Integral Ques: Show $\int_0^{\infty} x^4 \exp{-x^2} dx = \frac{3\sqrt{\pi}}{8}$

[John O' Meara]

Given that \int_{0}^{\infty} \exp{-x^2}dx =\frac{\sqrt{\pi}}{2} \\ \ \mbox{ show that } \ \int_0^{\infty} x^4 \exp{-x^2} dx= \frac{3\sqrt{\pi}}{8} \\. From the kinetic Theory the root mean square velocity of the molecules (\overline{v^2})^{\frac{1}{2}} is the square root of the integral 4\pi \frac{m}{2\pi kT} \int_0 ^{\infty} \exp{\frac{-mv^2}{2kT}}v^4 dv, where k is Boltzmann constant. T is the absolute temperature, m the mass of each molecule and v the speed of any molecule. Using the substitution x^2=\frac{mv^2}{2kT} \\ \ \mbox{ show that (\overline{v^2})^\frac{1}{2} = (\frac{3kT}{m})^\frac{1}{2} \\ Using integration by parts I get the following:<br /> \int x^4\exp{-x^2}dx =\lim_{x\rightarrow \infty} x^4|\frac{\sqrt{\pi}}{2} - \lim_{X\rightarrow \infty}\int 4x^3 dx \\ \ \mbox{ which does not give} \ 3\frac{\sqrt{\pi}}{8}. Any help would be welcome.<br /> <h2>Homework Statement </h2><br /> <h2>Homework Equations</h2><br /> <h2>The Attempt at a Solution</h2>

 

[John O' Meara]

The missing integral is: 4 \pi \frac{m}{2 \pi kT} \int_{0}^{\infty} \exp{\frac{-mv^2}{2kT}}v^4 dv

 

[arildno]

NO, NO, NO!

You are given that the definite integral:
\int_{0}^{\infty}e^{-x^{2}}=\frac{\sqrt{\pi}}{2}

When you do integration by parts, you need to find AN ANTI-DERIVATIVE of the integrand that you are to evaluate between the two limits.

The anti-derivative (a function!)\int{e}^{-x^{2}}dx is NOT equal to the value of the definite integral you were given!

Thus, you must proceed differently:
Let us rewrite:
\int_{0}^{\infty}x^{4}e^{-x^{2}}dx=\frac{1}{2}\int_{0}^{\infty}x^{3}(2xe^{-x^{2}})dx
Now, set:
u(x)=x^{3}, \frac{dv}{dx}=2xe^{-x^{2}}

See if you can use this hint properly..

 

[John O' Meara]

Thanks for the reply.
\mbox{ Let } \ v=\int2x\exp{-x^2}dx \ \mbox{ therefore } \ v=-\int\exp{u}du = -\exp{u}du \ \mbox{ but } u=-x^2 \ \mbox{therefore } \ v=-\exp{-x^2} \\
For the second part x=v(\frac{m}{2kT})^{\frac{1}{2}} \ \mbox{ therefore } \ dv=dx(\frac{2kT}{m})^{\frac{1}{2}} \ \mbox{ and also therefore } \ v^3=x^3(\frac{2kT}{m})^{\frac{3}{2}} \\ \mbox{ and } v=\sqrt{\frac{2kT}{m}}.The result I get then is \frac{1}{2}(\frac{1}{2})^{\frac{3}{2}}\frac{3kT}{m}\\ which is out by the constant factor \frac{1}{2}(\frac{1}{2})^{\frac{3}{2}}\\.Thanks for the help.

 

[John O' Meara]

Actually the missing integral is 4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}\int \mbox{ e.t.c.}