Fourier transform of the hyperbolic secant function

[Marie_Curie]

Homework Statement

Hi there!

I'm just trying to figure out the Fourier transform of the hyperbolic secant function... I already know the outcome:

4\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2)

But sadly, I cannot figure out how to work round to it! :( maybe one of you could help me...

Homework Equations

I was thinking of using the geometric series 1/(1+q) = \sum (-q)^n for q = e^(-2*x), as the hyperbolic secant is 2*e^(-x)/(1+e^(-2*x)) .
And then you need to multiply the hyperbolic secant with e^(iωt) and integrate from -∞
up to ∞.
But... frankly, that was it. I have never managed to come to the result
4\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2)

Maybe someone could give me a hint to to solve this problem...? :)

Lots of greetings,
Marie

 

[Ray Vickson]

Homework Statement

Hi there!

I'm just trying to figure out the Fourier transform of the hyperbolic secant function... I already know the outcome:

4\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2)

But sadly, I cannot figure out how to work round to it! :( maybe one of you could help me...

Homework Equations

I was thinking of using the geometric series 1/(1+q) = \sum (-q)^n for q = e^(-2*x), as the hyperbolic secant is 2*e^(-x)/(1+e^(-2*x)) .
And then you need to multiply the hyperbolic secant with e^(iωt) and integrate from -∞
up to ∞.
But... frankly, that was it. I have never managed to come to the result
4\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2)

Maybe someone could give me a hint to to solve this problem...? :)

Lots of greetings,
Marie

f(x) = sech(x) is an even function, so the FT of f is (g + conjugate g), where $g = \int_0^{\infty} f(x) e^{i \omega x} \, dx.$ Apply this term-by-term to
\text{sech}(x) = 2 \sum (-1)^n e^{-(2n+1)x}.

 

[Marie_Curie]

f(x) = sech(x) is an even function, so the FT of f is (g + conjugate g), where $g = \int_0^{\infty} f(x) e^{i \omega x} \, dx.$ Apply this term-by-term to
\text{sech}(x) = 2 \sum (-1)^n e^{-(2n+1)x}.

Aaahh, wonderful! Thank you!

Just one more question... where does the Term (-1)^n come from?

Thanks a lot!
MARIE

 

[Dick]

Aaahh, wonderful! Thank you!

Just one more question... where does the Term (-1)^n come from?

Thanks a lot!
MARIE

It comes from the expansion of 1/(1+a)=1-a+a^2-a^3+... It's the factor that alternates sign.

 

[Marie_Curie]

It comes from the expansion of 1/(1-a)=1-a+a^2-a^3+... It's the factor that alternates sign.

Oh yes, sure! Now I got it :)

Thank you very much! :) And have a nice evening!

_MARIE_

 

[Dick]

Oh yes, sure! Now I got it :)

Thank you very much! :) And have a nice evening!

_MARIE_

You're welcome. Nice evening to you too. Note I goofed on the post. I meant 1/(1+a) not 1/(1-a).