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Math Help - "Mathematica could not find a formula for your integral"...

  1. #1
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    "Mathematica could not find a formula for your integral"...

    this is the integral:



    what's the logic for starting the integration? what should I see that I don't...?

    thanks a lot!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dudinka View Post
    this is the integral:



    what's the logic for starting the integration? what should I see that I don't...?

    thanks a lot!
    Try telling Mathematica that \lambda, ~C are real.

    (attached you will find what Derive says it is - with the constant of
    integration set to 0)

    RonL
    Attached Thumbnails Attached Thumbnails "Mathematica could not find a formula for your integral"...-gash.jpg  
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Try telling Mathematica that \lambda, ~C are real.

    (attached you will find what Derive says it is - with the constant of
    integration set to 0)

    RonL
    (snort!) I'm proven wrong again!

    -Dan
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Try telling Mathematica that \lambda, ~C are real.

    (attached you will find what Derive says it is - with the constant of
    integration set to 0)

    RonL
    thank you!

    but could you please tell me how to start solving the integral?
    variables change? trigo. functions?
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Try telling Mathematica that \lambda, ~C are real.

    (attached you will find what Derive says it is - with the constant of
    integration set to 0)

    RonL
    In case nobody else has noticed, I note that I missed off a square root in the
    integrand.

    RonL
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  6. #6
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    Quote Originally Posted by dudinka View Post
    thank you!

    but could you please tell me how to start solving the integral?
    variables change? trigo. functions?
    First get the L out of under the integral by putting y=x/L.

    That won't necessarily help though, because there is no guarantee
    that this has a closed form integral in terms of elementary and not
    so elementary but known functions.

    RonL
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