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Math Help - nasty integral

  1. #1
    v71
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    nasty integral

    which is the integral for exp( -a/x ) dx ?

    i have tried every method

    thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by v71
    which is the integral for exp( -a/x ) dx ?

    i have tried every method

    thanks
    That's because the indefinite form is...ugly. I don't think it has a closed form. If this is supposed to be a definite integral, though, you may be able to calculate an exact value.

    -Dan
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  3. #3
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    Topsquark is correct.
    ---
    It is a famous case that,
    \int \frac{e^x}{x}dx is non-elementary.
    We shall convert this integral into this form.
    For simplicity sake we shall use a=1, thus,
    \int e^{-1/x}dx
    Manipulate it as,
    \int \frac{x^2e^{1/x}}{x^2}dx
    Let, (going to apply the substitution rule)
    u=-1/x then, u'=1/x^2
    Thus,
    \int \frac{e^u}{u^2}du
    If you use integration by parts with,
    v=e^u and w'=1/u^2
    Then,
    -\frac{e^u}{u}+\int \frac{e^u}{u}du
    We have converted this integral into an integral which has no elementary anti-derivative. This confirms topsquarks guess.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Topsquark is correct.
    ---
    It is a famous case that,
    \int \frac{e^x}{x}dx is non-elementary.
    We shall convert this integral into this form.
    For simplicity sake we shall use a=1, thus,
    \int e^{-1/x}dx
    Manipulate it as,
    \int \frac{x^2e^{1/x}}{x^2}dx
    Let, (going to apply the substitution rule)
    u=-1/x then, u'=1/x^2
    Thus,
    \int \frac{e^u}{u^2}du
    If you use integration by parts with,
    v=e^u and w'=1/u^2
    Then,
    -\frac{e^u}{u}+\int \frac{e^u}{u}du
    We have converted this integral into an integral which has no elementary anti-derivative. This confirms topsquarks guess.
    Yeah, what he said.

    -Dan
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