# Which is the easier way to do integration by parts when there is an exponential term?

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. –  user17762 Nov 21 '12 at 5:59
I didn't realize. Yes, the actual problem has the limits that I have added now. –  Joebevo Nov 21 '12 at 6:17
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. –  user17762 Nov 21 '12 at 6:40

What you want is $\Gamma(3/2)$, where $\Gamma(z)$ is the Gamma function. There are many ways to evaluate it. Below is one possible way. Let $$I = \int_0^{\infty}x^{1/2} e^{-x}dx$$ Let $\sqrt{x}=t$ i.e. $x=t^2 \implies dx = 2t dt$ We then get that $$I = \int_0^{\infty} t e^{-t^2} (2tdt) = 2 \int_{0}^{\infty} t^2 e^{-t^2} dt$$ Let $$K(\alpha) = \int_{0}^{\infty} e^{-\alpha t^2} dt$$ Note that $$\dfrac{d K(\alpha)}{d \alpha} = -\int_{0}^{\infty} t^2 e^{-\alpha t^2} dt$$ Hence, $$I = -2 \left. \dfrac{dK(\alpha)}{d \alpha} \right \vert_{\alpha=1}$$ Hence, all we need is to find $K(\alpha)$. But $K(\alpha)$ is a well-known integral once we make the susbtitution $z = \sqrt{\alpha} t$. Then $$K(\alpha) = \dfrac1{\sqrt{\alpha}} \int_0^{\infty}e^{-z^2} dz = \dfrac{\sqrt{\pi}}{2 \sqrt{\alpha}}$$ Look at the links below why $$\int_0^{\infty} e^{-z^2} dz = \dfrac{\sqrt{\pi}}2$$

Now our $$I = -2 \left. \dfrac{dK(\alpha)}{d \alpha} \right \vert_{\alpha=1} = \left. -2 \dfrac{\sqrt{\pi}}2 \times \dfrac{-1}2 \times \alpha^{-3/2} \right \vert_{\alpha=1} = \dfrac{\sqrt{\pi}}2$$

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