Inverse Laplace Transform Step by Step

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SUMMARY

The discussion focuses on performing an Inverse Laplace Transform using contour integrals, specifically for the frequency function 1/sqrt(s). The integral is defined as f(t) = (1/(2πi)) ∫(a-i∞)^(a+i∞) (e^(st)/√s) ds, with a chosen as 0 to avoid complications at zero. The transformation simplifies to f(t) = (1/√(πt)), utilizing known real integrals to derive the final result. The discussion provides a clear step-by-step approach to understanding the process of inverse transformation.

PREREQUISITES
  • Understanding of complex analysis and contour integrals
  • Familiarity with Laplace transforms and their properties
  • Knowledge of integral calculus, particularly improper integrals
  • Basic understanding of real integrals involving sine and cosine functions
NEXT STEPS
  • Study the properties of Laplace transforms in engineering applications
  • Learn about complex analysis techniques for evaluating contour integrals
  • Explore real integral evaluations, specifically integrals involving sine and cosine
  • Investigate alternative methods for Inverse Laplace Transforms, such as the residue theorem
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Students and professionals in mathematics, engineering, and physics who are looking to deepen their understanding of inverse Laplace transforms and contour integration techniques.

steve2k
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Hi - I really need someone to show me step by step how to do an Inverse Laplace transform using a contour integral. The one I would like to understand is the frequency function 1/sqrt(s)

Thank you if you can help me out.

Steve
 
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steve2k said:
Hi - I really need someone to show me step by step how to do an Inverse Laplace transform using a contour integral. The one I would like to understand is the frequency function 1/sqrt(s)

Thank you if you can help me out.

Steve
This is the contour integral that gives the inverse.
f(t)= \frac{1}{2{\pi}i} \int_{a-i\infty}^{a+i\infty} g(s)e^{st} ds
or for the specific function.
f(t)=\frac{1}{2{\pi}i}\int_{a-i\infty}^{a+i\infty} \frac{e^{st}}{\sqrt{s}} ds
We need to take "a" far enough to the right that we avoid problems.
Here we may take a=0, as even though the function has problems at zero, they are not major. You can consider a small right half circle and see the integral is small.
f(t)=\frac{1}{2{\pi}i}\int_{-i\infty}^{i\infty} \frac{e^{st}}{\sqrt{s}} ds
we can clean the integral up with a substitution i u=s t
f(t)=\frac{1}{2{\pi}\sqrt{it}}\int_{-\infty}^{\infty} \frac{e^{iu}}{\sqrt{u}} du
This integral can be written in terms of "know" real integrals.
\int_0^\infty \frac{sin(x)}{\sqrt{x}} dx=\int_0^\infty \frac{cos(x)}{\sqrt{x}} dx=\sqrt{\frac{\pi}{2}}
thus the answer
f(t)=\frac{2+2i}{2{\pi}\sqrt{it}}\sqrt{\frac{\pi}{2}}

f(t)=\frac{1}{\sqrt{{\pi}t}}
You can also do a real inversion.
f(t)=\lim_{k\rightarrow\infty}\frac{(-1)^k}{k!}g^{(k)}(\frac{k}{t})(\frac{k}{t})^{k+1}
 
Last edited:
Thank you.

Thanks for the reply I really appreciate it!

Steve
 
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