As the title suggests, I would like to show that the Laplace Transform, $\mathcal{L}\{\dfrac{1}{t}\sinh(at)\}=\dfrac{1}{2} \ln (\dfrac{s+a}{s-a})$, $s>|a|$ without evaluating the Laplace integrals explicitly.
I first expressed $\dfrac{1}{t}\sinh(at)$ as $\dfrac{1}{t}\big(\dfrac{e^{at}-e^{-at}}{2}\big)=\dfrac{1}{2t}e^{at}-\dfrac{1}{2t}e^{-at}=\dfrac{1}{2}t^{-1}e^{at}-\dfrac{1}{2}t^{-1}e^{-at}$.
Looking at the Laplace table, indeed, we have the result that $\mathcal{L}\{t^{n}e^{at} \}= \dfrac{n!}{(s-a)^{n+1}}$ if $n=1,2,3,\dots$ .
I am stuck here as there is no explicit formulas with $n=-1$ for my problem. I am thinking if I should use the convolution theorem to further simplify my expression but I am quite unsure. What should I do to manipulate the expression further any help will be appreciated!
The Laplace transform of $\frac{f(t)}{t}$ is naturally related to the primitive of the Laplace transform of $f(t)$.
We have $\mathcal{L}(\sinh(at))(s) = \frac{a}{s^2-a^2}$, whose primitive is $-\text{arctanh}\frac{s}{a}$, and we are essentially done.