Question

Evaluate the integral in equation $N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$numerically, to confirm the value quoted in the text.

Short answer

The integral in equation$N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$is evaluated in simpler form.

Step-by-step solution

Step 1. Given information 

The total number of atoms, $N$in Bose-Einstein distribution over all the states is given as:

$N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V{\int }_0^{\infty }\frac{\sqrt{\epsilon }d\epsilon }{e^{\frac{\epsilon }{kT}}-1}$

$=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V{\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx$

Where,

$h$= Planck's constant,

$k$= Boltzmann's constant,

$V$= volume of the box,

$\epsilon$= energy of the atom for higher energy level,

$T$= temperature,

$m$= mass of the atom,

$x=\frac{\epsilon }{kT}$is a new variable

Step 2. Calculating the integral ∫0∞xdxex-1

Now,

${\int }_0^{\infty }\frac{\sqrt{x}dx}{e^x-1}={\int }_0^{\infty }\frac{\sqrt{x}{e}^{-x}}{1-e^{-x}}dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-x}{\left(1-e^{-x}\right)}^{-1}$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-x}\left(1+e^{-x}+e^{-2x}+e^{-3x}+\dots \dots \right)dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}\left(e^{-x}+e^{-2x}+e^{-3x}+e^{-4x}+\dots \dots \right)dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}\sum _{k=0}^{\infty }{e}^{-(k+1)x}dx$

$=\sum _{k=0}^{\infty }{\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}dx$

Step 3. Solving the integral ∫0∞x12e-(k+1)xdx using the formula  ∫0∞xne-axdx=(n!)a-(n+1)

Therefore,

${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\left(\frac{1}{2}\right)!(k+1{)}^{-\left(\frac{1}{2}+1\right)}$

As we know $\left(\frac{1}{2}\right)!=\Gamma \left(\frac{1}{2}+1\right)$, and also $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

Now,

$\Gamma \left(\frac{1}{2}+1\right)=\frac{1}{2}\Gamma \left(\frac{1}{2}\right)$

$=\frac{1}{2}\sqrt{\pi }$

Substituting the value of $\left(\frac{1}{2}\right)!=\frac{1}{2}\sqrt{\pi }$in the equation ${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\left(\frac{1}{2}\right)!(k+1{)}^{-\left(\frac{1}{2}+1\right)}$we get,

${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\frac{\sqrt{\pi }}{2}(k+1{)}^{\frac{-3}{2}}$

Step 4.  Substituting the value of π2(k+1)-32=∫012x-(k+1)x in the equation

we get,

${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\sum _{k=0}^{\infty }\frac{\sqrt{\pi }}{2}(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{\pi }}{2}\sum _{k=0}^{\infty }(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{\pi }}{2}\left[1+\frac{1}{2^{\frac{3}{2}}}+\frac{1}{3^{\frac{3}{2}}}+\frac{1}{4^{\frac{3}{2}}}+\dots ..\right]$

$=\frac{\sqrt{\pi }}{2}\zeta \left(\frac{3}{2}\right)$

Step 5. Substitute the value of  ζ32=∑k=1∞1k32=2.612 in the equation 

we get,

${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\frac{\sqrt{\pi }}{2}(2.612)$

Step 6.  Substituting the value of π2(2.612)=∫0∞x12ex-1dx in the equation N=2π2πmkTh232V∫0∞xex-1dx

$N=\left(\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V\right)\left(\frac{\sqrt{\pi }}{2}(2.612)\right)$

$=2.612{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V$

The equation $N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$is evaluated in simpler form.