Derivation of BCS condensation energy Many textbook derive BCS condensation energy as
$$\begin{array}{l}
{\left\langle E \right\rangle _s} - {\left\langle E \right\rangle _n}\\
 = \sum\limits_{\left| {\bf{k}} \right| > {k_F}} {\left( {{\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)}  + \sum\limits_{\left| {\bf{k}} \right| < {k_F}} {\left( { - {\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)}  - \frac{{{\Delta ^2}}}{V}\\
 = 2\sum\limits_{\left| {\bf{k}} \right| > {k_F}} {\left( {{\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)}  - \frac{{{\Delta ^2}}}{V}\\
 = \left( {\frac{{{\Delta ^2}}}{V} - \frac{1}{2}N\left( 0 \right){\Delta ^2}} \right) - \frac{{{\Delta ^2}}}{V}\\
 =  - \frac{1}{2}N\left( 0 \right){\Delta ^2}
\end{array}$$
But I am so stupid to see why
$$\sum\limits_{\left| {\bf{k}} \right| > {k_F}} {\left( {{\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)}  = \sum\limits_{\left| {\bf{k}} \right| < {k_F}} {\left( { - {\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)} $$
and also why
$$2\sum\limits_{\left| {\bf{k}} \right| > {k_F}} {\left( {{\xi _k} - \frac{{\xi _k^2}}{{{E_k}}}} \right)}  = \left( {\frac{{{\Delta ^2}}}{V} - \frac{1}{2}N\left( 0 \right){\Delta ^2}} \right)$$
I can derive
$$\sum\limits_{ - \hbar {\omega _D} < {\xi _k} < \hbar {\omega _D}} {\frac{1}{{2{E_k}}}}  = \sum\limits_{ - \hbar {\omega _D} < {\xi _k} < \hbar {\omega _D}} {\frac{1}{{2\sqrt {\xi _k^2 + {\Delta ^2}} }} = } \frac{1}{V}$$
But I don't know how condensation energy is derived. Can somebody help?
 A: First, the result for the condensation energy it is derived at T=0, where the chemical potential is exactly equal to the Fermi energy $\mu=E_F=\frac{\hbar^2k_F^2}{2m}$. Recall that $\xi_k=\epsilon_k-\mu=\frac{\hbar^2}{2m}(k^2-k_F^2)$ and the energy of the quasiparticle spectrum in the superconductor is $E_k=\sqrt{\xi_k^2+\Delta_k^2}$. 
In the mean field approximation, the order parameter $\Delta_k$ is $$\Delta_k=\cases{\Delta\qquad |\xi_k|<\hbar\omega_D\\0\ \qquad
|\xi_k|>\hbar\omega_D}$$ where $\hbar\omega_D$ is the energy cutoff of the phonon spectrum ($\hbar\omega_D\ll E_F$). Thus it results $E_k=|\xi_k|$ for $|\xi_k|>\hbar\omega_D$, and the second equation you wrote is more explicitly written as $$\sum_{k_D+k_F>|{\bf k}|>k_F}\left(\xi_k-\frac{\xi_k^2}{E_k}\right)=\sum_{k_F>|{\bf k}|>k_F-k_D}\left(-\xi_k-\frac{\xi_k^2}{E_k}\right),$$ where I arbitrarily defined a momentum related to the energy $\hbar\omega_D$, namely $k_D^2=\frac{2m}{\hbar^2}\hbar\omega_D$. 
This last equation holds because $\xi_k=\frac{\hbar^2}{2m}(k^2-k_F^2)$ is odd around $k_F$, therefore $$\sum_{k_D+k_F>|{\bf k}|>k_F-k_D}\xi_k=0\rightarrow\sum_{k_D+k_F>|{\bf k}|>k_F}\xi_k+\sum_{k_F>|{\bf k}|>k_F-k_D}\xi_k=0$$ which it is equivalent at the previous equation.
Concerning your second question, it is more convenient to replace the sum with an integral over the energy $$ 2\sum_{|{\bf k}|>k_F}\left(\xi_k-\frac{\xi_k^2}{E_k}\right)\rightarrow 2N_0\int_0^{\hbar\omega_D}dE\left(E-\frac{E^2}{\sqrt{E^2+\Delta^2}}\right),$$ where $N_0$ is the density of states at the Fermi level in the normal state (it is customary to neglect its small variation on this energy scale). Try to work out this integral, then use the fact that $\hbar\omega_D\gg\Delta$ in the weak coupling limit. Note that the last equation you wrote can be rewritten as $$N_0\int_0^{\hbar\omega_D}\frac{dE}{\sqrt{E^2+\Delta^2}}=\frac{1}{V}.$$
