# Mellin transform-- complete Riemann function

• binbagsss
In summary, the problem is to show that the integral from 0 to infinity of the sum of e^-pi*n^2*t and t^(s-1) is equal to the sum of the integral from 0 to infinity of e^-t and t^(s-1). The suggested approach is to make a change of variables of the form u = pi*n^2*t.
binbagsss

## Homework Statement

Show that ## \int\limits^{\infty}_0 \sum\limits^{\infty}_{n=1}e^{-\pi n^{2} t} t^{s-1} dt = \sum\limits^{\infty}_{n=1}(\pi n^{2} )^{-s} \int\limits^{\infty}_{0} e^{-t} t^{s-1} dt ##

## Homework Equations

##\Phi (t)=\theta(it/2)=1+2\sum\limits^{\infty}_{n=1}e^{-\pi n^2 t}. ##

With the transformation formula : ##\Phi(1/t)=t^{1/2}\Phi(t) ##

## The Attempt at a Solution

I am unsure really where to start, a clue would be greatly appreciated.

The only thing I can really think of is transforming ##t## to get ##t^{s}## rather than ##t^{1/2}## on the RHS of (1), but I am a) unsure how to do this b) unsure whether it is the right thing to do

The other thing would be to expand out the exponential, writing it as two expressions multiplied together corresponding to the ##-t## in the exponent power and the ## \pi n^{2} ## seperately, collecting the sum to infinity to get ##e^{-t}## and leaving the bit corresponding to "##e^{\pi n^{2}}" ## as a summation. However the ##-s## means the choice of my variable of this summation would need to be ##s##, which doesn't really make sense?

I'm quite stuck.

Any help greatly appreciated,

Many thanks.

Last edited:
binbagsss said:

## Homework Statement

Show that ## \int\limits^{\infty}_0 \sum\limits^{\infty}_{n=1}e^{-\pi n^{2} t} t^{s-1} dt = \sum\limits^{\infty}_{n=1}(\pi n^{2} )^{-s} \int\limits^{\infty}_{0} e^{-t} t^{s-1} dt ##

## Homework Equations

##\Phi (t)=\theta(it/2)=1+2\sum\limits^{\infty}_{n=1}e^{-\pi n^2 t}. ##

With the transformation formula : ##\Phi(1/t)=t^{1/2}\Phi(t) ##

## The Attempt at a Solution

I am unsure really where to start, a clue would be greatly appreciated.

The only thing I can really think of is transforming ##t## to get ##t^{s}## rather than ##t^{1/2}## on the RHS of (1), but I am a) unsure how to do this b) unsure whether it is the right thing to do

The other thing would be to expand out the exponential, writing it as two expressions multiplied together corresponding to the ##-t## in the exponent power and the ## \pi n^{2} ## seperately, collecting the sum to infinity to get ##e^{-t}## and leaving the bit corresponding to "##e^{\pi n^{2}}" ## as a summation. However the ##-s## means the choice of my variable of this summation would need to be ##s##, which doesn't really make sense?

I'm quite stuck.

Any help greatly appreciated,

Many thanks.

Do a change of variables of the form ##u=\pi n^2 t##.

eys_physics said:
Do a change of variables of the form ##u=\pi n^2 t##.
cheers !

## 1. What is the Mellin transform?

The Mellin transform is a mathematical operation used to transform a function from one domain to another. It is similar to the more well-known Fourier transform, but instead of transforming a function from the time domain to the frequency domain, the Mellin transform transforms a function from the space domain to the logarithmic frequency domain.

## 2. What is the complete Riemann function?

The complete Riemann function, also known as the Riemann zeta function, is a mathematical function that is closely related to the distribution of prime numbers. It is defined as ζ(s) = ∑(n=1 to ∞) 1/n^s, where s is a complex number. It has many important properties and connections to other areas of mathematics, including the Mellin transform.

## 3. How are the Mellin transform and the Riemann function related?

The Mellin transform of a function f(x) is defined as M[f(x)](s) = ∫(0 to ∞) f(x) x^(s-1) dx. This integral can be rewritten in terms of the Riemann function, giving M[f(x)](s) = ∫(0 to ∞) f(x) ζ(s) x^(-s) dx. This connection allows for the use of the Mellin transform in the study of the Riemann function and its properties.

## 4. What are the applications of the Mellin transform and the Riemann function?

The Mellin transform and the Riemann function have many applications in mathematics and physics. They have been used to study the distribution of prime numbers, to solve differential and integral equations, to analyze power law behavior in various systems, and to characterize the fractal properties of certain functions. They are also important in number theory, complex analysis, and harmonic analysis.

## 5. Are there any real-world examples of the Mellin transform and the Riemann function?

Yes, the Mellin transform and the Riemann function have been applied to various real-world problems. For example, they have been used in signal processing to analyze the frequency content of signals or to compress data. They have also been used in economics to study the distribution of income and wealth. Additionally, the Riemann function has been used in cryptography to generate secure prime numbers for encryption algorithms.

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