# Integration with exponential

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 $2$
Alternatively, you can express the result as an (upper) incomplete gamma function: $\Gamma(0,|y|)/2$. This might be a more natural way of putting it, considering that OP encountered the integral in the context of probabilities... –  Ｊ. Ｍ. Oct 19 '11 at 13:13