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Thank you if you can help me out.

Steve

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Thank you if you can help me out.

Steve

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This is the contour integral that gives the inverse.steve2k said:

Thank you if you can help me out.

Steve

[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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Thanks for the reply I really appreciate it!

Steve

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