Hi guys,

I have a simple calculus question, (so simple it's actually embarrassing)

So I'm reading this article that says:

$\displaystyle \mbox{if we have: } p(D_i,D_i^* \mid \Psi) = \frac{1}{4w_i} sech^2(\frac{D_i^* - D_i}{2w_i}) $

$\displaystyle \mbox{then: }\sum\limits_{i=1}^{N}log \frac{\int_{d_i}^{\infty} p(D_i, D_i^* \mid \Psi) dD^* }{\int_{0}^{d_i} p(D_i, D_i^* \mid \Psi) dD^* } = \sum\limits_{i=1}^{N}w_i (d_i - D_i) $

But I can't get that result!!!

Can anyone show me the intermediate steps please?