Fourier transform of $f(x)=\frac{1}{e^x+e^{-x}+2}$

Solution 1:

Related techniques: (I), (II). First, we recall the Mellin transform of a function $f$

$$ F(s) =\int_{0}^{\infty} x^{s-1}f(x) dx .$$

Now, making the change of variables $u=e^{x}$ in the original integral gives

$$I = \int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{-iwx}}{{\rm e}^{x}}}{ \left( 1+{{\rm e}^{x}} \right) ^{2}}}{dx}= \int _{0}^{\infty }\!{\frac {{u}^{-iw}}{ \left( 1+u \right) ^{2}}}{du}. $$

Now, the last integral is nothing but the Mellin transform of the function $\frac{1}{(1+u)^2}$ with $s=1-iw$ which is given by

$$ I = \frac{\pi w}{\sinh(\pi w)} $$

Note: To find the Mellin transform of the function $\frac{1}{(1+u)^2}$, you can use the $\beta$ function. See here for the technique.