## How And Why To Seidr

First, it is known that if for a row there is a continuous, positive, monotonously decreasing function defined on a set such that, and has an antiderivative, the rest of a row is estimated so: where. Applying the aforesaid to a row (we will find that necessary function

Formula (it is important because it connects the natural row presented by a set of values of argument of dzeta-function with a set of prime numbers. We will take one more step in this direction, having estimated, namely having shown that where remains limited at.

It is easy to show that all formulas received for dzeta-function without changes are transferred to a case of complex argument. Proofs undergo the insignificant transformations connected with transition to absolute values.

Using absolute convergence of integral if, and limitation of function, we draw a conclusion that in the left part of equality (the integral too meets at. Means a formula (it is possible to continue dzeta-function and is one half-plane more right than a straight line.

We could already apply Mellin's formula, but then it would be very difficult to execute integration. Therefore before we will transform equality (as follows. Differentiating on s, we receive. Let's designate the left part through and we will put, (and we believe equal to zero at). Then, integrating in parts, we find at, or.

That the proof was strict, we have to prove term by term integration still. As a row (meets almost everywhere and its partial sums remain limited, term by term integration on any final piece is admissible. In view of for any, it is necessary to prove that at. But integrating internal integral in parts we have

(which is brought as follows. Using properties of integrals it is possible to write down. For any d at, means and, and. Therefore. The integral can be found integration in parts, accepting; then, and. As a result. Let's subtract from this integral previous and we will receive, from here equality easily follows (.

Let. Let's count absolute values of members of a row (. The first multiplier contains only real numbers and, as. To the second multiplier it is applicable the well-known formula of Euler, we will receive. Means. In view of convergence of a row at α> 1, we have absolute convergence of a row (.

Now let s> For research of convergence of a row (we will use an integrated sign of Cauchy. At each s we will consider function, where which is continuous on an interval, positive and monotonously decreasing. There are three various opportunities:

In this regard the remark becomes possible to use decomposition of dzeta-function in work, where s now any complex number, such that. Let's apply it to the proof of absence at function of roots.

To justify this result, it is enough to make sure that a row (evenly agrees on an interval and to use the theorem of differentiation of ranks. We use the same reception. Let's record any s0> 1 and we will present a row (in a look for s> s Multipliers, since n=2, monotonously decrease, remaining limited to number ln Therefore on the basis of Abel a row (meets evenly at s> s0, so and at any s> Whatever value s> 1 to take it it is possible to conclude between and where, and; the above theorem is applicable to an interval.

(. This integral has the necessary form, but will not affect an asymptotics. Really, as, the integral for meets evenly in half-plane that easily it is found comparison with in integral. Therefore, it is regular and limited in half-plane. The same fairly and relatively, as.