# Any simplifications possible for the following integral?

Here is an integral $$\kappa(a)=\int_0^a\mathop{\mathrm{d}z}f(z)\mathrm{e}^{\int_0^zf(z')\mathrm{d}z'+\int_0^zg(z')\mathrm{d}z'},$$ where all functions are real and positive and $a>0$. I am also given $$\xi(a)=\int_0^a\mathop{\mathrm{d}z}f(z)\mathrm{e}^{\int_0^zf(z')\mathrm{d}z'}.$$

My goal is to write a simple expression for $\kappa(a)$ in terms of $\xi(a)$, and some simple form that involves $g(z)$. Is there any simplifications possible to the above integral either by integration by parts or any other technique? My aim would be to get rid of integrals on the exponentials, or atleast to make the above integral more comprehensible.

The simplification I am able to make is $$\kappa(a)=\int_0^a\mathop{\mathrm{d}z}\frac{d}{dz}\left[\mathrm{e}^{\int_0^zf(z')\mathrm{d}z'}\right]\mathrm{e}^{\int_0^zg(z')\mathrm{d}z'}.$$

$$\kappa(a)=\int_0^a\mathop{\mathrm{d}z}f(z)\mathrm{e}^{\int_0^zf(z')\mathrm{d}z'+\int_0^zg(z')\mathrm{d}z'},$$ where $$\xi(a)=\int_0^a\mathop{\mathrm{d}z}f(z)\mathrm{e}^{\int_0^zf(z')\mathrm{d}z'}.$$
This means that $$\xi(a)=\int_0^a \mathrm{d}z \frac{d}{dz} e^{\int_0^zf(z')\mathrm{d}z'} = e^{\int_0^af(z')\mathrm{d}z'}-1.$$ Or rewritten as $$\int_0^af(z) dz = \log [\xi(a)+1]$$
As far as $\kappa(a)$ goes, I think the best you can do is integration by parts to get an expression involving $\xi$ and $g$ but no $f$.
$$\kappa(a) = \int_0^a dz f(z)e^{\int_0^zf(z') dz'}e^{\int_0^zg(z') dz'} \\ = \int_0^a dz \left( \frac{d}{dz}e^{\int_0^zf(z') dz'}\right) e^{\int_0^zg(z') dz'} \\ = e^{\int_0^af(z') dz'} e^{\int_0^ag(z') dz'} - \int_0^a dz e^{\int_0^zf(z') dz'} \left( \frac{d}{dz}e^{\int_0^zg(z') dz'}\right) \\ = \xi(a) e^{\int_0^ag(z') dz'} - \int_0^a dz \xi(z) \left( \frac{d}{dz}e^{\int_0^zg(z') dz'}\right)$$