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

Let $$f(x)=\large \frac{1}{e^x+e^{-x}+2}$$ Compute the Fourier transform of $f$.

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We can factor the denominator to get $$f(x)=\frac1{(\exp(x/2)+\exp(-x/2))^2}=\frac1{(2\cosh(x/2))^2}$$ I'm thinking of using residue from complex analysis. To find the singularity, we have $$\exp(x/2)=-\exp(-x/2)\iff\exp(x)=-1$$ We know $\exp(i\pi)=-1$. So the singularities are $i\pi+2\pi k i $.

Answer

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.