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With solving a problem in probability, I find an integral not sure how to do it.

$\int_{|y|}^{\infty}\frac{1}{2x}e^{-x}dx$

Because when I use the theorem \int f(x)g'(x)=f(x)g(x)-\int f'(x)g(x) .i.e.$\int_{|y|}^{\infty}\frac{1}{2x}e^{-x}=-\frac{1}{2x}e^{-x}-\frac{1}{2}\int_{|y|}^{\infty}(-x^{-2}e^{-x})dx$

But how to continue from here

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We can't integrate it in terms of elementary functions, so we give it a name. This is $E_1(|y|)$, the exponential integral divided by $2$

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    The problem with probabilit$y$ that the OP wants to solve is [here](http://math.stacke$x$change.com/q/70705/15941) and the marginal density that the OP is asking about here (without naming it as such) is not really needed to answer the question that he really wants to solve.2011-10-19