# Inverse Mellin transform

1. May 16, 2009

### Love*Physics

Can any one gave me a solved example on Inverse Mellin transform?

2. May 16, 2009

### squidsoft

Hi. I think Wikipedia has a good section on the Mellin transform pair. Although they go about it a little differently, the integral is just the sum of the residues inside the indicated contour.

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

That sum is obtained from taking the residues inside a square contour with right boundary on $$c\pm i\infty$$ 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 $$-n$$:

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

3. May 17, 2009

### Love*Physics

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

4. May 17, 2009

### squidsoft

Ok, approach this strictly in terms of contour integration: it's a rectangular contour with corners at $$c\pm i \infty$$, and $$-\infty\pm i\infty$$. Then that contour encloses the poles at $$n=0,-1,-2,-3,\cdots$$. And by the Residue Theorem, the integral, assuming the other legs go to zero, is just $$2\pi i$$ times the sum of the residues at that set of poles.

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

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

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

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