Sorry If this question does not belong here, I am trying for the last six days to prove the following integration but I am failing, and I will really appreciate a lit bit of help.
I have an expression for $\tilde{M}(\beta)$: $$ \tilde{M}(\beta)=\frac{1}{\hbar} \sum_n \left[ \left( \Omega_n \epsilon_n^2 - m_n\epsilon_n \right)f_n(\beta) - \frac{m_n}{2}\epsilon_n^2 f_n'(\beta) \right] \tag{1} $$ here $f_n(\beta)=(e^{\beta\epsilon_n}+1)^{-1}$ and $f_n'(\beta)\equiv \frac{\partial}{\partial \epsilon_n}f_n(\beta)=-\beta e^{\beta\epsilon_n}(e^{\beta\epsilon_n}+1)^{-2}$. From this equation, we want to calculate $M$ which is related to $\tilde{M}$ via relation $$ \frac{\partial}{\partial \beta}(\beta^2 M) = \beta \tilde{M} \tag{2} $$
According to this article, we get the following $M$ by integrating the above equation: $$ M=-\frac{1}{\hbar} \sum_n \left[ \Omega_n \left(\int_{\epsilon_n}^\infty dz f(z) z \right) + \frac{m_n}{2} f_n \epsilon_n \right] \tag{3} $$ here $f(z)=(e^{\beta z}+1)^{-1}$. I want to re-derive $M$ by integrating $(2)$ for my own understanding.
My attempt:
I start by re-writing $(2)$ as $\beta^2 M = \int_0^\beta d\beta (\beta \tilde{M}) $:
$$ \beta^2 M = \frac{1}{\hbar} \sum_n \left[ \left( \Omega_n \epsilon_n^2 - m_n\epsilon_n \right) \int_0^\beta d\beta (\beta f_n) - \frac{m_n}{2}\epsilon_n^2 \int_0^\beta d\beta (\beta f_n') \right] $$ $$ M= \frac{1}{\beta^2\hbar} \sum_n \left[ \left( \Omega_n \epsilon_n^2 - m_n\epsilon_n \right) \int_0^\beta d\beta \left( \frac{\beta}{e^{\beta\epsilon_n}+1} \right) - \frac{m_n}{2}\epsilon_n^2 \int_0^\beta d\beta \left(\frac{-\beta^2 e^{\beta\epsilon_n}}{(e^{\beta\epsilon_n}+1)^{2} } \right) \right]\tag{4} $$ I arrive at these strange integrals. Can someone please help me in proving that $(4)=(3)$?