content_image/360910.jpg content_image/360911.jpg content_image/360914.jpg content_image/360915.jpg content_image/360919.jpg content_image/360920.jpg content_image/360921.jpg content_image/360924.jpg
to arrive at 

The j = 1 term in the series is 

<img src='content_image/360910.jpg'>

<img src='content_image/360911.jpg'>

Writing the trace explicitly the x basis and using Eq. (21) shows that the j = 1 term is just the Fock exchange energy E$_{X}$ 

<img src='content_image/360914.jpg'>

Thus, we have 

This has naturally separated E$_{X}$ from the correlation energy Φ$_{c}$, 

<img src='content_image/360915.jpg'>

<img src='content_image/360919.jpg'>

The remaining frequency integrals with j ≥ 2 correspond to j th order autocorrelations of ¯ P ( t ) evaluated at t = 0, so the infinitesimal time shifts actually do not matter. Specifically, the j th term has an ω integral 

<img src='content_image/360920.jpg'>

We integrate over the times so that the infinitesimal region about t = 0 where ¯ P ( t ) ̸ = P ( t ) is of zero measure. Hence, we can safely replace ¯ P by P for the j ≥ 2 terms to get 

<img src='content_image/360921.jpg'>

In the second line, we added and subtracted the corresponding j = 1 term to the first line and summed up the series for the logarithm. In the third line, evaluating P ( t = 0) requires some care as plugging in t = 0 directly in Eq. (8) yields an indeterminate result. Rather, we take the limit t → 0 of P ( t ) in Eq. (8) to find 

<img src='content_image/360924.jpg'>

