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Determining Convergence/Divergence

1. Homework Statement
Determine whether each integral is convergent or divergent, if convergent find its value.
a)[tex] \int^2_1 \frac{dx}{x \ ln \ x}[/tex]

b)[tex] \int^3_0 (\frac{1}{\sqrt{x}})e^{-\sqrt{x}}\ dx[/tex]

3. The Attempt at a Solution
Hey Folks, I'm still at it. Both of these integrals have discontinuities at the left endpoint. I've completed the integration and applied the appropriate equation. However, I'm not sure how to interpret the results.

In this one ln(ln(1)) does not exist. Does that mean I should stop there and declare the inetgral divergent?
http://mcp-server.com/~lush/shillmud/int2.9a.JPG

There are no terms here that cannot be calculated but again I am getting a negative answer. Does this signify divergence? Thank you for clearing this up!
http://mcp-server.com/~lush/shillmud/int2.9b.JPG
 
Last edited:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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1. Homework Statement
Determine whether each integral is convergent or divergent, if convergent find its value.
a)[tex] \int^2_1 \frac{dx}{x \ ln \ x}[/tex]

b)[tex] \int^3_0 (\frac{1}{\sqrt{x}})e^{-\sqrt{x}}\ dx[/tex]

3. The Attempt at a Solution
Hey Folks, I'm still at it. Both of these integrals have discontinuities at the left endpoint. I've completed the integration and applied the appropriate equation. However, I'm not sure how to interpret the results.

In this one ln(ln(1)) does not exist. Does that mean I should stop there and declare the inetgral divergent?
http://mcp-server.com/~lush/shillmud/int2.9a.JPG[/img[/quote]
You have, with some typos, I think,
[tex]\lim_{x\rightarrow 1^+} ln|ln|t||- ln|ln|2||[/itex] and then "declare" that is equal to ln(ln(2)). How did you get that? Since ln(t) goes to 0 as t goes to 1 and ln(0) does not exist, as you say, that limit does not exist. The integral is divergent.

[quote] there are no terms here that cannot be calculated but again I am getting a negative answer. Does this signify divergence? Thank you for clearing this up!
[PLAIN]http://mcp-server.com/~lush/shillmud/int2.9b.JPG
You've lost a sign:
[tex]\int e^{-u} du= -e^{-u}+ C[/tex]
 

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