evaluate indefinite integral as an infinite series

**twilightstr** · May 4th 2009, 09:41 PM

integral of (e^x-1)/x dx

**TheEmptySet** · May 4th 2009, 09:47 PM

Originally Posted by twilightstr

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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