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Math Help - integration problem

  1. #1
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    integration problem

    hallo all,

    I have the following problem:


    Sorry, made an error and typed the wrong one.
    <br />
\int<br />
x^(-2 ). e^x dx is the one I mean.


    Is it even possible to solve this?
    Last edited by Coele; December 23rd 2008 at 01:07 PM. Reason: made an error
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  2. #2
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    Quote Originally Posted by Coele View Post
    hallo all,

    I have the following problem:

    x-2ex dx = ?

    Is it even possible to solve this?
    Sure. But let's begin by converting that \grave{o} symbol to the more familiar \int:

    \int x-2ex\;dx

    As written, the integral is simply performed:

    \int x-2ex\;dx=(1-2e)\int x\;dx=(\frac{1-2e}{2})x^2+C

    But I suspect that you actually mean this:

    \int x-2e^x\;dx

    For such an integral, we break it up into two portions:

    \int x\;dx-2\int e^x\;dx

    And hopefully you can take it from there.
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  3. #3
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    sorry,

    my bad,

    I made an error:

    I ment this one:

    e^x / x^2 dx (or as I wrote it before: x^-2 . e^x dx
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  4. #4
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    Quote Originally Posted by Coele View Post
    sorry,

    my bad,

    I made an error:

    I ment this one:

    e^x / x^2 dx (or as I wrote it before: x^-2 . e^x dx
    I don't think this has an elementary, closed form antiderivative.

    you could use the infinite series for y = e^x, divide by x^2, and integrate term for term.
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  5. #5
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    That is correct, it has no nice indefinite form.

    We can write it in terms of the Exponential Integral, though.

    \int\frac{e^{x}}{x^{2}}dx=\frac{-e^{x}}{x}-Ei(1,-x)

    Ei is the notation for the Exponential Integral.

    Ei(a,z)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt
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  6. #6
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    I see

    thanks a lot guys.

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