I would like to compute the Fourier transform of the product of $\tanh(x)$ and the Heaviside step function $H(x)$, i.e.
$$\int_{-\infty}^{\infty} H(x)\tanh(x)e^{-ikx}dx = \int_{0}^{\infty} \tanh(x)e^{-ikx}dx$$
For the Fourier transform of $\tanh(x)$ alone, I have read that the use of differentiation:
$$ik\mathcal{F}\left[ \tanh(x)\right](k) = \mathcal{F}\left[ \text{sech}^2(x)\right](k)$$
is a possible way of proceding, as the integral does not exist as a classical Riemann integral. I have then read derivations of the Fourier transform of $\text{sech}^2(x)$ using contour integration.
Adapting for the new problem I have supposed:
$$ik\mathcal{F}\left[ H(x)\tanh(x)\right](k) = \mathcal{F}\left[H(x) \text{sech}^2(x)\right](k)$$
However, I cannot work out how to adapt contour integration to the half line integral from 0 to $\infty$ instead.
Can someone explain how to compute the Fourier transform either by the method outlined above or a different method?
Thank you!
notice that: $$I=\int_{-\infty}^\infty \text{H}(x)\tanh(x)e^{-ikx}dx=\int_0^\infty\frac{e^{2x}-1}{e^{2x}+1}e^{-ikx}dx=\int_0^\infty\frac{u^{-ik}(u-1)}{u^2+1}$$ now notice that: $$I=\int_0^\infty\frac{u^{1-ik}}{u^2+1}-\frac{u^{-ik}}{u^2+1}du$$ if we let: $$J(a)=\int_0^\infty\frac{u^a}{u^2+1}du=\frac{1}{2}\int_1^\infty(v-1)^{\frac{a-1}{2}}v^{-1}dv=\frac{(-1)^{\frac{a-1}{2}}}{2}\int_1^\infty(1-v)^{\frac{a-1}{2}}v^{-1}dv$$ $$J(a)=\lim_{z\to\infty}\frac{(-1)^{\frac{a-1}{2}}}{2}\left[B\left(z;0,\frac{a+1}{2}\right)-B\left(1;0,\frac{a+1}{2}\right)\right]$$ and we know that: $$I=J(1-ik)-J(-ik)$$ where $B$ is the incomplete beta function