The Riemann zeta function is an important function in mathematics.
The Riemann zeta function for with can be written as
Dirichlet series is used here.
We know that such that the summation converges and is a completely multiplicative function. Since is completely multiplicative, we have
Since is multiplicative, we take the product of the sum over all prime powers: The LHS's sum is just the geometric progression sum. We have it equal to
(The addition of this proof in this section is intended to simplify the discussion of Euler's proof on the infinitude of number of primes. This is actually a detailed form of Proof 2.)
This proof will deal only with real number . We'll also ignore technicalities about convergence. But the absolute convergence (not just conditional convergence) of the sum and product can be used to justify our manipulations.
Recall the formula for the sum of a geometric series
where . Letting , where is a prime, we obtain
Now, consider the product
When we expand this, we obtain
where denotes the set of all such integers including whose prime factorizations involve only primes in , which is a set of primes. Note that each integer from occurs exactly once in the sum. For example, , and the term comes from . The uniqueness of factorization says that this is the only way it can occur.
When we multiply by , we get
In general, if is the prime, we have
Let . The left side converges to the product over all primes. Since every positive integer has a prime factorization, each positive integer lies in for a sufficiently large integer . Therefore, the right side converges to the sum over all positive integers . This gives the identity in the theorem. That is,
An interesting result that comes from this is the fact that there are infinite prime numbers. As at it diverges, the product must be taken over infinite numbers, and hence there are infinite primes.
The zeta function can be represented as
Consider the gamma function We substitute and thus : We take the sum over all the positive integers as We apply the formula for the sum of a geometric progression:
Consider Let . So Summing over all positive integers, We recall the Jacobi theta function: So, let Now split up in two parts Consider Putting this in the first integral, We can easily integrate the terms independent of to get the integral to be Change with to get So, Note that So
Consider Using Lagrange duplication formula and Euler reflection formula, Replacing with gives
For any positive even integer
where denotes the Bernoulli number.
Consider Taking log and d.w.r.s Interchanging the summations, Now consider The Taylor series of is used. Since for being an odd positive integer : Since in a power series the coefficients must be the same, we have
where is a negative integer.
Consider First notice if is for positive integer, the is zero and hence the LHS is zero. So This is also known as the trivial zeroes of the zeta function.
Now replace with then the becomes and So, we have
Special Case of Hurwitz Zeta Function
Hurwitz zeta function is defined as
If we take , we'll get
Now on rearranging the limits, we get
Hence, we can conclude that the Riemann zeta function is the special case of the Hurwitz zeta function.
Special Case of Polygamma Function
Polygamma function is defined as
It is also represented as
Now for we get
Hence, we can conclude that the Riemann zeta function is the special case of Polygamma function.
Special Case of Dirichlet Series
View Dirichlet Series.
We define a Dirichlet series as
where is an arithmetic function. At the special case of , it is the zeta function. Also, many other Dirichlet series give result in terms of the zeta function, as you can see in the wiki or these notes.
These relations are very useful while solving problems.
Now, we deviate from what we have obtained at (1). W'll evaluate the series expansion of .
Consider the series . Here is an unknown real valued function:
On using the Taylor series of expansion of and , we get
On comparing coefficients in (2), we get
From (1), we can deduce that .
Hence (3) becomes
This equation is true for with being an integer.
From the recursive relation we get