Inverse Laplace Transform Step by Step

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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
 

lurflurf

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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.
[tex] f(t)= \frac{1}{2{\pi}i} \int_{a-i\infty}^{a+i\infty} g(s)e^{st} ds [/tex]
or for the specific function.
[tex] f(t)=\frac{1}{2{\pi}i}\int_{a-i\infty}^{a+i\infty} \frac{e^{st}}{\sqrt{s}} ds [/tex]
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.
[tex] f(t)=\frac{1}{2{\pi}i}\int_{-i\infty}^{i\infty} \frac{e^{st}}{\sqrt{s}} ds [/tex]
we can clean the integral up with a substitution i u=s t
[tex] f(t)=\frac{1}{2{\pi}\sqrt{it}}\int_{-\infty}^{\infty} \frac{e^{iu}}{\sqrt{u}} du [/tex]
This integral can be written in terms of "know" real integrals.
[tex]\int_0^\infty \frac{sin(x)}{\sqrt{x}} dx=\int_0^\infty \frac{cos(x)}{\sqrt{x}} dx=\sqrt{\frac{\pi}{2}}[/tex]
thus the answer
[tex]f(t)=\frac{2+2i}{2{\pi}\sqrt{it}}\sqrt{\frac{\pi}{2}}[/tex]

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

Thanks for the reply I really appreciate it!

Steve
 

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