Laplace transform of a function defined by integral

Question

I have a function of the form $f(u) = ke^{-au}\int_{0}^{\infty}x^{z-1}(1+x)^{-z-1}e^{-bxu}dx$ where $a,b,u,z>0$ are all real positive numbers and $k$ is a positive normalization factor. I need to calculate the Laplace transform of $f(u)$. We have

$$F(s) = \int_{0}^{\infty} f(u)e^{-su}du = k\int_{0}^{\infty} e^{-au} \left(\int_{0}^{\infty}x^{z-1}(1+x)^{-z-1}e^{-bxu}dx\right) e^{-su} du$$

Since Fobini's theorem holds (Does it really hold?) we can change the order of integration

$$F(s) = k\int_{0}^{\infty} x^{z-1}(1+x)^{-z-1}\left(\int_{0}^{\infty} e^{-(a+s+bx)u}du\right) dx= k\int_{0}^{\infty} \frac{x^{z-1}(1+x)^{-z-1}}{a+s+bx} dx $$

I can't go any further! What should I do next? Also I'm contemplating the change of integration order, is it valid and correct?!? What about the region of convergence for Laplace transform, how to find it?

Thanks in advance!

==============================Corrections============================= The values that $z$ can assume are positive real numbers $z>0$ however if integral diverges, then we can restrict it to other values!

Answer 1

You probably have to use $$ _2F_1(a,b;c; z)=\frac{1}{B(b,c-b)}\int_0^1\frac{x^{b-1}(1-x)^{c-b-1}}{(1-xz)^a}dx$$ by a suitable change of variable. Your $a$ can be taken as $0$ without loss of generality.

Answer 2

I don't see Kummer functions. From the Svatovslav presention we get at least formally

$$F(s)=\frac{1}{a+s}\times B(z,z+1)\ _2F_1(1,z;z+1; 1-\frac{b}{a+s}).$$ Discussion when $z$ is not a positive number seems complicated.