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I am trying to calculate the following integral, and I would like to know if there is a general rule where we set either $u(x)$ equal to the exponential term or $v'(x)$ equal to the exponential term.

Assuming there is such a general rule, does it matter whether the coefficient of $x$ in the exponential term is positive or negative?

The integral is : $$I=\int\limits_{0}^{\infty} x^{1/2}e^{-x}dx$$

Where the integration by parts formula is:$$\int u(x)\frac{dv(x)}{dx} = uv -\int v\frac{du(x)}{dx}$$


EDIT:

The actual integral I am trying to evaluate is $$I_0=\int\limits_{0}^{\infty}x^2 e^{[-\frac{x^2}{\sigma_0^2}]}dx$$I have obtained $I$ above by using the substitution $y=\frac{x^2}{\sigma_0^2}$ which was what the professor said I should do. Unfortunately, it doesn't seem to make things any easier.

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    Do you have limits for your integral i.e. for instance do you want to evaluate $\displaystyle \int_0^{\infty} x^{1/2} e^{-x} dx$? If not, the integral as such cannot be expressed in terms of elementary functions.2012-11-21
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    I didn't realize. Yes, the actual problem has the limits that I have added now.2012-11-21
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    For this problem, integration by parts along won't take you anywhere, you need to evaluate the Gaussian integral i.e. $\displaystyle \int_0^{\infty} e^{-x^2} dx$ at some stage.2012-11-21

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