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

$N=\frac{2}{\sqrt{蟺}}{\left(\frac{2蟺m}{h^2}\right)}^{\frac{3}{2}}V{鈭�}_0^{鈭�}\frac{\sqrt{蔚}d蔚}{e^{\frac{蔚}{kT}}-1}$

$=\frac{2}{\sqrt{蟺}}{\left(\frac{2蟺mkT}{h^2}\right)}^{\frac{3}{2}}V{鈭�}_0^{鈭�}\frac{\sqrt{x}}{e^x-1}dx$

Where,

$h$= Planck's constant,

$k$= Boltzmann's constant,

$V$= volume of the box,

$蔚$= energy of the atom for higher energy level,

$T$= temperature,

$m$= mass of the atom,

$x=\frac{蔚}{kT}$is a new variable

Now,

${鈭�}_0^{鈭�}\frac{\sqrt{x}dx}{e^x-1}={鈭�}_0^{鈭�}\frac{\sqrt{x}{e}^{-x}}{1-e^{-x}}dx$

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

$={鈭�}_0^{鈭�}{x}^{\frac{1}{2}}{e}^{-x}\left(1+e^{-x}+e^{-2x}+e^{-3x}+鈥�鈥�\right)dx$

$={鈭�}_0^{鈭�}{x}^{\frac{1}{2}}\left(e^{-x}+e^{-2x}+e^{-3x}+e^{-4x}+鈥�鈥�\right)dx$

$={鈭�}_0^{鈭�}{x}^{\frac{1}{2}}\underset{k=0}{\overset{鈭�}{鈭�}}{e}^{-(k+1)x}dx$

$=\underset{k=0}{\overset{鈭�}{鈭�}}{鈭�}_0^{鈭�}{x}^{\frac{1}{2}}{e}^{-(k+1)x}dx$

Therefore,

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

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

Now,

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

$=\frac{1}{2}\sqrt{蟺}$

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

${鈭�}_0^{鈭�}{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\frac{\sqrt{蟺}}{2}(k+1{)}^{\frac{-3}{2}}$

we get,

${鈭�}_0^{鈭�}\frac{\sqrt{x}}{e^x-1}dx=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{\sqrt{蟺}}{2}(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{蟺}}{2}\underset{k=0}{\overset{鈭�}{鈭�}}(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{蟺}}{2}\left[1+\frac{1}{2^{\frac{3}{2}}}+\frac{1}{3^{\frac{3}{2}}}+\frac{1}{4^{\frac{3}{2}}}+鈥�..\right]$

$=\frac{\sqrt{蟺}}{2}味\left(\frac{3}{2}\right)$

we get,

${鈭�}_0^{鈭�}\frac{\sqrt{x}}{e^x-1}dx=\frac{\sqrt{蟺}}{2}(2.612)$

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

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

The equation $N=\frac{2}{\sqrt{蟺}}{\left(\frac{2蟺m}{h^2}\right)}^{3/2}V{鈭�}_0^{鈭�}\frac{\sqrt{系}d系}{e^{系/kT}-1}$is evaluated in simpler form.