- #1

- 7

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Can anyone gave me a solved example on Inverse Mellin transform?

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- #1

- 7

- 0

Can anyone gave me a solved example on Inverse Mellin transform?

- #2

- 56

- 0

[tex]\mathcal{M}^{-1}\left\{\frac{\pi}{\sin(\pi s)}\right\}=\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}\frac{\pi}{x^s\sin(\pi s)}ds=\sum_{n=0}^{\infty}(-x)^n=\frac{1}{1+x};\quad |x|<1.[/tex]

That sum is obtained from taking the residues inside a square contour with right boundary on [tex]c\pm i\infty[/tex] and allowed to expand to infinity. The integral tends to zero on the remaining three legs and we're left with (via the Residue Theorem) an infinite sum of residues corresponding to the poles at [tex]-n[/tex]:

[tex]\mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots[/tex]

- #3

- 7

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ok but Why you choose the pole -n ? i can take the residue at n and 0

- #4

- 56

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Also, I though it was a bit of a challenge to show:

[tex]

\mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots

[/tex]

for me anyway. Maybe an easier way to do though.

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