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anonymous

  • one year ago

Primitive !

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  1. anonymous
    • one year ago
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    \[\int\limits_{}^{}e^\frac{ -1 }{ t }\]

  2. anonymous
    • one year ago
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    dt

  3. anonymous
    • one year ago
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    |dw:1436757696187:dw|

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt }\) \(\large\color{black}{ \displaystyle u=-1/t }\) \(\large\color{black}{ \displaystyle du=1/t^2~dt }\) \(\large\color{black}{ \displaystyle t^2~du=dt }\) \(\large\color{black}{ \displaystyle (-1/u)^2~du=dt }\) \(\large\color{black}{ \displaystyle 1/u^2~du=dt }\) \(\large\color{black}{ \displaystyle \int \frac{e^{u}}{u^2}~du }\) I don't think this has an answer in term of a simple function. What I see to do instead is: \(\large\color{black}{ \displaystyle e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!} }\) \(\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1/t)^n}{n!} }\) \(\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} }\) \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\int \sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} dt }\) \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n+1}}{~(-n+1)~n!~} +C }\) and that you can simplify as: \(\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^n}{~~t^{n-1}~(-n+1)~n!~} +C }\)

  5. SolomonZelman
    • one year ago
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    I mean if you haven't learned this technique using a series, then disregard my reply....

  6. anonymous
    • one year ago
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    I agree that there's no antiderivative in terms of elementary functions. This seems relevant: http://mathworld.wolfram.com/ExponentialIntegral.html You should be able to get that if you integrate by parts. (I haven't bothered checking for myself, though.)

  7. SolomonZelman
    • one year ago
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    Oh, I have seen a post with this idea of integration by parts a certain number of times after which you arrive at an infinite series. I have done a similar thing when integrated e^x/x. (of course differentiating the 1/x) This is how I got to know the series technique, but without proving/showing using by parts every time is more convinient:D

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