I have a couple questions on the mathematics behind some general value at risk (VaR) and expected shortfall (ES) calculations. $$ P(L > VaR(\alpha))=\alpha $$ $$ ES(\alpha)=E[L|L > VaR(\alpha)] $$
First, I need to find that, for distribution $R$ with pdf $f$ , $$ ES(\alpha) = \frac{-S_0}{\alpha}\int^{q_\alpha}_{-\infty}rf(r)dr $$ I have done the following $$ ES(\alpha) = E[L|L>VaR(\alpha)] = E[-S_0R|L>VaR(\alpha)] $$ $$ = \int^{q_\alpha}_{-\infty}-S_0r\frac{P(R=r)}{P(L>VaR(\alpha))}dr = -S_0\int^{q_\alpha}_{-\infty}r\frac{f(r)}{\alpha}dr = \frac{-S_0}{\alpha}\int^{q_\alpha}_{-\infty}rf(r)dr $$ I understand this for the most part, but I am somewhat confused as to how the $q_\alpha$ comes into the fray?
Second, I need to find that, for distribution $R\sim N(\mu, \sigma^2)$, $$ ES(\alpha)=-S_0\left(\mu-\frac{\sigma}{\alpha\sqrt{2\pi}}e^{\frac{-z^2_\alpha}{2}}\right) $$ where $z_\alpha$ is the $\alpha$th quantile of $N(0,1)$. I assume that I follow from where the first part left off and use the normal distribution pdf in place of $f(r)$ such that $$ ES(\alpha)= \frac{-S_0}{\alpha}\int^{q_\alpha}_{-\infty}r\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(r-\mu)^2}{2\sigma^2}}dr $$ where $\frac{(r-\mu)^2}{\sigma^2}=z^2$. However, how this would connect back to the given $ES(\alpha)$ is very unclear to me.
For the first part, you are conditioning on on $L>VaR(\alpha),$ which means $R<q_\alpha.$ The density of $R,$ conditional on $R<q_\alpha$ is zero for $R>q_\alpha,$ so when you integrate over it to get the conditional expectation, the integral cuts off at $q_\alpha.$ You are correct that the conditional density is given by $\frac{f(r)}{\alpha}$ for $r<q_\alpha,$ and your line of reasoning to derive the formula looks mostly right, except that $P(R=r)$ is not the same thing as the density $f(r)$ (in fact $P(R=r)=0$ for any $r$ assuming the distribution is continuous).
For the second part, you are right that you just need to plug the normal distribution. As for where to go from there, you have to compute the integral. To do this, substitute $u=(r-\mu)/\sigma$ and get $$ \frac{-S_0}{\alpha}\int^{q_\alpha}_{-\infty}r\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(r-\mu)^2}{2\sigma^2}}dr = \frac{-S_0}{\alpha}\int^{z_\alpha}_{-\infty}(\mu+\sigma u)\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}du\\=-\frac{S_0\mu}{\alpha}\int_{-\infty}^{z_\alpha}\frac{1}{\sqrt{2\pi}}e^{-u^2/2}du -\frac{S_0\sigma}{\alpha\sqrt{2\pi}}\int_{-\infty}^{z_\alpha} ue^{-u^2/2}du$$ where we used the fact that $z_\alpha=(q_\alpha-\mu)/\sigma.$ The first integral is $\alpha$ by definition and the second can be done with a substitution.