Contour Integration for Inverse Mellin Transform

In summary, the conversation discusses the Inverse Mellin transform and its application in finding the sum of residues inside a contour. The integral is equal to the sum of residues at the poles, which are chosen to be -n as the contour encloses them. The conversation also mentions that it can be challenging to show the formula for the residues at -n.
  • #1
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Can anyone gave me a solved example on Inverse Mellin transform?
 
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  • #2
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.

[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
ok but Why you choose the pole -n ? i can take the residue at n and 0
 
  • #4
Ok, approach this strictly in terms of contour integration: it's a rectangular contour with corners at [tex]c\pm i \infty[/tex], and [tex]-\infty\pm i\infty[/tex]. Then that contour encloses the poles at [tex]n=0,-1,-2,-3,\cdots[/tex]. And by the Residue Theorem, the integral, assuming the other legs go to zero, is just [tex]2\pi i[/tex] times the sum of the residues at that set of poles.

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.
 

What is an inverse Mellin transform?

An inverse Mellin transform is a mathematical operation that takes a function in the complex domain and transforms it back to its original form in the real domain. It is the inverse of the Mellin transform and is typically used in mathematical physics and engineering applications.

How is the inverse Mellin transform calculated?

The inverse Mellin transform is calculated using a contour integral in the complex plane. The integral is taken along a contour that encloses the singularities of the function being transformed. The resulting expression is then simplified using techniques such as residue calculus.

What is the relationship between the inverse Mellin transform and the Laplace transform?

The inverse Mellin transform and the Laplace transform are closely related. The Laplace transform is a special case of the Mellin transform, and the inverse Mellin transform is a special case of the inverse Laplace transform. Both transforms are used to solve differential equations and have applications in signal processing and control theory.

What are some common uses of the inverse Mellin transform?

The inverse Mellin transform has many applications in engineering and physics. It is used to solve differential equations, analyze signals and systems, and study the behavior of functions in the complex plane. It is also used in probability theory and number theory.

What challenges are associated with using the inverse Mellin transform?

One challenge in using the inverse Mellin transform is determining the appropriate contour for the integral. This can be a complex task, as it often involves understanding the behavior of the function being transformed in the complex plane. Additionally, the inverse Mellin transform can produce complicated expressions, which can be difficult to interpret and manipulate.

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