Integration of exponential function times polynomial of fractional degree

andyyee
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Homework Statement



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

Homework Equations



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

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

The Attempt at a Solution



I plug it into the equation, and get \frac{\Gamma(\frac{3}{2})}{s^\frac{3}{2}} = \frac{\int^\infty_0 {t^\frac{1}{2}e^{-t}dt}}{s^{3/2}}. 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 \int{vdu}, so that won't work. I looked it up on Wolfram|Alpha, and it gave me \frac{1}{2}\sqrt{\pi}\text{erf}{(\sqrt{t})} - e^{-t}\sqrt{t} for the indefinite form and \frac{\sqrt{\pi}}{2} 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 \infty) for other coefficient values in the power of the polynomial?

Thanks
Andrew
 
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But you don't have to do the integral. You have the LaPlace transform from your formula;

\frac{\Gamma(\frac 3 2)}{s^\frac 3 2}

\Gamma(3/2) is just a constant. Look in your text and see if it doesn't give you \Gamma(1/2) and a relationship between the Gamma value at 3/2 and 1/2.
 
Thanks for the help :-). I completely overlooked the part in my book that mentioned an appendix, which happened to show some basic integration (for gamma of 1/2 using double integrals), as well as ways to solve similar gamma values without integration (simple algebra transformations using the already computed, given gamma of 1/2 and some other formulas). These forums are really great, because although Wolfram|Alpha gives me solutions, it does not tell me why (or to look closer at my book :-).

- Andrew
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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