- #1

andyyee

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## Homework Statement

I'm working out a differential equation problem that I am supposed to solve with the formula [itex]\mathcal{L}\{t^\alpha\} = \frac{\Gamma{(\alpha + 1)}}{s^{\alpha+1}}[/itex]. The problem is [itex]\mathcal{L}\{t^{\frac{1}{2}}\}[/itex] (finding the Laplace transform of the given function).

## Homework Equations

[itex]\mathcal{L}\{t^\alpha\} = \frac{\Gamma(\alpha + 1)}{s^{\alpha+1}}, \alpha > -1[/itex]

[itex]\Gamma(\alpha) = \int^\infty_0{t^{\alpha-1}e^{-t}dt}, \alpha > 0[/itex]

## The Attempt at a Solution

I plug it into the equation, and get [itex]\frac{\Gamma(\frac{3}{2})}{s^\frac{3}{2}} = [/itex] [itex]\frac{\int^\infty_0 {t^\frac{1}{2}e^{-t}dt}}{s^{3/2}}[/itex]. That's where I run into a problem, I have no idea how to solve that integral. I can't use integration by parts, because one term will never disappear or the original integral will not appear again as [itex]\int{vdu}[/itex], so that won't work. I looked it up on Wolfram|Alpha, and it gave me [itex]\frac{1}{2}\sqrt{\pi}\text{erf}{(\sqrt{t})} - e^{-t}\sqrt{t}[/itex] for the indefinite form and [itex]\frac{\sqrt{\pi}}{2}[/itex] for the definite form. It also cited some stuff about integrating the normal distribution and error form, but I don't understand what it is talking about. What I am unsure about is the steps involved in solving the integral, and is there a generalized solution (for the definite integral from 0 to [itex]\infty[/itex]) for other coefficient values in the power of the polynomial?

Thanks

Andrew