Question about convergence of this improper integral

[trickslapper]

This is for a real analysis class, i already turned in this problem and i got marked wrong for it and got into a little argument with my professor about it:

The integral sinx/lnx from 2 to infinity. I used the Abel-Dirichlet test that my professor showed us in class:

1. the integral of f(x) must be bounded
2. as x approaches infinity Limit of g must be zero
3. g'(x) must be continous on [a,infinity)

So here is my work:

Let f=sin(x); g=1/ln(x)

1. integral of f(x)=-cos(x) which is indeed bounded
2. the limit as x goes to infinity of g is indeed zero.
3. g'(x)= -1/xln(x)ln(x) which is continuous on [2,infinity)

Ok so from this test i came to the conclusion that the integral of sinx/lnx converges. My professor disagrees:

Here is my professors argument:

Consider lnx<x this implies that 1/lnx>1/x which implies that sin(x)/ln(x)>sin(x)/x

By the p-test we know that sin(x)/x converges and then that means that sin(x)/ln(x) also diverges by comparison test.

I'm not some stuck up student who thinks he's always right, but i'm pretty sure that i'm right but... i also see my professors argument so before i bring it up again i want to make sure that i'm correct.

Can anyone see what might be going wrong here?

thanks!

 

[Deleted member 4993]

This is for a real analysis class, i already turned in this problem and i got marked wrong for it and got into a little argument with my professor about it:

The integral sinx/lnx from 2 to infinity. I used the Abel-Dirichlet test that my professor showed us in class:

1. the integral of f(x) must be bounded
2. as x approaches infinity Limit of g must be zero
3. g'(x) must be continous on [a,infinity)

So here is my work:

Let f=sin(x); g=1/ln(x)

1. integral of f(x)=-cos(x) which is indeed bounded
2. the limit as x goes to infinity of g is indeed zero.
3. g'(x)= -1/xln(x)ln(x) which is continuous on [2,infinity)

Ok so from this test i came to the conclusion that the integral of sinx/lnx converges. My professor disagrees:

Here is my professors argument:

Consider lnx<x this implies that 1/lnx>1/x which implies that sin(x)/ln(x)>sin(x)/x

By the p-test we know that sin(x)/x converges and then that means that sin(x)/ln(x) also diverges by comparison test.

I am not following this part of the arguement.

1/2[sup:32jfb20p]n[/sup:32jfb20p] > 1/3[sup:32jfb20p]n[/sup:32jfb20p]

We know 1/3[sup:32jfb20p]n[/sup:32jfb20p] converges - but so does 1/2[sup:32jfb20p]n[/sup:32jfb20p]

both of those converge!!!

In my opinion - you cannot say anything (diverge/converge) of sin(x)/ln(x) from the above comparison test



I'm not some stuck up student who thinks he's always right, but i'm pretty sure that i'm right but... i also see my professors argument so before i bring it up again i want to make sure that i'm correct.

Can anyone see what might be going wrong here?

thanks!

 

[trickslapper]

oops i meant that by the p test that sin(x)/x diverges and then that implies that sin(x)/ln(x) also diverges.

He is pretty adamant that the comparison test does work.. I've asked other professors and they say that i'm right but i'm still not sure

 

[galactus]

\(\displaystyle \int_{2}^{\infty}\frac{sin(x)}{ln(x)}dx\) does converge.

Actually, it converges to \(\displaystyle -.0964377.............\)

 

[trickslapper]

thats what i got.. now how to present this to my professor without enraging him lol...

*actually can anyone see why the comparison test wouldn't work for this?

 

[galactus]

\(\displaystyle \int_{1}^{\infty}\frac{sin(x)}{x}dx\) does converge.

Your professor said it diverged?.

Let \(\displaystyle dv=sin(x)dx, \;\ v=-cos(x), \;\ u=\frac{1}{x}, \;\ du=\frac{-1}{x^{2}}dx\)

\(\displaystyle \int_{1}^{\infty}\frac{sin(x)}{x}dx=\lim_{b\to 0}\left[\frac{-cos(x)}{x}\right]_{1}^{b}-\int_{2}^{\infty}\frac{cos(x)}{x^{2}}dx=cos(1)-\int_{2}^{\infty}\frac{cos(x)}{x^{2}}dx\) converges.

 

[trickslapper]

shouldn't the b approach infinity since we're doing an improper integral and your limits of integration changed from 1 to 2. I'm guessing those are typos but when i did it i got what you got. We are assuming that the integral of cos(x)/x^2 converges by p-test right?

 

[trickslapper]

Ok, so in class today we discussed this problem and he said that he agreed that it converged conditionally BUT it absolutely diverges and he said that absolute divergence/convergence supercedes conditional convergence... so i guess i was wrong ;_;