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

[dRic2]

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:

 

[Dewgale]

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.

 

[Ray Vickson]

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.

 

[dRic2]

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.

(...) 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!

 

[Ray Vickson]

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".

 

[dRic2]

Thanks! Btw I still can't solve it myself... I'm so stressed out

 

[Dewgale]

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.

 

[vela]

$$ 2 \pi i \left[ \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {-\pi i} 4}} + \left( \frac {e^{-2 \pi i x ξ}} {4x^3} \right)_{x = e^{\frac {- 3\pi i} 4}} \right]$$

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.

 

[Ray Vickson]

Thanks! Btw I still can't solve it myself... I'm so stressed out

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.

 

[dRic2]

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