$$\int_{-\infty}^\infty e^{i \omega x-i b \tanh( x)}dx$$
I have to solve this integral for my research project .I substituted $ \tanh{x} = y $ and $ \tanh^{-1}{x} = \frac{1}{2}\log{\frac{1+y}{1-y}} $, and finally got the integral
$$ \int_{-1}^{+1} dy \, (1+y)^{\frac{\iota \omega}{2}-1} \, (1-y)^{\frac{-\iota \omega}{2}-1} \, e^{-\iota \omega y} $$
where $ \omega $ is the frequency and is positive and continuous.
and then here we have Closed form of $\frac1s\int_0^\infty e^{-s(\tanh(t)+at)}dt=\frac1s\int_0^1e^{-st}(1-t)^{\frac{as}2-1}(1+t)^{-\frac{as}2-1}dt$ to invert $\tanh(x)+ax$? . My question is does this provide answer to my question .