Arising out of my analytical solution to the 2-dimensional transient diffusion equation, I have to invert the following Laplace transform to get a surface flux in physical space (I have greatly simplified the notation for the purpose herein) I'd specifically like to know if the logic that I'm using to invoke the Residue Theorem is sound, and how to overcome an issue with the outcome of the evaluation of the residues.
$$ F(s)=\frac{q(s)\sinh[bq(s)]\sinh[ap(s)]}{g(s)}\frac{1}{p(s)}\Biggl[\frac{1}{s+cw}-\frac{1}{s+w}\Biggr]\hspace 2 in [\text{EQN 1}] $$
where
$$ p(s)=\sqrt{\frac{s}{c}+w}\hspace 1.5in q(s)=\sqrt{s+w} $$
and
$$ g(s)=q(s)\cosh[ap(s)]\sinh[bq(s)]+\phi p(s)\sinh[ap(s)]\cosh[bq(s)]\hspace 2 in [\text{EQN 2}] $$
In the above equations, $a,b,c,w$, and $\phi$ are positive real constants with $c>1$ (if it were equal to $1$, I'd be home for an early lunch!)
It is clear from the definitions of $p(s)$ and $q(s)$ that $s=-w$ and $s=-cw$ (corresponding to which $p(s)=0$) are branch points and not poles. Therefore these terms can be combined with the numerator of EQN. $1$, so that the Laplace transform can be written in the form
$$ F(s)=\frac{N(s)}{g(s)}\hspace 2 in [\text{EQN 3}] $$
where $$ N(s)=\frac{q(s)\sinh[bq(s)]\sinh[ap(s)]}{p(s)}\Biggl[\frac{1}{s+cw}-\frac{1}{s+w}\Biggr]\hspace 2 in[\text{EQN 4}] $$
For $s>-w$, there are no poles of $g(s)$ since the function is strictly positive.
Between the branch points, i.e., for $-cw<s<-w$, we have
$$ g(s)=\bar{q}(s)\cosh[ap(s)]\sin[b\bar{q}(s)]+i\Big[\phi p(s)\sinh[ap(s)]\cos[b\bar{q}(s)]\Bigr]\hspace 2 in [\text{EQN 5}] $$
where $\bar{q}(s)=\sqrt{-(s+w)}$, and there are no poles in this interval either, since there are no values of $s$ for which $\sin[b\bar{q}(s)]$ and $\cos[b\bar{q}(s)]$ are simultaneously zero, so that both real and imaginary parts vanish.
For $s<-cw$, the denominator of the Laplace Transform is
$$ g(s)=i\Biggl[\bar{q}(s)\cos[a\bar{p}(s)]\sin[b\bar{q}(s)]+\phi \bar{p}(s)\sin[a\bar{p}(s)]\cos[b\bar{q}(s)]\Biggr]\hspace 2 in [\text{EQN 6}] $$
where
$$\bar{p}(s)=\sqrt{-\Bigl(\frac{s}{c}+w\Bigr)}$$
For $s<-cw$, the denominator function $g(s)$ has an infinitum of poles along the negative real axis all the way between $s=-cw$ and $s=-\infty$. I could now create a closed contour by setting the $\gamma$ offset in the Bromwich Contour to $0$ and cuts around the branch points as shown below
PROBLEMS:
Assuming that everything above is logically sound, I could then invoke the Cauchy Residue Theorem and evaluate the Laplace Transform by subtracting the contour integrals along the negative real axis. This seems like an inordinate amount of work.
The integration around the branch cut at $s=-cw$ appears problematic as the radius of the circle around it shrinks to zero.
Is there an alternative contour that will let me sidestep the contour integration part, or some other clever workaround? Any advice on a simpler approach will be deeply appreciated.
One thing that helps is that the real part of the sum of the residues is zero, so that I can skip the Residue Theorem altogether!