Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to do this integral:

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


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

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 :)

share|cite|improve this question
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
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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.