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

Question

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

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 $.

You meant the singularities are $i\pi + 2k\pi i$, didn't you? — Daniel Fischer, Dec 14 '13 at 23:08
Thank you, Daniel. My question is: what theorem should I apply next? Does Jordan's lemma apply here? I'm thinking of a semicircular contour. — user3100957, Dec 15 '13 at 02:57

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

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.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 15 '13 at 04:32

Mhenni Benghorbal

47,431

1

that is Mellin? or Laplace? – obataku Dec 15 '13 at 04:48
1

@oldrinb: You are right. It is corrected. It happens a lot. One has something in his mind and writes something else. Thank you very much for your comment. – Mhenni Benghorbal Dec 15 '13 at 05:01
Mhenni Benghorbal: How did you arrive at $\sinh(\pi w)$? That strongly resembles Euler's Gamma reflection formula, but it's $\sin(\pi w)$ in the denominator. Also should I make the substitution $x=1/u-1$? Thanks. – user3100957 Dec 18 '13 at 05:41
@user3100957: put $x=\frac{1}{1+u}$. Yes, you will need the reflection formula. – Mhenni Benghorbal Dec 18 '13 at 13:45

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

1 Answers1

Linked