Assume that part c) has been proved and ignore parts c) & d).
To invert the Laplace transform we would do $\displaystyle u(x,t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{{\gamma+i\infty}}\overline{u}(x,p)e^{pt}dp$
So how would you find the poles (and their orders) and the residues. My solution suggests using approximation but I am unsure as to how this works.
The answer is there are simple poles at $p=i\pi(2n+1)$, $n\in \mathbb{Z}$.
Thanks
Let $f(x,p)=(\frac{2}{p^3}(\frac{\sinh xp+\sinh p(1-x)}{\sinh p })-1)+\frac{x(x-1)}{p})e^{pt}$,
$=(\frac{2}{p^3}(\frac{\sinh xp+\sinh p(1-x)-p^3\sinh p+x(x-1)p^2\sinh p}{\sinh p }))e^{pt}$
I believe now that we have a pole of order $1$ at $(x,0)$ for $x\ne 0,\ne 1$, i.e a simple pole