On The Integral Expression for Converging Series

On The Integral Expression for Converging Series and its Possible Application to the Zeta Function. J.Pillay, R. Iyer.

Institute and Faculty of Actuaries Indian Institute of Technology

Abstract. We establish an integral-expression Aζ for series of the form:

∞ X

f (n) where:

1

limx→∞ f (x) = 0 with a margin of error less than or equal to f (1). We then use the result obtained to minimize the approximation of ζ(s). The result will show that IM (Aζ(x+it) ) = 0 at x = 21 . Introduction. The origins and details surrounding series-analysis can be found in almost any serious treatise on Real-Analysis. A primary concern is ascertaining whether a given series is convergent or not and for this purpose alone there exists a set of well known tests and analytic techniques and are covered in most undergraduate courses in mathematical analysis. The field of infinite sum analyses alone is quite vast and though techniques and tests have been formulated over the centuries; only the ’well behaved’ or the monotone subset of such series is fairly well understood. The Wikipedia article Convergence tests. https://en.wikipedia.org/wiki/Convergence_tests provides a good summery of available tests. X An infinite sum, written: aj expresses in shorthand the sum of terms: ∀j

X1 . n ∀n This is an example of a well behaved monotone series, and one of the tests that can be used to prove its divergence is the Integral-Test: a0 + a1 + ... + an + .... For instance, the harmonic series is expressed as:

Integral Test For a continuous function f defined over [N, ∞) that is monotone-decreasing; Z ∞ ∞ X f (n) converges to a real number if and only if the improper integral f (x)dx N

n=N

exists1. Going back to the harmonic series and applying the integral test we obtain: 1

Z

∞

1 dx = [ln(x)]∞ 1 which is clearly divergent. x For non-monotone, oscillating functions, the task of ascertaining whether the associated series converges or not can be difficult to seemingly impossible. Taking for instance the alternating series test: 1

Alternating Series Test If for all n, an is positive, non-increasing (i.e. 0 < an+1 ≤ an ), and approaching ∞ ∞ X X zero, then the alternating series (−1)n an and (−1)n−1 an both converge1. 1

1

This test explicitly requires that an be monotone convergent, so one can’t for ∞ X Sin(x) using this test alone. However instance, establish convergence of: x 1 we are saved by the Squeeze-Theorem which can be used to establish converSin(x) gence by means of the following argument: −1 ≤ x1 , since the series x ≤ x associated with either ends of the inequality are both absolutely convergent, it ∞ X Sin(x) follows for: as well. x 1 The following proposal which can be perceived as somewhat of a low lying fruit actually stemmed from a subtle but persistent feeling that an integral expression exists that is closely coupled in some manner to converging series, much like the integral test but for both monotone and non-monotone series. Proposition 1 (Part I). Given a single valued function f continuous over [0, ∞), the series

∞ X

fn is con-

1 h

Z

xf 0 (x)dx exists.

vergent if and only if the integral : Limh→∞ 1

Proposition 1 (Part II). Z h ∞ X fn = Limh→∞ xf 0 (x)dx + O where O < |f1 |. 1

1

Proof Z ∞ Z f (x)dx = − c

c

f (x)dx

∞

From the above, we have that: − Z ∞ f 0 (x)dx + ...

∞ X

Z fn =

0

Z

f (x)dx + 1

1

∞

∞

f 0 (x)dx + .. +

2

k

It is easy to see that over each interval [1,2], [2,3] etc. The integrals may be Z 2 Z 3 0 f (x)dx + 2 f 0 (x)dx + .. and so on, as there is a repetire-written as: 1 1

2

tion of the area in proportion with each integral value over which the original

2

sum is evaluated. Given that this is the Z ∞case, one may re-write the above in approximation by use of the formula: xf 0 (x)dx. 1

In the way of establishing part two Zof the proposition, we note that the difZ ∞ ∞ 0 f 0 (x)dx + ... (an upper-bound) and: f (x)dx + 2 ference between: 1 2 1 Z ∞ Z ∞ Z ∞ 0 f 0 (x)dx + 1 f 0 (x)dx + ... (a lower-bound) is simply f 0 (x)dx. 1

2

1

There is a lot of narration left to further clarify the above and is unfortunately the most difficult portion; to convince an audience of. We firstly note trivially Z h that: Limh→∞ − f 0 (x)dx = (−1)Limh→∞ [f (h)]hc = f (c). Further to this we have that:

∞ X

c

fn = f (1) + f (2) + .. + f (n) + .. Which can be expressed as the

1

sum of the entries in the last column of the matrix that follows:  Z     ∞ X  fn =    1   

2

f 0 (x)dx

1

Z

3

Z2 3 0

f 0 (x)dx f 0 (x)dx

2

0

Z

4

Z3 4 Z3 4

0

f 0 (x)dx f 0 (x)dx f 0 (x)dx

Z

Znn+1 . Znn+1 .

3

0

0

n+1

.

Znn+1 0

. n

Z Resulting in the approximation:

f 0 (x)dx f 0 (x)dx f 0 (x)dx f 0 (x)dx

Z

∞

f 0 (x)dx

...| Z1 ∞ ...| Z2 ∞ ...| Z3

f 0 (x)dx

∞

...| n

∞

xf 0 (x)dx.

1

An example of Proposition part 1. in practice is to prove the divergence X1 ; with the obvious substitutions, we have of the harmonic series. Taking n Z Z Z 1 1 0 xf (x) = −x( 2 ) which becomes: − , the integral of which forms: x x −ln(x) + C which clearly diverges as: x → ∞. 3

f 0 (x)dx

f 0 (x)dx

           

X 1 √ . Again with substitution; Another simple example is the divergence of n Z Z √ 1 0 − 32 xf (x) = − x(x ) = − x + C which again clearly diverges as: x → ∞. 2 Calculations on the Zeta-Function Its natural to provide a basic introduction to the origins of the zeta function. It’s significance is best seen with the aide of the following equation: ∞ X 1 1 = Πprime s n 1 − p−s n=1

(1)

3 One of the most significant results on the zeta function ζ(s) due to Hardy is that there are an infinite number of zeros on the critical strip x = 21 4. The outlline of this fantastic result follows with Ξ(t) having real zeros for zeros of ζ.

 Ξ(t) := ξ

       1 it 1 1 2 1 1 it 1 + it = − + + it t + π− 4 − 2 Γ ζ 2 2 4 4 2 2

(2)

  s s−1 s 1 1 1−s s(s − 1)π − 2 Γ ζ(s) = s(s − 1)π 2 Γ ζ(1 − s) = ξ(1 − s) 2 2 2 2 (3) Using a result by Ramanujan on integrals involving Ξ(t), we have: Z ∞

x π x Ξ(t) 2 − 2e− 2 ψ e−2x cos(xt) dt = e (4) 1 2 t2 + 4 0

ξ(s) =

where ψ(s) :=

∞ X

2

e−n

πs

is the theta function. Now setting x := −iα, we get:

n=1

Z lim

α→ π 4

0

∞

(−1)n π cos Ξ(t) 2n t cosh(αt) dt = 4n t2 + 41

The following integral for 0 ≤ α ≤ Z 0

∞

π 8


(5)

π 4

is uniformly convergent with respect to α:
  (−1)n π cos π8 πt Ξ(t) 2n t cosh dt = (6) 4 4n t2 + 14

With signs alternating infinitely often in the resulting algebra along with the left-hand side
having the same sign for sufficiently large values of n, we can infer that ζ 21 + it has an infinite number of zeros5. Calculations on the Zeta function. 4

The aim from here on is to leverage off of our proposal allowing us to approxi∞ X 1 mate to a significant level of accuracy (i.e. Within f (1)) and to ascertain ns 1 when the approximate sum is minimal. We begin with the simple expression: n−x−iy = n−x [Cos(−yln(n)) + iSin(−yln(n))]

(7)

Given trivially that: n(−x)n−x−1 = (−x)n−x we wish to obtain an expression for: nζ 0 (s) which we do by evaluating the derivative w.r.t n and multiplying the resulting expression by n: −y )[−Sin(−yln(n))+iCos(−yln(n))] n (8) To make use of Proposition(part II), we need to integrate the terms respectively Z

n−x (−x)[Cos(−yln(n))+iSin(−yln(n))]+n1−x (

∞

nζ 0 (s)dn

so as to find 1

−y n dn

Let u = −yln(n). We than have that: du = −u(−x) y

and e−u = ny ; e

−u y

= n

−x

= n . Finally given the previous; the expression for the real and e portion of the integrand follows as: Z Z u(x−1) −u(1−x) x y e Cos(u)du − e y Sin(u)du (9) y Substituting

(x−1) y

with a Z Z x eau Cos(u)du − eau Sin(u)du y

(10)

in the imaginary case : x y

Z

eau Sin(u)du +

Z

eau Cos(u)du

(11)

Using integral tables to integrate the above general form2, we have for the real part: u(x−1)

u(x−1)

x e y (x − 1) e y (x − 1) [ (x−1) ]( Cos(u)+Sin(u))−[ (x−1) ]( Sin(u)−Cos(u)) 2 2 y ( y y ) +1 ( ) +1 y

y

(12) In the imaginary case: u(x−1)

u(x−1)

e y (x − 1) e y (x − 1) x [ (x−1) ]( Sin(u)+Cos(u))+[ (x−1) ]( Cos(u)−Sin(u)) y ( y y )2 + 1 ( )2 + 1 y

y

(13) x 1 (x − 1) 1 − [ (x−1) ]( ) + [ (x−1) ](−1) 2 y ( y ) +1 ( )2 + 1 y

y

5

(14)

1 x 1 (x − 1) ) − [ (x−1) ](1) − [ (x−1) ]( y ( y )2 + 1 ( )2 + 1 y

(15)

y

1

−[

2 ( (x−1) y )

−

(16)

(2x − 1) y2 (( )) (x − 1)2 + y 2 y

(17)

x(x − 1) + y 2 y2 ( ) (x − 1)2 + y 2 y2

(18)

−

y(2x − 1) (x − 1)2 + y 2

(19)

−

x(x − 1) + y 2 (x − 1)2 + y 2

(20)

− Real:

x (x − 1) ]( ( ) + 1) y +1 y

Imaginary:

Real:

−

x+

y2 x−1

(x − 1) +

y2 x−1

f 0 (x) −2x + 1 =− 0 g (x) 2(x − 1) 1

(21)

(22)

−y 2

At x = 12 the sum evaluates to: 41 +y2 , which is less than one (under obvious 4 conditions) for the real part, and evaluates to 0 for its imaginary counterpart. It is significant to note that f (1) = 1.

References [1] Wikipedia. (2018 February 08). Convergence tests. https://en.wikipedia. org/wiki/Convergence_tests [2] mathportal.org. 3. Integrals of Exponential Functions https://www. mathportal.org/formulas/integration_formulas.pdf [3] Wikipedia. (2018 February 23). Riemann zeta function. https://en. wikipedia.org/wiki/Riemann_zeta_function [4] G. H. Hardy. Sur les zros de la fonction ζ(x) Riemann. French. In:Comptes Rendusde lAcadmie des Sciences158(1914), pp.101214.issn:00014036. [5] E. C. Titchmarsh. (1986). The Theory of The Riemann Zeta-Function. Retrieved from http://plouffe.fr/simon/math/The%20Theory%20Of% 20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf.

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