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Math Help - infinite integral

  1. #1
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    Exclamation infinite integral

    The homework question states:

    Determine whether the infinite integral \int_{1}^{\infty}\frac{1}{x}cos(lnx) dx converges or diverges. Explain the outcome of your calculation, using a sketch (representing a domain of the form [1,a] for a suitable large value a) of the graph of sin(lnx).

    My question(s):

    How does a graph help explain, better put, what am I looking for in this graph and how am I supposed to use it? (I've read my notes over and over, but they still make no sense...)

    I understand that the integral = sin(ln(x))....

    With regards to [1,a], isn't a=lower bound of the integral? i.e. a=1?

    Thanks in advance

    tsal15
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  2. #2
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    Quote Originally Posted by tsal15 View Post
    The homework question states:

    Determine whether the infinite integral \int_{1}^{\infty}\frac{1}{x}cos(lnx) dx converges or diverges. Explain the outcome of your calculation, using a sketch (representing a domain of the form [1,a] for a suitable large value a) of the graph of sin(lnx).

    My question(s):

    How does a graph help explain, better put, what am I looking for in this graph and how am I supposed to use it? (I've read my notes over and over, but they still make no sense...)

    I understand that the integral = sin(ln(x))....

    With regards to [1,a], isn't a=lower bound of the integral? i.e. a=1?

    Thanks in advance

    tsal15
    Yes, using a u-substitution, you should get [\sin{(\ln{x})}]_1^\varepsilon as the integral.

    So now find \lim_{\varepsilon \to \infty}[\sin{(\ln{\varepsilon})}] - [\sin{(\ln{1})}]

    The second part is easy...

    \ln{1} = 0, \sin{0} = 0.

    So, what happens as \varepsilon \to \infty?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    So, what happens as \varepsilon \to \infty?
    I believe it doesn't approach L, so thus it approaches infinity. (correct?)

    Also, how does the graph come into play?
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    Quote Originally Posted by tsal15 View Post
    I believe it doesn't approach L, so thus it approaches infinity. (correct?)

    Also, how does the graph come into play?
    Not quite...

    You're right that \varepsilon \to \infty, \ln{\varepsilon} \to \infty.

    What happens to the sine function as x tends to infinity?

    Sine oscillates between -1 and 1...

    How can you say what it converges to? You can't. Therefore it diverges...


    You should be able to see the oscillatory behaviour in the graph... That's what it's asking for.
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    Quote Originally Posted by Prove It View Post
    You should be able to see the oscillatory behaviour in the graph... That's what it's asking for.
    I plugged sin(ln(x)) into my graphing calculator, but i didn't see any oscillation on the graph. It looks to start off like a log graph which then dives below the x-axis... have i used the wrong function? should i have used sin(x)?

    Thank you for your help
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  6. #6
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    Quote Originally Posted by tsal15 View Post
    I plugged sin(ln(x)) into my graphing calculator, but i didn't see any oscillation on the graph. It looks to start off like a log graph which then dives below the x-axis... have i used the wrong function? should i have used sin(x)?

    Thank you for your help
    Try zooming out. Also, is your calculator in radian mode?
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