-2x) dx = 0 thus converging.

 

[JustaNickname]

Quoting my notes:

If ∫[1,∞[ f(x) is convergent <=> Ʃ∞n=1 an is convergent where an = f(x)

So the goal is to prove the integral is convergent to find out if the serie is converging.