# 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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I found the answer to the general question in this helpful webpage: http://math.ucsd.edu/~wgarner/math20b/int_by_parts.htm

The relevant sentences are

"Typically, one will see a function that involves two distinct functions. There are five major classes of functions that we will consider. The first is Logarithmic Functions. The second is Inverse Functions, such as tan-1 and so forth. The third is Polynomial Functions, those are the functions that we first integrated. The fourth is Exponential Functions. The last is Trigonometric Functions.

This order also helps to answer the second question: what is u and dv. Think of the classes of functions as a list of desirability for choosing u. That is, we would really like to have a function with logs in it to be u. The last thing we would want to be u is a function with sine and cosine in it. The following pneumonic may help. It is called LIPET (lie-pet)."

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