Inverse fourier transform of sech(mt/2)^(1/m)

**AlexisM** · July 19th 2010, 03:15 AM

Hi,

could someone help me find a closed form for the inverse fourier transform of $\text{sech}(m\omega/2)^{1/m}$

Namely $\frac{1}{2\pi}\int_{-\infty}^{\infty} \exp(it\omega) \text{sech}(m\omega/2)^{1/m} d\omega$

sech is the hyperbolic secant :

$<br /> \text{sech}(x) = \frac{2}{(\exp(x) + \exp(-x))}<br />$

and m>0

Thanks for you help

AlexisM

**Ackbeet** · July 19th 2010, 05:46 AM

Could you please be more explicit about what is in the argument of the sech function? Do you mean

$\text{sech}^{1/m}(mt/2),$ or
$\text{sech}((mt/2)^{1/m})?$

You have to be very careful with parentheses!

**AlexisM** · July 19th 2010, 07:43 AM

Originally Posted by Ackbeet

Could you please be more explicit about what is in the argument of the sech function? Do you mean

$\text{sech}^{1/m}(mt/2),$ or
$\text{sech}((mt/2)^{1/m})?$

You have to be very careful with parentheses!

it's $\text{sech}^{1/m}(mt/2) = (\text{sech}(mt/2))^{1/m}$

and i want $\frac{1}{2\pi}\int_{-\infty}^{\infty} \text{exp}(itx) (\text{sech}(mt/2))^{1/m}dx$

which is, I think, the same as :

$\frac{1}{\pi}\int_{0}^{\infty} \text{cos}(tx) (\text{sech}(mt/2))^{1/m}dx$

**Ackbeet** · July 19th 2010, 08:05 AM

WolframAlpha balks at that integral (as does my local copy of Mathematica, as you'd expect). Are you sure the inverse FT has a closed-form solution? It sure doesn't look like it.

Also, I should point out that the two integrals you claim are equal are only equal if the first one is real. If you take the real part of the first integral, you get the second one.

[EDIT]: I think you also inadvertently changed the limits of integration in the second integral without multiplying by two to take advantage of the symmetry.

**AlexisM** · July 19th 2010, 08:22 AM

Originally Posted by Ackbeet

WolframAlpha balks at that integral (as does my local copy of Mathematica, as you'd expect). Are you sure the inverse FT has a closed-form solution? It sure doesn't look like it.

Also, I should point out that the two integrals you claim are equal are only equal if the first one is real. If you take the real part of the first integral, you get the second one.

[EDIT]: I think you also inadvertently changed the limits of integration in the second integral without multiplying by two to take advantage of the symmetry.

yes i forgot the factor 2 (i'll edit the previous post to correct that). This, I think possible, because the function sech is even. So if i split the integral on both side of zero, replaces x with -x, dx with -dx and invert the sens of integration I can factorize and get a factor exp(itx)+exp(-itx) = 2*cos(tx).

I'm not sure it has a closed form (may be it's not the good expression but I am french so please forgive me if I misuse some expressions) , but i'm about sure you can give an answer in the form of a serie or something like that.

**Ackbeet** · July 19th 2010, 08:38 AM

[EDIT]: You can try a term-by-term series expansion integration. However, perhaps a more fundamental question is this: does the inverse FT integral you've given us even converge?

**AlexisM** · July 19th 2010, 10:24 AM

it converges (it is computable with closed form) for m->0, m=1, m->infty.

Here is a demonstration for the FT of $\text{sech}(\alpha x)$ using residues.

I am not familiar with residues but I wonder if the same method can't be used to compute the inverse FT of $(\text{sech}(m \omega/2))^{1/m}$ using the similarity between
$\int_{-\infty}^{\infty} exp(- i t \omega) \text{sech}(\alpha \omega) d \omega$ and
$\int_{-\infty}^{\infty} (exp(i m t \omega) \text{sech}(m \omega /2 ))^{1/m} d \omega$

**Ackbeet** · July 19th 2010, 12:58 PM

[EDIT]: This post is mostly bosh.

Computable with closed form is not the same thing as convergent. In this case, by convergent, I mean that the integral is finite and "settles down" to one specific value.

Let's look at the asymptotics here. Suppose we look at
$\omega\to\infty$ in the integral $\int_{-\infty}^{\infty}\cos(t\omega)\,\text{sech}^{1/m}(m\omega/2)\,d\omega.$

Then $\text{sech}(m\omega/2)\to e^{m\omega/2}/2.$
This implies that

$\displaystyle{\text{sech}^{1/m}(m\omega/2)\to \frac{e^{\omega/2}}{2}}.$

If we take half the integral, namely,

$\int_{0}^{\infty}\cos(t\omega)\,\frac{e^{\omega/2}}{2}}\,d\omega,$

we see that this integral does not converge.

These are not rigorous arguments. However, the problem I see with your inverse FT is that you've got oscillatory behavior (which is good for convergence, usually) multiplied by an exponential function with a positive exponent (not good). So it's not merely that you've got a very large function: you've got huge alternating areas "canceling" each other out. Basically, it's a divergent integral with no defined value, so far as I can see.

Are you sure that you can even take the inverse FT of this function?

**AlexisM** · July 19th 2010, 01:52 PM

I think we have a confusion about the définition of $\text{sech}(x)$

$\text{sech}(x) = \frac{2}{\text{exp}(-x) + \text{exp}(x)}$

This function "behaves" well, since it decreses exponentially.

And when $\omega\to\infty$ , then $\text{sech}(m\omega/2)\to 2 e^{-m\omega/2}$
and
$\displaystyle{\text{sech}^{1/m}(m\omega/2)\to 2 e^{-\omega/2}.$

So we have an oscillatory function damped with an exponentially decreasing function. I can't see why it would not converge.

**Ackbeet** · July 19th 2010, 03:23 PM

Oops. You're right. I was thinking cosh. Sech is a very nicely behaved function, I agree. And I agree that the integral should converge. However, it does not seem to have a nice antiderivative. I would probably go with the series solution.

**AlexisM** · July 20th 2010, 01:01 AM

Someone pointed to me this result from "A course on modern analysis, of Whittaker and Watson" :
$<br /> \text{B}(t+r,t-r) = \frac{1}{4^{t-1}} \int_0^{\infty} \frac{\text{cosh}(2ru)}{\text{cosh}^{2t}(u)}du$
for $|r| < t$ where $\text{B}$ is the beta function.

If I do the subsitutions :
$t = \frac{1}{2m}$

$u = \frac{m \omega}{2}$

$r = i \frac{\lambda}{m}$

this gives :

$<br /> \text{B} (\frac{1}{2m}+i \frac{\lambda}{m},\frac{1}{2m}-i \frac{\lambda}{m}) = \frac{m}{2}\frac{1}{4^{\frac{1}{2m}-1}} \int_0^{\infty} \frac{\text{cos}(\lambda \omega)}{\text{cosh}^{\frac{1}{m}}(\frac{m \omega}{2})}\ d \omega$
for $|\lambda| < 1/2$

Thus :
$\displaystyle \int_0^{\infty} \frac{\text{cos}(\lambda \omega)}{\text{cosh}^{\frac{1}{m}}(\frac{m \omega}{2})}\ d \omega = \frac{2^{\frac{1}{m}-1}}{m}\text{B} (\frac{1}{m}(\frac{1}{2}+i \lambda}),\frac{1}{m}(\frac{1}{2}-i \lambda}))$

This holds for a limited range of values of $\lambda$ . Is there any reason that this integral could not exists for other values of $\lambda$ ? Or does it mean that we can have different forms of the solution for different domains ?

**Ackbeet** · July 20th 2010, 05:34 AM

Wow. That's some pretty heavy analysis you have there. Good work. On the face of it, it seems like that approach could work just fine. If you look here, you'll see that the real parts of both arguments to the beta function must be positive. You have that here, since in the OP you mentioned that m>0. There doesn't seem to me any a priori reason why you couldn't have any values for $\lambda$ .

**AlexisM** · July 22nd 2010, 02:28 AM

Yes, I think that the $|r|<t$ is just there to ensure that the real part of arguments is positive. So here I don't have to care about it. Would you know where I can find the demonstration for the formula, or some hints to do it ?

$\displaystyle B(t+r,t-r) = \frac{1}{4^{t-1}}\int_0^\infty \frac{\text{cosh}(2ru)}{\text{cosh}^{2t}(u)} du }$

**Ackbeet** · July 22nd 2010, 02:36 AM

Other than the Whittaker and Watson book you mentioned, I wouldn't know for certain where you could find that result and its proof. Some options you could try, though, would be books on mathematical methods for physicists. Those kinds of books always have significant sections on special functions. You could try Boas, or Arfken and Weber, or Courant and Hilbert.

Good luck!

**AlexisM** · July 22nd 2010, 08:04 AM

Ok, I've finally found out myself :

I use mainly these two equivalences :

$B(x,y)=B(y,x)$
$B(x,y)=\int_0^\infty \frac{t^{x-1}}{(t+1)^{x+y}}dt,\quad Re(x),Re(y)>0$

The second equality can be symmetrized using :
$B(x,y)=\frac{1}{2}B(x,y)+B(y,x)$
$B(x,y)=\frac{1}{2} \int_0^\infty \frac{t^{x-1}}{(t+1)^{x+y}}dt + \frac{1}{2} \int_0^\infty \frac{t^{x-1}}{(t+1)^{x+y}}dt$

$B(x,y)=\diplaystyle\frac{1}{2} \int_0^\infty \frac{t^{x-1} + t^{y-1}}{(t+1)^{x+y}}dt$

With a notation change (t->x) and applying this formula to our problem we get :

$B(t+r,t-r)=\diplaystyle\frac{1}{2} \int_0^\infty \frac{x^{t+r-1} + x^{t-r-1}}{(x+1)^{2t}}dx$

$B(t+r,t-r)=\diplaystyle\frac{1}{2} \int_0^\infty \frac{x^r + x^{-r}}{(x+1)^{2t}x^{-t+1}}dx$

$B(t+r,t-r)=\diplaystyle \int_0^\infty \frac{x^r + x^{-r}}{(x^{\frac{1}{2}}+x^{-\frac{1}{2}})^{2t}}\frac{1}{2x}dx$

Now if i make the variable change :
$x = e^{2u}$
$dx =2e^{2u}du = 2xdu$

we have

$B(t+r,t-r)=\diplaystyle \int_{-\infty}^\infty \frac{e^{2ur} + e^{-2ur}}{(e^{u}+e^{-u})^{2t}}du$

$B(t+r,t-r)=\diplaystyle \frac{2}{4^t} \int_{-\infty}^\infty \frac{cosh(2ru)}{cosh^{2t}(u)}du$

And we use the fact that the function we integrate is symmetric in u :
$B(t+r,t-r)=\diplaystyle \frac{1}{4^{t-1}} \int_{0}^\infty \frac{cosh(2ru)}{cosh^{2t}(u)}du$

And the condition is slightly different (less strict) from the one in Whittaker & Watson, namely $|Re(r)|<t$ which gives no constraint on the imaginary part.

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