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QUESTION

Use the complex inversion formula to show that
$${\cal{L}}^{-1}\frac{\sinh sx}{s^2\sinh
sa}=\frac{xt}{a}+\frac{2a}{\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\sin\frac{n
\pi x}{a}\sin \frac{n \pi t}{a}.$$

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ANSWER
$$f(s)=\frac{\sinh sx}{s^2\sinh sa}$$
 This has simple poles at $s=\frac{n \pi i}{a}$ and a double pole
 at $s=0$
 \begin{eqnarray*}
 f(s)e^{st}&=&\frac{\left(sx+\frac{(sx)^3}{3!}+\ldots\right)
 \left(1+st+\frac{(st)^2}{2!}+\ldots\right)}{s^2\left(sa+\frac{(sa)^3}{s!}\right)}\\
 &=&\frac{x}{as^2}\frac{\left(1+\frac{(3x)^2}{6}+\ldots\right)
 \left(1+st+\frac{(st)^2}{2!}+\ldots\right)}{\left(1+\frac{(sa)^2}{6}\right)}\\
 &=&\frac{x}{as^2}(1+st+O(s^2))\\
 &\Rightarrow&\textrm{Res}(0)=\frac{xt}{a}
 \end{eqnarray*}
 To find $\textrm{Res}\left(\frac{n \pi i}{a}\right)$ use l'H\^opital's rule for simple
 poles
 \begin{eqnarray*}
  f(s)e^{st}&=&\lim_{s \to \frac{n \pi i}{a}}\frac{\sinh sx
  e^{st}}{s^2a\cosh sa +2s\sinh a}\\
  &=&\frac{\sinh \frac{n \pi i}{a}xe^{\frac{n \pi
  i}{a}t}}{\left(\frac{n\pi i}{a}\right)^2a \cosh n \pi i}\\
  &=&\frac{i \sin \frac{n \pi x}{a}e^{\frac{n \pi i}{a}t}}{-\left(\frac{n
  \pi}{a}\right)^2a(-1)^n}\\
  F(t)&=&\frac{xt}{a}+\sum_{n=1}^\infty \textrm{Res}\left(\frac{n \pi
  i}{a}\right)+\textrm{Res}\left(-\frac{n \pi i}{n}\right)\\
  &=&\frac{xt}{a}+\sum_{n=1}^\infty \frac{2a}{n^2 \pi
  ^2}(-1)^n\sin \frac{n \pi x}{a} \sin \frac{n \pi t}{a}
 \end{eqnarray*}


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