Integral relating to Fourier transform of tanh

Question

I'm trying to find the Fourier transform of $\tanh(x) - \operatorname{sgn}(x)$, and I ended up with the following integral:

$$\int_0^{\infty} \frac{\sin kx}{e^{2x} + 1} d x$$

Any help is appreciated.

Answer 1

We can also use complex integration. $$I(k)=\int_0^\infty \frac{\sin kx}{e^{2x} + 1} dx=\frac12\int_0^\infty (1-\tanh x)\sin kx\,dx$$ $$\overset{IBP}{=}-\frac{\cos kx}{2k}(1-\tanh x)\,\bigg|_{x=0}^\infty-\frac1{4k}\int_{-\infty}^\infty\frac{e^{ikx}}{\cosh^2x}dx=\frac1{2k}-\frac1{4k}J(k)$$ Integrating along the rectangular contour $C: -R\to R\to R+\pi i\to-R+\pi i\to-R$ $$J(k)\big(1-e^{-\pi k}\big)=\oint_C\frac{e^{ikz}}{\cosh^2z}dz=2\pi i\underset{z=\frac{\pi i}2}{\operatorname{Res}}\frac{e^{ikz}}{\cosh^2z}=2\pi k\,e^{-\frac{\pi k}2}$$ $$\Rightarrow\,\, J(k)=\frac{\pi k}{\sinh\frac{\pi k}2}$$ $$\boxed{\,\,I(k)=\frac1{2k}-\frac\pi 4\frac1{\sinh\frac{\pi k}2}\,\,}$$

Answer 2

I suggest another approach. For $x>0$ we have $$ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{1 - e^{-2x}}{1 + e^{-2x}} = 1 + 2 \sum_{n=1}^{\infty} (-e^{-2x})^n = 1 + 2 \sum_{n=1}^{\infty} (-1)^n e^{-2nx} $$ and $\operatorname{sgn}x=1$ so $$ \tanh x - \operatorname{sgn}x = 2 \sum_{n=1}^{\infty} (-1)^n e^{-2nx}. $$ Calculate the one-sided Fourier transform of each term and then sum them. Then do the same for $x<0$ and add that result.

Answer 3

Another approach.

$$\int \frac{e ^{ikx}}{e^{2x} + 1}\, d x=-\frac i {k}\,e^{i k x} \, _2F_1\left(1,\frac{i k}{2};\frac{2+i k}{2};-e^{2 x}\right)$$ $$\int_0^\infty \frac{e ^{ikx}}{e^{2x} + 1}\, d x=\frac{1}{4} \left(H_{-\frac{i k}{4}}-H_{-\frac{i k}{4}-\frac{1}{2}}\right)$$ $$\int \frac{\sin(kx)}{e^{2x} + 1}\, d x=\frac{1}{2 k}-\frac{\pi}{4} \text{csch}\left(\frac{k\pi }{2}\right)$$