How to use the Bromwich Integral

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In summary, the speaker converted the function {t^n}{e^{-t}} from the time domain to the s-domain using a unilateral Laplace transform. They then integrated the function in the s-domain and obtained the result \frac{n!}{(s+1)^{n+1}}. The purpose of this was to find \int}{t^n}{e^{-t}}dt and add some variety to the integration process. The speaker is not sure how to convert the s-domain result back to the time domain using the Bromwich Integral, and is seeking help. They mention that the integral must be computed along a straight line parallel to the imaginary axis and intersecting the real axis in a specific point. They also mention the use of
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flouran
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I converted the function, [tex]{t^n}{e^{-t}}[/tex] where t is the variable (time domain) and n is any real whole integer greater than or equal to 0, to the s-domain using a unilateral Laplace transform. In the end, I got f(s) = [tex]\frac{n!}{(s+1)^{n+1}}[/tex]. The reason why I did this was because I wanted to find [tex]\int}{t^n}{e^{-t}}dt[/tex]. I know how to do this with integration by parts, but I wanted to "spice things up a bit". Once I had converted the function to the s-domain, I integrated it with respect to s, in other words, I thought [tex]\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& \int}{{t^n}{e^{-t}}}dt[/tex], correct me if I am wrong. Thus, I got [tex]\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& {\frac{n!}{-(n+2){{(s+1)^{n+2}}}}[/tex]. Thus, in order to find the [tex]\int}{t^n}{e^{-t}}dt[/tex], I just have to convert [tex]{\frac{n!}{-(n+2){{(s+1)^{n+2}}}}[/tex] back to the time domain using the Bromwich Integral, but I don't know how to do this (again, correct me if I am wrong). Could somebody help me?
 
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Mustn't it be integrated along a straight line parallel to the imaginary axis and intersecting the real axis in the point [tex]$ \gamma$[/tex] which must be chosen so that it is greater than the real parts of all singularities of F(s)? I just don't know how do this. I've only done math until BC Calculus. Also, in practice, computing the complex integral can be done by using the Cauchy residue theorem, right?
 

FAQ: How to use the Bromwich Integral

What is the Bromwich Integral and what is its purpose?

The Bromwich Integral is a mathematical tool used in complex analysis to evaluate integrals of the form f(z)/z, where f(z) is a function of a complex variable z. Its purpose is to calculate the values of these integrals by using a contour integral in the complex plane.

How do you use the Bromwich Integral to solve complex integrals?

To use the Bromwich Integral, you first choose a contour in the complex plane that encloses the singularities of the function f(z)/z. Then, you use the Cauchy Integral Theorem to evaluate the contour integral, which will give you the value of the original integral.

What are some common applications of the Bromwich Integral?

The Bromwich Integral has many applications in physics and engineering, particularly in the fields of signal processing, control theory, and electromagnetic theory. It is also used in the study of Laplace transforms and their inverse transforms.

Are there any limitations to using the Bromwich Integral?

One limitation of the Bromwich Integral is that it can only be used to evaluate integrals of the form f(z)/z, where f(z) is a function of a complex variable. It cannot be used for integrals involving real variables. Additionally, the contour chosen for the integral must be carefully selected to avoid crossing any singularities of the function.

What are some tips for using the Bromwich Integral effectively?

One tip for using the Bromwich Integral is to carefully choose the contour for the integral, ensuring that it encloses all of the singularities of the function. It is also important to be familiar with the Cauchy Integral Theorem and its applications. Additionally, using substitution and partial fraction decomposition can often simplify the integrand and make the integral easier to evaluate.

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