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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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1 Answer 1

up vote 2 down vote accepted

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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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
    
The problem with probability that the OP wants to solve is here 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. –  Dilip Sarwate Oct 19 '11 at 16:01

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