Fourier Transform of 1/(1+x^4)

dRic2
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Homework Statement


Calculate ##F(\frac 1 {1+x^4})##.

Homework Equations


##\hat f (ξ) = \int_ℝ \frac 1 {1+x^4} e^{-2\pi i ξ x} dx##
and Residue Theorem

The Attempt at a Solution


I know the function has to be real and even because ##\frac 1 {1+x^4}## is real and even, but I can't work out the calculations:

I will start to evaluate for ##ξ > 0## and then exploit symmetry to extend the result to ##ξ < 0##

for ##ξ > 0## I have ## -2\pi i ξ x < 0## so

$$\hat f (ξ) = \int_ℝ \frac 1 {1+x^4} e^{-2\pi i ξ x} dx = 2 \pi i [ Res(..., e^{\frac {-\pi i} 4}) + Res(..., e^{\frac {-3 \pi i} 4}) ] = 2 \pi i \left[ \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {-\pi i} 4}} + \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {- 3\pi i} 4}} \right]$$

I've tried everything but I can't figure out a single way to make this a Real function... Even Wolfram Alpha can't do it. Also google doesn't help me :( :(

Here's the smooth result from my book:

Schermata 2018-12-17 alle 20.01.46.png
 

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Remember that ##x## is a real number in this case. You want a complex number, so consider the analogous case
$$\int_\mathbb{C} \frac{1}{1+z^4} e^{-2\pi i \xi z}$$
where the integral follows a semi-circle that goes from ##-R\to R## along the real axis, then goes either up to ##R i## or down to ##-R i## in an arc, closing the semi-circle. The part of the curve along the real axis is the same as your integral above, except it's from ##-R \to R## instead of over all the reals. So if you can solve for that part of the curve, and take ##\lim_{r\to\infty}##, you should get the same integral as your situation.

You need to do both cases, not just up or down.
 
dRic2 said:

Homework Statement


Calculate ##F(\frac 1 {1+x^4})##.

Homework Equations


##\hat f (ξ) = \int_ℝ \frac 1 {1+x^4} e^{-2\pi i ξ x} dx##
and Residue Theorem

The Attempt at a Solution


I know the function has to be real and even because ##\frac 1 {1+x^4}## is real and even, but I can't work out the calculations:

I will start to evaluate for ##ξ > 0## and then exploit symmetry to extend the result to ##ξ < 0##

for ##ξ > 0## I have ## -2\pi i ξ x < 0## so

$$\hat f (ξ) = \int_ℝ \frac 1 {1+x^4} e^{-2\pi i ξ x} dx = 2 \pi i [ Res(..., e^{\frac {-\pi i} 4}) + Res(..., e^{\frac {-3 \pi i} 4}) ] = 2 \pi i \left[ \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {-\pi i} 4}} + \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {- 3\pi i} 4}} \right]$$

I've tried everything but I can't figure out a single way to make this a Real function... Even Wolfram Alpha can't do it. Also google doesn't help me :( :(

Here's the smooth result from my book:

View attachment 235886

If
##r_1 = \frac{1}{\sqrt{2}} (1+i), r_2 = \frac{1}{\sqrt{2}} (-1+i), r_3 = \frac{1}{\sqrt{2}} (1-i), r_4 = \frac{1}{\sqrt{2}} (-1-i)## we have
$$z^4+1 = (z-r_1)(z-r_2)(z-r_3)(z-r_4)$$ Thus, with ##f(z) = \exp(-2 \pi i z \xi)/(z^4+1)## we have
$$\text{res}(f)|_{z = r_3} = \frac{\exp(-2 \pi i r_3 \xi)}{(r_3-r_1)(r_3-r_2)(r_3-r_4)}$$ etc. Theoretically, this should match the expressions you gave.

BTW: for ##\xi > 0##, you complete clockwise in the lower ##z##-plane, so you need ##-2 \pi i## instead of ##+2 \pi i## in front.

When I do the computations using Maple, I get exactly same result as your book's.
 
Last edited:
Ray Vickson said:
BTW: for ξ>0ξ>0\xi > 0, you complete clockwise in the lower zzz-plane, so you need −2πi−2πi-2 \pi i instead of +2πi+2πi+2 \pi i in front.

yes, my mistake. Thank you.

Ray Vickson said:
(...) Theoretically, this should match the expressions you gave.

Yes, I don't think this is going to help much since they are the same expression in the end (after some calculations), but thanks anyway :)

Btw if you know some trick to make Wolfram Alpha (or maybe Mathematica) solve this puzzle, I'm all ears!
 
dRic2 said:
yes, my mistake. Thank you.
Yes, I don't think this is going to help much since they are the same expression in the end (after some calculations), but thanks anyway :)

Btw if you know some trick to make Wolfram Alpha (or maybe Mathematica) solve this puzzle, I'm all ears!

I don't have access to Mathematica, and Wolfram Alpha is pretty inconvenient to use on a problem like this one. Here are the steps in Maple; perhaps Mathematica has equivalent commands (I would be surprised if not). Note: in Maple, an input command ending in a ";" prints out the results on-screen, but a line ending in ":" suppresses that print. The command "evalc(Z)" means "compute the real x and y in Z = x + iy" (used if that form has not already been given). Finally, I is the imaginary unit in Maple.

Oh: I am also using ##\exp(-i k x)## in the transform, instead of your ##\exp(-i 2 \pi \xi x).##

Finally, here is also a direct Fourier transform computation in Maple, letting the program do all the work. However, to get it to work I needed to write the denominator in factored form. It uses the command "%", which just means "the final expression above".
 

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Thanks! Btw I still can't solve it myself... I'm so stressed out :mad::mad:
 
Given a particular semi-circle choice (with the properties I described in my previous post), we can divide the path (call it ##\gamma##) into the straight part (##\gamma_1##) along the real line, and the arc (##\gamma_2##). This would give (replacing ##-2\pi i \xi x## with just ##-i \xi x##, and ignoring differences in normalization for now)

$$\int_\gamma \frac{e^{-i\xi x}}{1+x^4} = \int_{\gamma_1} \frac{e^{-i\xi x}}{1+x^4} + \int_{\gamma_2} \frac{e^{-i\xi x}}{1+x^4} \implies \int_{\gamma_1} \frac{e^{-i\xi x}}{1+x^4} = \int_\gamma \frac{e^{-i\xi x}}{1+x^4} - \int_{\gamma_2} \frac{e^{-i\xi x}}{1+x^4},$$
where ##x \in C##. You can find ##\int_\gamma## using the residue theorem, and as ##R \to \infty##, ##\int_{\gamma_1}## just becomes an integral over the reals. All you need to do at that point is just show that as ##R \to \infty##, ##\int_{\gamma_2} \to 0##. Then you're done.
 
dRic2 said:
$$ 2 \pi i \left[ \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {-\pi i} 4}} + \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {- 3\pi i} 4}} \right]$$
Take the first residue:
\begin{align*}
\left. \frac{e^{-2 \pi i x \xi}}{x^3} \right|_{x=e^{-i \pi/4}}
&= \frac {e^{-2 \pi i (e^{-i \pi/4}) \xi}}{(e^{-i \pi/4})^3} = e^{i 3\pi/4} e^{-2 \pi i (e^{-i \pi/4}) \xi} \\
&= \left(-\frac 1{\sqrt 2} + i \frac 1{\sqrt 2}\right) e^{-2 \pi i (\frac 1{\sqrt 2} - i \frac 1{\sqrt 2}) \xi} \\
&= \left(-\frac 1{\sqrt 2} + i \frac 1{\sqrt 2}\right) e^{-i \sqrt 2 \pi \xi } e^{-\sqrt 2 \pi \xi} \\
&= \frac {e^{-\sqrt 2 \pi \xi}}{\sqrt{2}} (-e^{-i \sqrt 2 \pi \xi } + i e^{-i \sqrt 2 \pi \xi } )
\end{align*} Do the same thing with the second term, and you'll get terms involving ##e^{i \sqrt 2 \pi \xi }##. Use the identities ##e^{i\theta}+e^{-i\theta} = 2 \cos\theta## and ##e^{i\theta}-e^{-i\theta} = 2i \sin\theta## to combine terms. As long as you avoid making sign mistakes somewhere along the way, everything should work out so that you end up with a real result.
 
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dRic2 said:
Thanks! Btw I still can't solve it myself... I'm so stressed out :mad::mad:

The best way is to take it slowly, in pieces, then put things together after the parts are as simple as possible.

Let's use the notation ##y## instead of ##\xi##, and let ##c = 1/\sqrt{2}##. For real ##\theta## we have ##1/\exp(-i \theta) = \exp(i \theta),## obtained by multiplying both the numerator and denominator by ##\exp(i \theta).##

(1) For ## r_1 =\exp(-i \pi/4) = c(1-i)## we have ##1/(4 r_1^3) = 1/(4 \exp(-i \pi 3/4) )= (1/4) \exp(i \pi 3/4) = (c/4) (-1+i).## We also have
$$\begin{array}{rl}\exp(-2 i \pi y r_1) &=& \exp(-i 2 \pi y c(1-i)) = \exp((-i)^2 2 \pi c y) \exp( - i 2 \pi y c) \\
&=& E (C-iS)
\end{array}$$
where ##E = e^{-2 \pi c y}##, ##C = \cos(2 \pi c y)## and ##S = \sin(2 \pi c y).## Thus
$$\text{res}(f)|_{z=r_1} = T_1 = (c/4) (-1+i) E (C-iS)$$

(2) For ##r_2 = \exp(-i 3 \pi/4) = c(-1-i)## we have ##1/(4 r_2^3) = 1/(4 \exp(-i \pi 9/4) )= (1/4) \exp(i \pi 9/4)= (1/4)\exp(i \pi/4) = (c/4)(1+i).## We also have
$$\begin{array}{rl} \exp(-2 i \pi y r_2) &=& \exp(-i 2 \pi y c(-1-i)) = \exp((-1)^2 2 \pi c y) \exp(i 2 \pi c y)\\
&=& E (C+iS) \end{array}$$ Thus
$$\text{res}(f)|_{z=r_2} = T_2 = (c/4)(1+i) E (C+IS)$$

The sum of the residues is ##T_1 + T_2 = (i c/2) E (C+S),## so ##-2 \pi i (T_1+T_2) = \pi c E (C+S)## is real.

I could not see where you made your errors because you did not lay out the work step-by-step in modest chunks that were each easy to check.
 
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Thank you very much! Maybe my mistake was keeping also the denominator in exponential form because I know the result I got is Real (I checked it with wolfram alpha), but I can't simplify it. I will try out your methods! Thanks again
 
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