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I need to do this integral:

$\int{\left( -\frac{\partial f}{\partial \varepsilon } \right)}\,{{\varepsilon }^{3/2}}d\varepsilon$,

where

$f\left( \varepsilon \right)={{e}^{-\varepsilon /{{k}_{B}}T}}$

As you can see, this is not a totally easy integral since the second term is ^$(3/2)$. So, my idea was to substitute $\varepsilon$ with ${{u}^{2}}$.

That would make it a gaussian function times an integer power of u, which has a common solution, and is very doable. But my question is, when I do the substitution, do I also change $d\varepsilon$, and in that case to what ? $duu$, or...?

Thanks in advance :)

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Are there integration limts, like from $[0,\infty]$? –  Ron Gordon May 1 '13 at 20:18
    
Yes, that is the limits. –  Denver Dang May 1 '13 at 20:39
    
See the Gamma function. –  Antonio Vargas May 1 '13 at 20:50
    
I think you might have misunderstood me. I know how to do the integral when I get it on the form of $\int_{0}^{\infty }{{{e}^{-a{{x}^{2}}}}{{x}^{m}}dx}$ My problem is, that currently the 2nd function is a function with half-integer powers. And that is not easily done, not even with gamma function as far as I know. My question was how to change it to $\varepsilon ={{u}^{2}}$ so that it becomes a normal integer power function. That would make the exponential function a gaussian function, and of course, the right function a integer power function - and that is doable via gamma function. –  Denver Dang May 1 '13 at 20:56
    
Note that you can get displayed equations by using double dollar signs instead of single dollar signs. That centres the equations and makes things like fractions look less cramped. –  joriki May 2 '13 at 2:05

1 Answer 1

up vote 0 down vote accepted

I'll write $x$ instead of $\varepsilon$ since it's easier on my hands.

If $f(x) = e^{-a x}$ then $\frac{df}{dx} = -a e^{-ax}$, so that

$$ \int_0^\infty \left(-\frac{df}{dx}\right) x^{3/2}\,dx = a \int_0^\infty e^{-ax} x^{3/2}\,dx. $$

Make the change of variables $ax = y$ to get

$$ \begin{align} a \int_0^\infty e^{-ax} x^{3/2}\,dx &= \int_0^\infty e^{-y} (y/a)^{3/2}\,dy \\ &= \frac{1}{a^{3/2}}\int_0^\infty e^{-y} y^{3/2}\,dy \\ &= \frac{1}{a^{3/2}} \Gamma(5/2) \\ &= \frac{3\sqrt{\pi}}{4a^{3/2}}, \end{align} $$

where we have used the fact that

$$ \Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \Gamma\left(\frac{3}{2}\right) = \frac{3}{4} \Gamma\left(\frac{1}{2}\right) = \frac{3\sqrt{\pi}}{4}. $$

You have $a = \frac{1}{k_B T}$, so your integral is

$$ \frac{3\sqrt{\pi} (k_B T)^{3/2}}{4}. $$

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Can you really use the gamma function that way ? In my book it seems that the n in $x^{n}$ has to be an integer ? But maybe that's just me ? –  Denver Dang May 1 '13 at 21:54
    
@DenverDang the most important property of the Gamma function is that $$\Gamma(1+z) = z\Gamma(z),$$ just like the factorial function. The Gamma function is defined for all complex $z$ with the exception of the nonpositive real integers. –  Antonio Vargas May 1 '13 at 21:55
    
Ahhh, I think I got it now :) Thank you very much :D –  Denver Dang May 1 '13 at 22:21
    
@DenverDang You're very welcome. –  Antonio Vargas May 1 '13 at 22:23

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