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Math Help - Showing convergence of Integral

  1. #1
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    Showing convergence of Integral

    Ok, I just want to make sure I got it right:

    I need to show if the Integral \int_0^{\Pi/2} \frac {1} {sinx} is convergent.
    So I solved the Integral: = ln[tan \frac {x} {2}]
    then I put in the values \Pi/2 and \alpha lim ->0

    ln[tan \frac {\Pi} {4}] - ln[tan \frac {\alpha} {2}] =1

    so the Integral is convergent and the limit is 1. Am I right? Is that the right way to do it?
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  2. #2
    Super Member PaulRS's Avatar
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    That integral is divergent.

    Just see that \sin(x)\sim{x} as x\rightarrow{0} and the function \frac{1}{\sin(x)} is contiunous in (0, \frac{\pi}{2}]

    Thus, since \int_0^{\delta}{\frac{dx}{x}} is divergent ( \delta>0), then so is: \int_0^{\pi/2}{\frac{dx}{\sin(x)}}

    And apart from that: \int\frac{dx}{\sin(x)}=\ln\left(\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\right)+k but I'd use the method above
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JanW View Post
    Ok, I just want to make sure I got it right:

    I need to show if the Integral \int_0^{\Pi/2} \frac {1} {sinx} is convergent.
    So I solved the Integral: = ln[tan \frac {x} {2}]
    then I put in the values \Pi/2 and \alpha lim ->0

    ln[tan \frac {\Pi} {4}] - ln[tan \frac {\alpha} {2}] =1

    so the Integral is convergent and the limit is 1. Am I right? Is that the right way to do it?
    PaulRS's Response was very good, but if I was doing this I would have broken it up first

    \int_0^{\frac{\pi}{2}}\frac{dx}{\sin(x)}=\int_0^{1  }\frac{dx}{\sin(x)}+\int_1^{\frac{\pi}{2}}\frac{dx  }{\sin(x)}

    Now as Paul stated as x\to{0} we have that \sin(x)\sim{x}

    and since

    \int_0^{1}\frac{dx}{x} is divergent by the special p-value test

    and \int_1^{\frac{\pi}{2}}\frac{dx}{\sin(x)} is convergent

    The sum of a convergent and divergent integral is a divergent one
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  4. #4
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    Great, thanks for the detailed explenation. Just one question, what do you mean with the p-value test? Sorry about my ignorance..
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JanW View Post
    Great, thanks for the detailed explenation. Just one question, what do you mean with the p-value test? Sorry about my ignorance..
    It is the test that says an integral of the form

    \int_0^{1}\frac{dx}{x^{\beta}}

    Converges iff \beta<1
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  6. #6
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    Wait do you mean \beta >1
    And what about if I have the function : \int_0^{1}\frac{dx}{1+x^{\beta}} with \beta = 1 ?
    Is it now convergent or divergent?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JanW View Post
    Wait do you mean \beta >1
    And what about if I have the function : \int_0^{1}\frac{dx}{1+x^{\beta}} with \beta = 1 ?
    Is it now convergent or divergent?
    There is no longer a discontinuity there, look where on [0,1] is there a problem for \frac{1}{1+x^{\beta}}

    and no, \beta>1 is for integrals of the form

    \int_1^{\infty}\frac{dx}{x^{\beta}}
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  8. #8
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    Ok, thanks a lot for your help!
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