Does laplace transform tanht exists?

Question

I am trying to calculate laplace transform of tanht. My question is that does its laplace transform exists?

And if exists then what will be its laplace transform?

Answer 1

We're interested in $L(s):=\int_0^\infty\tanh t\exp(-st)dt$. The integrand is $\sim t$ for small $t$, and $\sim\exp(-st)$ for large $t$, so $L$ converges for all $s>0$. For such $s$,$$\begin{align}L(s)&=\int_0^\infty\frac{1-\exp(-2t)}{1+\exp(-2t)}\exp(-st)dt\\&\stackrel{\ast}{=}\sum_{k\ge0}(2-\delta_{k0})(-1)^k\int_0^\infty\exp(-(s+2k)t)dt\\&=-\frac1s+\sum_{k\ge0}\frac{(-1)^k}{s/2+k}\\&=-\frac1s+\Phi(-1,\,1,\,s/2)\end{align}$$in terms of the Lerch zeta. Note $\stackrel{\ast}{=}$ uses the dominated convergence theorem.

Answer 2

$$\tanh x=1-2\frac{e^{-x}}{e^x+e^{-x}}=1+2\sum_{k=1}^\infty (-1)^ke^{-2kx}$$

so that, term by term,

$$\mathcal L\tanh x=\frac1s+2\sum_{k=1}^\infty\frac{(-1)^k}{s+2k}.$$