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Math Help - evaluate indefinite integral as an infinite series

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    evaluate indefinite integral as an infinite series

    integral of (e^x-1)/x dx
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    integral of (e^x-1)/x dx
    e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}

    so

    \frac{-1+e^{x}}{x}=\frac{-1+ \sum_{n=0}^{\infty}\frac{x^n}{n!} }{x}=

    Now kick out the first term of the series

    \frac{-1+1+\sum_{n=1}^{\infty}\frac{x^n}{n!} }{x}=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}

    so

    \int \frac{e^{x}-1}{x}dx=\int \sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}dx=\sum_{n=1}^{\infty}\int \frac{x^{n-1}}{n!}dx

    =C+\sum_{n=1}^{\infty}\frac{x^{n}}{n\cdot n!}
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