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

• Poirot
In summary, the conversation discusses the use of Cauchy's residue theorem in determining the residue at a pole inside a contour. It also covers the checks for the integral on the contour and the need for a different contour for negative values of x. The summary concludes with a clarification on the conditions for closing the contour in the upper half plane for positive values of x and in the lower half plane for negative values of 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.

Poirot said:
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##?

vela said:
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 :)!

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##…

## 1. What is a complex contour integral?

A complex contour integral is a type of integral used in complex analysis to calculate the value of a function over a specific path or contour in the complex plane. It involves integrating a complex-valued function over a complex curve, which can be represented by parametric equations in terms of a complex variable.

## 2. How is a complex contour integral different from a regular integral?

A regular integral involves integrating a real-valued function over a real interval, while a complex contour integral involves integrating a complex-valued function over a complex curve. Additionally, the path of integration in a complex contour integral can be any curve in the complex plane, whereas in a regular integral it is limited to a straight line.

## 3. What is the purpose of using complex contour integrals to find a function?

The purpose of using complex contour integrals to find a function is to extend the concept of integration to complex-valued functions and to evaluate these functions over specific paths in the complex plane. This allows us to solve problems in complex analysis, such as finding the value of a function at a point or calculating residues.

## 4. How do you solve a complex contour integral?

To solve a complex contour integral, you first need to parameterize the contour in terms of a complex variable. Then, you substitute this parameterization into the complex-valued function and perform the integration using techniques such as the Cauchy Integral Formula or the Residue Theorem. Finally, you can evaluate the integral to find the value of the function.

## 5. What are some applications of complex contour integrals?

Complex contour integrals have various applications in mathematics, physics, and engineering. They are used to solve problems in complex analysis, to evaluate complex-valued functions, to find the values of real integrals, and to calculate residues. They are also used in the study of differential equations, Fourier and Laplace transforms, and in the analysis of electric and magnetic fields.