Seja a uma constante. Mostre que
\begin{aligned} & \mathscr{L}^{-1}\left\{\frac{\operatorname{senh} a \sqrt{s}}{s \operatorname{senh} \sqrt{s}}\right\}= \\\ & \quad \sum_{n=0}^{\infty}\left[\operatorname{erf}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)-\operatorname{erf}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)\right] \end{aligned}

Passo 1

Resumo do enunciado: queremos calcular
\mathcal{L}^{-1}\Bigl\{\frac{\operatorname{sinh}(a\sqrt{s})}{s\,\operatorname{sinh}(\sqrt{s})}\Bigr\}.
O que vamos usar:
  1. A expansão em série
\frac{1}{\operatorname{sinh}(x)}=2\sum_{n=0}^{\infty}e^{-(2n+1)x},
  1. A transformada inversa conhecida
\mathcal{L}^{-1}\Bigl\{\frac{1}{s}e^{-b\sqrt{s}}\Bigr\} =\operatorname{erfc}\!\Bigl(\tfrac{b}{2\sqrt{t}}\Bigr).

Passo 2

Aplicamos a expansão com x=\sqrt{s}:
\frac{1}{\operatorname{sinh}(\sqrt{s})} =2\sum_{n=0}^{\infty}e^{-(2n+1)\sqrt{s}}.

Passo 3

Escrevemos \operatorname{sinh}(a\sqrt{s}) em termos de exponenciais:
\operatorname{sinh}(a\sqrt{s}) =\frac{e^{a\sqrt{s}}-e^{-a\sqrt{s}}}{2}.
Logo
\frac{\operatorname{sinh}(a\sqrt{s})}{s\,\operatorname{sinh}(\sqrt{s})} =\frac{1}{s}\sum_{n=0}^{\infty}\Bigl(e^{-(2n+1-a)\sqrt{s}} -e^{-(2n+1+a)\sqrt{s}}\Bigr).

Passo 4

Invertemos cada termo usando
\mathcal{L}^{-1}\Bigl\{\frac{1}{s}e^{-b\sqrt{s}}\Bigr\} =\operatorname{erfc}\!\Bigl(\tfrac{b}{2\sqrt{t}}\Bigr).
Assim
\mathcal{L}^{-1}\Bigl\{\tfrac{1}{s}e^{-(2n+1\mp a)\sqrt{s}}\Bigr\} =\operatorname{erfc}\!\Bigl(\tfrac{2n+1\mp a}{2\sqrt{t}}\Bigr),
e por linearidade
\mathcal{L}^{-1}\Bigl\{\tfrac{\operatorname{sinh}(a\sqrt{s})}{s\,\operatorname{sinh}(\sqrt{s})}\Bigr\} =\sum_{n=0}^{\infty}\Bigl[\operatorname{erfc}\!\bigl(\tfrac{2n+1-a}{2\sqrt{t}}\bigr) -\operatorname{erfc}\!\bigl(\tfrac{2n+1+a}{2\sqrt{t}}\bigr)\Bigr].

Passo 5

Usamos a identidade \operatorname{erfc}(u)=1-\operatorname{erf}(u) em cada termo:
\operatorname{erfc}\!\bigl(\tfrac{2n+1-a}{2\sqrt{t}}\bigr) -\operatorname{erfc}\!\bigl(\tfrac{2n+1+a}{2\sqrt{t}}\bigr) =\operatorname{erf}\!\bigl(\tfrac{2n+1+a}{2\sqrt{t}}\bigr) -\operatorname{erf}\!\bigl(\tfrac{2n+1-a}{2\sqrt{t}}\bigr).
Isso coincide com a série pedida.

Resposta

\sum_{n=0}^{\infty}\Bigl[\operatorname{erf}\!\bigl(\tfrac{2n+1+a}{2\sqrt{t}}\bigr) -\operatorname{erf}\!\bigl(\tfrac{2n+1-a}{2\sqrt{t}}\bigr)\Bigr].

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