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I need to solve the following integral:

$$x\int\frac{e^{\frac{-x^2}{2}}}{x^2}\,dx$$

I'm told that by using the Taylor expansion of $e^z$ and "integrating term by term" show that it's equal to:

$$\alpha x + \sum_{n=0}^{\infty}\frac{(-1)^n}{2^n(2n-1)n!}x^{2n}.$$

Really not sure what to do.

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    why is the $x$ before the integral? what is the meaning of that? – 2017-02-26
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    there is an $x$ before the integral in the question. The whole integral appears to be multiplied by $x$ for some reason. – 2017-02-26
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    Also what is with the title?When did you get erf? – 2017-02-26
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    the solution containes the error function – 2017-02-26
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    $$\int\frac{e^{-\frac{x^2}{2}}}{x^2}\,dx = - \frac{\sqrt{\pi}erf\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2}} - \frac{e^{-\frac{x^2}{2}}}{x} + C$$ – 2017-02-26
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    @M.Turner First replace $e^{-x^2/2}=\sum_{n=0}^\infty\frac{(-x^2/2)^n}{n!}$ then integrate the sum term by term. – 2017-02-26
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    Ohh yes, That helped a lot! I think I've shown it now. Thank you. – 2017-03-01

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