Help with Fourier integral using contour

In summary: C$. Since the pole of $f(\omega)$ is at $\omega = a$, which is inside the contour $C_R$, the residue at $\omega = a$ is equal to $e^{-iat}$. Therefore, the integral around $C_R$ is equal to $2\pi i e^{-iat}$, which is equal to zero since $t>0$.In summary, to find $F=\frac{\hbar}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{E_0-\frac{i\Gamma}{2} -\hbar \omega} \,d\omega$ using contour integration, we can
  • #1
ognik
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Find $ F=\frac{\hbar}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{E_0-\frac{i\Gamma}{2} -\hbar \omega} \,d\omega $ using contour integration. I have a couple questions I'd like some help with please...

Taking out the $\hbar$ in the denominator, which cancels with the $\hbar$ outside the integral, the denominator becomes $(a-\omega)$ where $a=\frac{1}{\hbar}(E_0-\frac{i\Gamma}{2}) $

Then $F = \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{a-\omega} \,d\omega $ with $f(\omega) = \frac{e^{-i \omega t}}{a-\omega}=\frac{cos \omega t}{a-\omega}-i\frac{Sin \omega t}{a-\omega} $

Q1: so if $f(\omega)$ is analytic, a is a simple pole and I can use residues, but I'm not sure how to prove $f(\omega)$ is analytic, ie. how to write $\omega$ in terms of x and y, so I can use the Cauchy Riemann conditions?

Assuming analytic, I can use Cauchy's residue theorem (with a simple pole): $ \oint f(\omega) d\omega=2\pi i \: res(f, a)$, where $res(f, a)=\lim_{{\omega }\to{a}} (\omega - a)f(\omega) = e^{-iat}$

$ \therefore F= \frac{\hbar}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{E_0-\frac{i\Gamma}{2} -\hbar \omega} \,d\omega = e^{-it \left[ \frac{1}{\hbar}(E_0-\frac{i\Gamma}{2}) \right] }$ $= {e}^\left[ {\frac{-iE_{0 }t}{\hbar}} \right] {e}^\left[ {\frac{-t \Gamma}{2\hbar}}\right] $

but Q2: I am not sure about the contour. I assumed a semi-circle, but am not sure where the pole is? $f(\omega)$ looks like 4th quadrant? I also need to consider t>0 and t<0 ...

Q3: I think the contour would be $ \oint_C = \lim_{{R}\to{\infty}}\int_{-R}^{R} +\int_{2\pi}^{\pi} $, but for this function I have no idea how to show $\int_{2\pi}^{\pi} = 0$
 
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  • #2


Hi there! It's great to see that you are using contour integration to solve this problem. To answer your questions:

Q1: To prove that $f(\omega)$ is analytic, you can show that it satisfies the Cauchy-Riemann conditions. The Cauchy-Riemann conditions state that a function $f(x+iy)$ is analytic if and only if it satisfies the following conditions:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

where $u$ and $v$ are the real and imaginary parts of $f(x+iy)$, respectively. In this case, $f(\omega)$ can be written as $f(\omega) = u(\omega) + iv(\omega)$, where $u(\omega) = \frac{cos \omega t}{a-\omega}$ and $v(\omega) = -\frac{sin \omega t}{a-\omega}$. You can then calculate the partial derivatives of $u$ and $v$ with respect to $x$ and $y$ and show that they satisfy the Cauchy-Riemann conditions.

Q2: The pole of $f(\omega)$ is at $\omega = a$, which is given by $a=\frac{1}{\hbar}(E_0-\frac{i\Gamma}{2})$. To determine the contour, you can use a semi-circle in the lower half plane, since $f(\omega)$ has a pole in the lower half plane. This is because the integral along the semi-circle will tend to zero as the radius goes to infinity, as long as $t>0$.

Q3: Yes, the contour would be $\oint_C = \lim_{{R}\to{\infty}}\int_{-R}^{R} +\int_{C_R}$, where $C_R$ is the semi-circle in the lower half plane. To show that $\int_{C_R} = 0$, you can use the Cauchy integral formula, which states that for a function $f(z)$ that is analytic in a region containing a closed contour $C$, the integral around $C$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at its poles inside
 

FAQ: Help with Fourier integral using contour

What is a Fourier integral?

A Fourier integral is a mathematical representation of a function that decomposes it into its constituent frequencies. It is used to analyze periodic functions and signals in various fields, such as physics, engineering, and mathematics.

What is the purpose of using a contour in a Fourier integral?

A contour is used in a Fourier integral to deform the original integration path into a more convenient shape, making it easier to evaluate the integral. This is especially useful when dealing with complex functions or infinite intervals.

How do you choose the contour for a Fourier integral?

The contour chosen for a Fourier integral depends on the specific function and the desired outcomes. Generally, it should enclose all the singularities of the function and be chosen in a way that simplifies the integral and preserves its convergence.

What are some common techniques for evaluating a Fourier integral using contour?

Some common techniques for evaluating a Fourier integral using contour include the Residue Theorem, Cauchy's Integral Formula, and the Method of Steepest Descent. These techniques use properties of complex functions and contours to simplify the integral and make it more manageable to evaluate.

Can a Fourier integral using contour be used in real-world applications?

Yes, Fourier integrals using contour are widely used in various real-world applications, such as signal processing, image reconstruction, and solving differential equations. They are also used in fields such as physics, engineering, and mathematics to analyze and manipulate periodic functions and signals.

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