What is the value of the Mellin transform ?

In summary, the integral \int_{0}^{\infty}dtt^{s-1} has no value for s\geq 0 and s\leq 0. It is only defined for 0<s<1, where its value is √(π/2a) for s = 1/2. It cannot be extended to values bigger than 1 using analytic continuation, as it is divergent.
  • #1
zetafunction
391
0
my question is what is the value of

[tex] \int_{0}^{\infty}dtt^{s-1} [/tex]

if s >1 the integral is divergent as x -->oo

if s<1 the integral would become divergent as x-->0

then what is the value of this integral, or the value (regularized i presume) of


[tex] \int_{0}^{\infty}dtt^{s-1}cos(at) [/tex] fro every real value of 'a' thanks.
 
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  • #2
It has meaning only for 0<s<1. For s = 1/2, the value is √(π/2a). Since cos is even, a >0 can be assumed.
 
  • #3
thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

for example if we consider in a regularized sense that [tex] \int_{0}^{\infty}t^{s-1}dt=0 [/tex] for every 's' then

[tex] \int_{0}^{\infty}frac(1/t)t^{s-1}dt= -\zeta(s)/s [/tex] for every s except s=1
 
  • #4
zetafunction said:
my question is what is the value of

[tex] \int_{0}^{\infty}dtt^{s-1} [/tex]

if s >1 the integral is divergent as x -->oo

if s<1 the integral would become divergent as x-->0

It is divergent on the limit [itex]x\to\infty[/itex] when [itex]s\geq 0[/itex], and on the limit [itex]x\to 0^+[/itex] when [itex]s\leq 0[/itex]. The whole integral is thus divergent for any [itex]s[/itex].

zetafunction said:
thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

for example if we consider in a regularized sense that [tex] \int_{0}^{\infty}t^{s-1}dt=0 [/tex] for every 's' then

What do you mean by this "regularization". You have given value 0 to a divergent integral!
 

1. What is the Mellin transform and how is it different from other transforms?

The Mellin transform is a mathematical tool used to convert a function from one domain to another, similar to the Fourier or Laplace transforms. However, unlike these transforms, the Mellin transform is defined over the complex plane and can handle a wider range of functions, including those with singularities or oscillatory behavior.

2. What is the purpose or application of the Mellin transform?

The Mellin transform has various applications in mathematics, physics, signal processing, and engineering. It is commonly used to simplify and solve integrals, differential equations, and other mathematical problems. It is also used in image and signal processing to analyze and manipulate data.

3. How is the Mellin transform calculated and what is its formula?

The Mellin transform is calculated by multiplying the function by the variable raised to a power and integrating over the positive real line. Its formula is F(s) = ∫f(x)x^(s-1)dx, where s is a complex variable and f(x) is the original function.

4. What is the inverse Mellin transform and how is it used?

The inverse Mellin transform is the process of converting a Mellin-transformed function back to its original form. It is calculated by integrating over a contour in the complex plane and using the residue theorem. It is used to solve inverse problems and to convert from one domain to another for analysis or manipulation.

5. Are there any limitations or drawbacks to using the Mellin transform?

Although the Mellin transform has a wide range of applications, it is not suitable for all types of functions. It requires some level of mathematical expertise to understand and use effectively. Additionally, the inverse Mellin transform may not always exist or be unique, making it difficult to calculate. Overall, the Mellin transform is a powerful tool, but it should be used with caution and in conjunction with other methods.

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