Deriving the extinction cross section

Reading through Scattering, Absorption, and Emission of Light by Small Particles by Mishchenko, I came across a formula to calculate the extinction and scattering cross section (Book page 57, PDF page 77, Link to the book, it’s publicly available).
Sadly there are not many proofs of the formulas and usually they are thrown with the words ‘… we derive after some algebra’.

So I sat down and wanted to derive the expression for the power caused by the extinction Poynting vector, but struggled to get the same result as in the book, and in other literature for that matter. Just for notice, the book says:
$$ W^{ext} = frac{2pi}{k_1}sqrt{frac{epsilon_1}{mu_0}}ImBig[E^{sca}_1(hat{n}^{inc})cdot E^{inc ast}_0Big]
C^{ext} = frac{W^{ext}}{I^{inc}} = frac{frac{2pi}{k_1}sqrt{frac{epsilon_1}{mu_0}}ImBig[E^{sca}_1(hat{n}^{inc})cdot E^{inc ast}_0Big]}{frac{1}{2}sqrt{frac{epsilon_1}{mu_0}}mid E^{inc}_0mid^2} = frac{4pi}{k_1mid E^{inc}_0mid^2}ImBig[E^{sca}_1(hat{n}^{inc})cdot E^{inc ast}_0Big] $$

Having that mind we have that
$$ langle S^{ext}(pmb{r}) rangle = frac{1}{2}ReBig{E^{inc}(pmb{r})times[H^{sca}(pmb{r})]^ast + E^{sca}(pmb{r})times[H^{inc}(pmb{r})]^ast Big}$$
$$E^{inc}(pmb{r}) = E^{inc}_0 exp(ik_1hat{pmb{n}}cdotpmb{r}) = frac{i2pi}{k_1}Big[delta(hat{pmb{n}} + hat{pmb{r}})frac{e^{-ik_1r}}{r} – delta(hat{pmb{n}} – hat{pmb{r}})frac{e^{ik_1r}}{r}Big]E^{inc}_0 quadtext{for}quad k_1rrightarrowinfty $$
$$H^{inc}(pmb{r}) = sqrt{frac{epsilon_1}{mu_0}} hat{pmb{n}} times E^{inc}(pmb{r})$$
$$E^{sca}(pmb{r}) = frac{e^{ik_1r}}{r}E^{sca}_1(hat{pmb{r}})$$
$$H^{sca}(pmb{r}) = sqrt{frac{epsilon_1}{mu_0}} hat{pmb{r}} times E^{sca}(pmb{r})$$
where $E^{inc}$ and $H^{inc}$ is the incidence field and $E^{sca}$, $H^{sca}$ is the scattered field, $hat{pmb{r}} = pmb{r} / r$, and $hat{pmb{n}}=hat{n}^{inc}$ is the direction of the incidence field.

Putting this in the definition of the power gives
$$W^{ext} = -r^2int_{4pi}dhat{pmb{r}} langle S^{ext}(pmb{r}) rangle cdot hat{pmb{r}} = …$$
$$… = frac{pi}{k_1}sqrt{frac{epsilon_1}{mu_0}} ImBig{int_{4pi}dhat{pmb{r}}Big[delta(hat{pmb{n}} + hat{pmb{r}})e^{-2ik_1r} – delta(hat{pmb{n}} – hat{pmb{r}})Big]Big[E^{incast}_0 cdot E^{sca}_1(hat{pmb{r}})^astBig] – int_{4pi}dhat{pmb{r}}Big[delta(hat{pmb{n}} + hat{pmb{r}})e^{2ik_1r} – delta(hat{pmb{n}} – hat{pmb{r}})Big]Big[E^{sca}_1(hat{pmb{r}})^astcdot E^{incast}_0Big]Big}$$
I am stuck here, especially since I do have both fields complex conjugated and not alternating. I am using the fact, that $a^asttimes(btimes c) = (acdot c)b – (acdot b)c$ and $E^{inc}_0 cdot hat{pmb{n}} = 0$.

I wont expect someone to dive into the material too deep and help me solve this problem I have. A suggestion on where to find a solution to this would be more than appriciated or some hints on how to solve it. I am sorry if the post is too lengthy, so I skipped the part in me deriving the formula. If longer posts are not a problem here, I can also write down the whole procedure. Sorry for any mistakes, I am new in this community 🙂