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Das ideale Bosegas Eq: math.2561.9

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TeX (original user input):

\begin{align}
  & \ln Y=\prod\limits_{j=1}^{l}{{}}\ln {{Y}_{j}}=-\sum\limits_{j}^{{}}{{}}\ln \left( 1-\zeta {{e}^{-\beta {{E}_{j}}}} \right) \\ 
 & \approx -\left( 2s+1 \right)\frac{4\pi V}{{{h}^{3}}}\int_{0}^{\infty }{{}}dp{{p}^{2}}\ln \left( 1-\zeta {{e}^{-\beta \frac{{{p}^{2}}}{2m}}} \right) \\ 
 & =-\left( 2s+1 \right)\frac{4\pi V}{{{h}^{3}}}\left[ \frac{{{p}^{3}}}{3}\ln \left( 1-\zeta {{e}^{-\beta \frac{{{p}^{2}}}{2m}}} \right)_{0}^{\infty }-\int_{0}^{\infty }{{}}dp\frac{{{p}^{3}}}{3}\frac{\beta \frac{p}{m}\zeta {{e}^{-\beta \frac{{{p}^{2}}}{2m}}}}{\left( 1-\zeta {{e}^{-\beta \frac{{{p}^{2}}}{2m}}} \right)} \right] \\ 
 & \frac{{{p}^{3}}}{3}\ln \left( 1-\zeta {{e}^{-\beta \frac{{{p}^{2}}}{2m}}} \right)_{0}^{\infty }=0 \\ 
 & \Rightarrow \ln Y=\frac{2}{3}\beta \left( 2s+1 \right)\frac{4\pi V}{{{h}^{3}}}\int_{0}^{\infty }{{}}dp{{p}^{2}}\frac{\beta \frac{{{p}^{2}}}{2m}}{\left( \frac{1}{\zeta }{{e}^{\beta \frac{{{p}^{2}}}{2m}}}-1 \right)}=\frac{2}{3}\beta \left( 2s+1 \right)\frac{V}{{{h}^{3}}}\int_{0}^{\infty }{{}}dp4\pi {{p}^{2}}\left\langle N(p) \right\rangle E(p) \\ 
 & \Rightarrow \ln Y=\frac{2}{3}\beta \left( 2s+1 \right)\frac{V}{{{h}^{3}}}\int_{0}^{\infty }{{}}dp4\pi {{p}^{2}}\left\langle N(p) \right\rangle E(p)=\frac{2}{3}\beta U \\ 
\end{align}

TeX (checked):

{\begin{aligned}&\ln Y=\prod \limits _{j=1}^{l}{}\ln {{Y}_{j}}=-\sum \limits _{j}^{}{}\ln \left(1-\zeta {{e}^{-\beta {{E}_{j}}}}\right)\\&\approx -\left(2s+1\right){\frac {4\pi V}{{h}^{3}}}\int _{0}^{\infty }{}dp{{p}^{2}}\ln \left(1-\zeta {{e}^{-\beta {\frac {{p}^{2}}{2m}}}}\right)\\&=-\left(2s+1\right){\frac {4\pi V}{{h}^{3}}}\left[{\frac {{p}^{3}}{3}}\ln \left(1-\zeta {{e}^{-\beta {\frac {{p}^{2}}{2m}}}}\right)_{0}^{\infty }-\int _{0}^{\infty }{}dp{\frac {{p}^{3}}{3}}{\frac {\beta {\frac {p}{m}}\zeta {{e}^{-\beta {\frac {{p}^{2}}{2m}}}}}{\left(1-\zeta {{e}^{-\beta {\frac {{p}^{2}}{2m}}}}\right)}}\right]\\&{\frac {{p}^{3}}{3}}\ln \left(1-\zeta {{e}^{-\beta {\frac {{p}^{2}}{2m}}}}\right)_{0}^{\infty }=0\\&\Rightarrow \ln Y={\frac {2}{3}}\beta \left(2s+1\right){\frac {4\pi V}{{h}^{3}}}\int _{0}^{\infty }{}dp{{p}^{2}}{\frac {\beta {\frac {{p}^{2}}{2m}}}{\left({\frac {1}{\zeta }}{{e}^{\beta {\frac {{p}^{2}}{2m}}}}-1\right)}}={\frac {2}{3}}\beta \left(2s+1\right){\frac {V}{{h}^{3}}}\int _{0}^{\infty }{}dp4\pi {{p}^{2}}\left\langle N(p)\right\rangle E(p)\\&\Rightarrow \ln Y={\frac {2}{3}}\beta \left(2s+1\right){\frac {V}{{h}^{3}}}\int _{0}^{\infty }{}dp4\pi {{p}^{2}}\left\langle N(p)\right\rangle E(p)={\frac {2}{3}}\beta U\\\end{aligned}}

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\begin{aligned} \ln Y = \prod_{j = 1}^{l} \ln Y_{j} = - \sum_{j}^{} \ln (1 - ζ e^{- β E_{j}}) \\ \approx - (2 s + 1) \frac{4 π V}{h^{3}} \int_{0}^{\infty} d p p^{2} \ln (1 - ζ e^{- β \frac{p^{2}}{2 m}}) \\ = - (2 s + 1) \frac{4 π V}{h^{3}} [\frac{p^{3}}{3} \ln {(1 - ζ e^{- β \frac{p^{2}}{2 m}})}_{0}^{\infty} - \int_{0}^{\infty} d p \frac{p^{3}}{3} \frac{β \frac{p}{m} ζ e^{- β \frac{p^{2}}{2 m}}}{(1 - ζ e^{- β \frac{p^{2}}{2 m}})}] \\ \frac{p^{3}}{3} \ln {(1 - ζ e^{- β \frac{p^{2}}{2 m}})}_{0}^{\infty} = 0 \\ \Rightarrow \ln Y = \frac{2}{3} β (2 s + 1) \frac{4 π V}{h^{3}} \int_{0}^{\infty} d p p^{2} \frac{β \frac{p^{2}}{2 m}}{(\frac{1}{ζ} e^{β \frac{p^{2}}{2 m}} - 1)} = \frac{2}{3} β (2 s + 1) \frac{V}{h^{3}} \int_{0}^{\infty} d p 4 π p^{2} ⟨ N (p) ⟩ E (p) \\ \Rightarrow \ln Y = \frac{2}{3} β (2 s + 1) \frac{V}{h^{3}} \int_{0}^{\infty} d p 4 π p^{2} ⟨ N (p) ⟩ E (p) = \frac{2}{3} β U \end{aligned}

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