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}