# Find, using complex contour integrals, the function f(x)...

## Homework Statement

whose Fourier transform is f~(p) = 1/(a2 + p2)

## Homework Equations

f(x) = 1/√2π ∫-∞ eipx f~(p)

## The Attempt at a Solution

First of all I let f(z) = eixz/(z2 + a2)
and γ = γ1 + γ2
with the ϒ's parametrised by:

γ1 : {z=t, -R<t<R}
γ2 : {z=Reit, 0<t<π}
(So a semicircle of radius R)

In this contour the only pole that lies inside is the z= +ia
so using Cauchy's residue theorem:
ϒf(z)dz = 2πi (Res(f, z=ia))
I found the residue of z=ia to be Rez(f,z=ia)= -i e-ax/2a
So ∫ϒf(z)dz= π e-ax/a

And I have something in my notes about the fact you have to check that the integral of f(z) on γ2 goes to zero as R goes to infinity, which I vaguely understand because we actually want the integral from -∞ to +∞.
So the checks I have in my notes are:
Suppose γ has length L and on γ |f(z)|<M
Then |∫γf(z)dz|≤M⋅L
With γ ≡ {z=γ(t), a≤t≤b}
L = ∫abdt|γ'(t)|
On γ2: |z| =R
Using ||a|-|b|| ≤ |a+b| ≤ |a|+|b|
R2-a2 ≤ |z2+a2| ≤ R2+a2
So 1/R2-a2 ≥ 1|z2+a2| ≥ 1R2+a2
Therefore |∫γ2f(z)dz|≤πR/(R2-a2)
So ∫γ2f(z)dz → 0 as R → ∞
as you have |∫γ2 f(z)dz| ≤ e-Rxsin(t)/(R2 - a2)

and if x ≥0 the exponent is, at most, equal to 1, but if x<0, it blows up and we need a different contour for x<0.

Let γ = γ1 + γ3
with ϒ3 : {z=Reit, π<t<2π}

so a semicircle in the negative part, and this time it's a clockwise contour so Cauchy's residue must carry a minus sign on the right hand side.

So doing everything as before I found ∫ϒ = πeax/a

But I'm having trouble with the checks, I found that the inequality for the modulus of the integral over ϒ3 comes out the same? So effectively it still blows up?

But I know this is the right answer I just can't follow through with the checks.

I appreciate this is quite long, but a lot of it is just background to set the scene, and I think I just have a lack of understanding of the checks and just know the procedure.

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vela
Staff Emeritus
Homework Helper
So the checks I have in my notes are:
Suppose γ has length L and on γ |f(z)|<M
Then |∫γf(z)dz|≤M⋅L
With γ ≡ {z=γ(t), a≤t≤b}
L = ∫abdt|γ'(t)|
On γ2: |z| =R
Using ||a|-|b|| ≤ |a+b| ≤ |a|+|b|
R2-a2 ≤ |z2+a2| ≤ R2+a2
So 1/R2-a2 ≥ 1|z2+a2| ≥ 1R2+a2
Therefore |∫γ2f(z)dz|≤πR/(R2-a2)
So ∫γ2f(z)dz → 0 as R → ∞
as you have |∫γ2 f(z)dz| ≤ e-Rxsin(t)/(R2 - a2)

and if x ≥0 the exponent is, at most, equal to 1, but if x<0, it blows up and we need a different contour for x<0.
Looks like your notes have some mistakes. Assume $R>a$ since you're going to make the circle big in th end. You showed $\left\lvert \frac{1}{z^2+a^2} \right\rvert \le \left\lvert \frac{1}{R^2-a^2} \right\rvert$, so you can say
$$\left\lvert \frac{e^{izx}}{z^2+a^2} \right\rvert \le \left\lvert \frac{e^{izx}}{R^2-a^2} \right\rvert.$$ On the semicircle, you have $z=R\cos\theta + iR\sin\theta$, so $\lvert e^{izx} \rvert = \lvert e^{ixR\cos\theta}e^{-xR\sin\theta} \rvert = e^{-xR\sin\theta}$. Hence, an upper bound on the integrand is $M = \frac{e^{-xR\sin\theta}}{R^2-a^2}$. With $L=\pi R$, you get
$$\left\lvert \int_{\gamma_2} \frac{e^{izx}}{z^2+a^2} \right\rvert \le \frac{\pi R e^{-xR\sin\theta}}{R^2-a^2}$$ where $0<\theta<\pi$.

Consider the exponential. What combinations of $\theta$ and $x$ cause it to go to 0 in the limit $R \to \infty$?

Looks like your notes have some mistakes. Assume $R>a$ since you're going to make the circle big in th end. You showed $\left\lvert \frac{1}{z^2+a^2} \right\rvert \le \left\lvert \frac{1}{R^2-a^2} \right\rvert$, so you can say
$$\left\lvert \frac{e^{izx}}{z^2+a^2} \right\rvert \le \left\lvert \frac{e^{izx}}{R^2-a^2} \right\rvert.$$ On the semicircle, you have $z=R\cos\theta + iR\sin\theta$, so $\lvert e^{izx} \rvert = \lvert e^{ixR\cos\theta}e^{-xR\sin\theta} \rvert = e^{-xR\sin\theta}$. Hence, an upper bound on the integrand is $M = \frac{e^{-xR\sin\theta}}{R^2-a^2}$. With $L=\pi R$, you get
$$\left\lvert \int_{\gamma_2} \frac{e^{izx}}{z^2+a^2} \right\rvert \le \frac{\pi R e^{-xR\sin\theta}}{R^2-a^2}$$ where $0<\theta<\pi$.

Consider the exponential. What combinations of $\theta$ and $x$ cause it to go to 0 in the limit $R \to \infty$?
The only thing I can think of at the moment is if θ is -θ then the sin(-θ)= -sin(θ) so we'll have the extra minus sign and for x<0 this works? but I'm not sure how to word it and/or if that's right?
Thank you for taking the time to help by the way, greatly appreciated :)!

vela
Staff Emeritus
I think you have it. If $x>0$, you need $\sin\theta>0$ to make the sign of the exponent negative. That's why you close the contour in the upper half plane. If $x<0$…