Fourier transform of Fermi function As an alternative approach to the Sommerfeld-expansion, my lecturer tries to motivate properties of free fermions, such as temperature dependencies of the chemical potential $\mu(T)$, electron number $N_e(T)$, energy density $U(T)$, etc. by expanding the Fourier transform of the Fermi function for low temperatures:
$$\int d\epsilon g(\epsilon)f(\epsilon)=\int d\epsilon g(\epsilon)\int dt \tilde{f}(t)e^{-i\epsilon t}\\
\text{where}~ \tilde{f}(t)=\frac{e^{i\mu t}}{2\pi i}\left(\pi i \delta (t)+\frac{1}{t}\frac{\pi t /\beta}{\sinh(\pi t /\beta)}\right)\\
\text{and} ~ g(\epsilon) ~\text{some arbitrary well-behaved function}.
$$
I have only come so far to calculate the Fourier transform $\tilde{f}(t)$ of $f(\epsilon)=\frac{1}{(e^{\beta(\epsilon-\mu)}+1)}$:
$$\tilde{f}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} d\epsilon \frac{e^{i\epsilon t}}{e^{\beta(\epsilon-\mu)}+1}=\frac{e^{i\mu t}}{2\pi \beta}\int dx \frac{e^{ixt/\beta}}{e^x+1}=\dots= \frac{e^{i\mu t}}{2\pi i}\left(\pi i \delta (t)+\frac{1}{t}\frac{\pi t /\beta}{\sinh(\pi t /\beta)}\right)$$
I have used $x=\beta(\epsilon-\mu)$.
Can anybody give me hints for the calculation steps in between?
 A: I have in my notes a  related Laplace transform: 
$$
I=\int_{-\infty}^{\infty} \frac{d\epsilon}{2\pi} e^{\tau\epsilon/2\pi} \left\{ \frac{1}{1+e^{\beta(\epsilon-\mu)}}-\theta(-\epsilon)\right\}= \frac 1{\tau}\left\{ \frac{(\frac{\tau T}{2})}{\sin(\frac{\tau T}{2})} e^{\tau\mu/2\pi}-1\right\}, \quad 0<\tau T/2\pi< 1.
$$
I evaluated it  goes as follows:
$$
I=\int_0^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau\epsilon/2\pi}}{1+e^{\beta(\epsilon-\mu)}}+\frac{e^{-\tau\epsilon/2\pi}}{1+e^{\beta(-\epsilon-\mu)}}-
e^{-\tau\epsilon/2\pi}\right\}\nonumber\\
= \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau\epsilon/2\pi}}{1+e^{\beta(\epsilon-\mu)}}\right\}-\frac 1\tau\\
= e^{\mu\tau/2\pi} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau(\epsilon-\mu)/2\pi}}{1+e^{\beta(\epsilon-\mu)}}\right\}-\frac 1\tau\\
=  e^{\mu\tau/2\pi} T\int_{-\infty}^\infty \frac{d\xi }{2\pi} \left\{\frac{e^{\xi T\tau/2\pi}}{1+e^\xi}\right\}-\frac 1\tau\\
= e^{\mu\tau/2\pi} T  \int_{0}^\infty \frac{dx }{2\pi}\frac{x^{T\tau/2\pi-1}}{1+x} -\frac 1\tau\\
=  \frac 1{\tau}\left\{ \frac{(\frac{\tau T}{2})}{\sin(\frac{\tau T}{2})} e^{\tau\mu/2\pi}-1\right\}.
$$
I set $x=\exp\{\beta(\epsilon-\mu)\}$ and at the last step   used the standard integral
$$
\int_0^\infty dx \frac{x^{\alpha-1}}{1+x}= \frac{\pi}{\sin\pi \alpha}, \quad 0<\alpha<1.
$$
There may be an easier way!
The integral is interesting because it seems to be related to the generating function for the  $\hat A$ genus that appears in the Dirac index theorem.  I learned this from a paper of Loganayagam and Surówka: arXiv:1201.2812
