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Math Help - Definite Integration Problem

  1. #1
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    Definite Integration Problem

    Given that \int\limits_{0}^{\infty} e^{-x^{2}}dx = <br />
\frac{\sqrt{\pi}}{2}<br />
;
    find <br />
\int\limits_{0}^{\infty} x^{- \frac{1}{2}} e^{-2}dx<br />
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  2. #2
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    Quote Originally Posted by zorro View Post
    Given that \int\limits_{0}^{\infty} e^{-x^{2}}dx = <br />
\frac{\sqrt{\pi}}{2}<br />
;
    find <br />
\int\limits_{0}^{\infty} x^{- \frac{1}{2}} e^{-2}dx <br />
Mr F says: The integrand is obviously incomplete ....
    The complete question probably requires you to make the substitution u = x^{1/2}.
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  3. #3
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    can to provide me with the first few steps..

    Quote Originally Posted by mr fantastic View Post
    The complete question probably requires you to make the substitution u = x^{1/2}.
    Thanks for ur reply , but can u please provide me with the first few steps...
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  4. #4
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    Quote Originally Posted by zorro View Post
    Thanks for ur reply , but can u please provide me with the first few steps...
    Read what I said in post #2 (the red writing). You will need to post the correct and complete integrand before further headway can be made.
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  5. #5
    MHF Contributor chisigma's Avatar
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    The term e^{-2} is a constant so that the integral you propose is...

    \int_{0}^{\infty} \frac{e^{-2}}{\sqrt{x}}\cdot dx = e^{-2}\cdot \int_{0}^{\infty} \frac{dx}{\sqrt{x}} (1)

    The problem is that the integral in (1) diverges ... May be that the integral you intended to propose us was...

    \int_{0}^{\infty} \frac{e^{-2x}}{\sqrt{x}}\cdot dx (2)

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by chisigma View Post
    The term e^{-2} is a constant so that the integral you propose is...

    \int_{0}^{\infty} \frac{e^{-2}}{\sqrt{x}}\cdot dx = e^{-2}\cdot \int_{0}^{\infty} \frac{dx}{\sqrt{x}} (1)

    The problem is that the integral in (1) diverges ... May be that the integral you intended to propose us was...

    \int_{0}^{\infty} \frac{e^{-2x}}{\sqrt{x}}\cdot dx (2)

    Kind regards

    \chi \sigma


    thanks for ur reply but the question the i have posted is correct please can u provide me the solution to the question?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    thanks for ur reply but the question the i have posted is correct please can u provide me the solution to the question?
    Quote Originally Posted by zorro View Post
    Given that \int\limits_{0}^{\infty} e^{-x^{2}}dx = <br />
\frac{\sqrt{\pi}}{2}<br />
;
    find <br />
{\color{red}\int\limits_{0}^{\infty} x^{- \frac{1}{2}} e^{-2}dx}<br />
    As chisigma pointed out (and mr fantastic 2 months ago...), the integral you posted (highlighted in red) is divergent (since it appears to be incomplete)! There is a major mistake in the problem because we can't incorporate the fact that \int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}!!
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by zorro View Post
    thanks for ur reply but the question the i have posted is correct please can u provide me the solution to the question?
    You have been told that if the question you posted is correct, then the integral is divergent. It has also been suggested that the question as posted is very unlikely to be correct.

    So there is nothing further do to until as with another post you eventually decide that the question is not correct and post what it is supposed to be, when the odds are on that the method of doing it has already been posted above.

    CB
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  9. #9
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    I would guess that our friend meant to write either \int_0^\infty x^{-1/2}\mathrm e^{-x}\mathrm dx or \int_0^\infty x^{-1/2}\mathrm e^{-2x}\mathrm dx. Assume the former, as the latter is very similar.

    Put x=u^2, u>0. Then ...
    Last edited by CaptainBlack; July 5th 2009 at 08:00 AM. Reason: spoon feeding deleted
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  10. #10
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    Quote Originally Posted by halbard View Post
    I would guess that our friend meant to write either \int_0^\infty x^{-1/2}\mathrm e^{-x}\mathrm dx or \int_0^\infty x^{-1/2}\mathrm e^{-2x}\mathrm dx. Assume the former, as the latter is very similar.

    Put x=u^2, u>0. Then ...
    Which is exactly what MrF's reply (the first one in this thread) suggested, but left the detail for the OP to finish themselves.

    CB
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  11. #11
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    Oops, sorry CaptainBlack, for treading on toes. Mea culpa.
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