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On The Integral Expression for Converging Series and
its Possible Application to the Zeta Function
J Pillay
To cite this version:
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
1
f (n) where: 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 = 12.
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.
An infinite sum, written: X
∀j
aj expresses in shorthand the sum of terms:
a0+ a1+ ... + an+ .... For instance, the harmonic series is expressed as:
X
∀n
1 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:
Integral Test
For a continuous function f defined over [N, ∞) that is monotone-decreasing;
∞
X
n=N
f (n) converges to a real number if and only if the improper integral Z ∞
N
f (x)dx exists1.
Z ∞
1
1
xdx = [ln(x)]
∞
1 which is clearly divergent.
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:
Alternating Series Test
If for all n, an is positive, non-increasing (i.e. 0 < an+1≤ an), and approaching
zero, then the alternating series
∞ X 1 (−1)nan and ∞ X 1 (−1)n−1anboth converge1.
This test explicitly requires that an be monotone convergent, so one can’t for
instance, establish convergence of:
∞
X
1
Sin(x)
x using this test alone. However we are saved by the Squeeze-Theorem which can be used to establish conver-gence by means of the following argument: −1x ≤ Sin(x)x ≤ 1
x, since the series
associated with either ends of the inequality are both absolutely convergent, it follows for: ∞ X 1 Sin(x) x as well.
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 expres-sion 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
1
fnis
con-vergent if and only if the integral : Limh→∞
Z h
1
xf0(x)dx exists. Proposition 1 (Part II).
∞ X 1 fn= Limh→∞ Z h 1 xf0(x)dx + O where O < |f1|. Proof Z ∞ c f (x)dx = − Z c ∞ f (x)dx From the above, we have that: −
∞ X 1 fn = Z ∞ 1 f0(x)dx + Z ∞ 2 f0(x)dx + .. + Z ∞ k f0(x)dx + ...
It is easy to see that over each interval [1,2], [2,3] etc. The integrals may be re-written as: 1 Z 2 1 f0(x)dx + 2 Z 3 2
sum is evaluated. Given that this is the case, one may re-write the above in approximation by use of the formula:
Z ∞
1
xf0(x)dx.
In the way of establishing part two of the proposition, we note that the dif-ference between: 1 Z ∞ 1 f0(x)dx + 2 Z ∞ 2
f0(x)dx + ... (an upper-bound) and:
0 Z ∞ 1 f0(x)dx + 1 Z ∞ 2 f0(x)dx + ... (a lower-bound) is simply Z ∞ 1 f0(x)dx.
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 that: Limh→∞−
Z h
c
f0(x)dx = (−1)Limh→∞[f (h)]hc = f (c). Further to this we
have that:
∞
X
1
fn = f (1) + f (2) + .. + f (n) + .. Which can be expressed as the
sum of the entries in the last column of the matrix that follows:
∞ X 1 fn= Z 2 1 f0(x)dx Z 3 2 f0(x)dx Z 4 3 f0(x)dx . Z n+1 n f0(x)dx ...| Z ∞ 1 f0(x)dx 0 Z 3 2 f0(x)dx Z 4 3 f0(x)dx . Z n+1 n f0(x)dx ...| Z ∞ 2 f0(x)dx 0 0 Z 4 3 f0(x)dx . Z n+1 n f0(x)dx ...| Z ∞ 3 f0(x)dx 0 0 0 . Z n+1 n f0(x)dx ...| Z ∞ n f0(x)dx
Resulting in the approximation: Z ∞
1
xf0(x)dx.
An example of Proposition part 1. in practice is to prove the divergence of the harmonic series. TakingX1
n; with the obvious substitutions, we have Z xf0(x) = Z −x(1 x2) which becomes: Z −1
Another simple example is the divergence ofX√1
n. Again with substitution; Z xf0(x) = Z −1 2x(x −3 2) = − √
x + C which again clearly diverges as: x → ∞. 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 n=1 1 ns = Πprime 1 1 − p−s (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 =124.
The outlline of this fantastic result follows with Ξ(t) having real zeros for zeros of ζ. Ξ(t) := ξ 1 2+ it = −1 2 t2+1 4 π−14− it 2 Γ 1 4 + it 2 ζ 1 2+ it (2) ξ(s) =1 2s(s − 1)π −s 2Γ s 2 ζ(s) = 1 2s(s − 1)π s−1 2 Γ 1 − s 2 ζ(1 − s) = ξ(1 − s) (3) Using a result by Ramanujan on integrals involving Ξ(t), we have:
Z ∞ 0 Ξ(t) t2+1 4 cos(xt) dt = π 2 e x 2 − 2e−x2ψ e−2x (4) where ψ(s) := ∞ X n=1
e−n2πs is the theta function. Now setting x := −iα, we get:
lim α→π 4 Z ∞ 0 Ξ(t) t2+1 4 t2ncosh(αt) dt = (−1) nπ cos π 8 4n (5)
The following integral for 0 ≤ α ≤π4 is uniformly convergent with respect to α:
Z ∞ 0 Ξ(t) t2+1 4 t2ncosh πt 4 dt = (−1) nπ cos π 8 4n (6)
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 12+ it has an infinite number of zeros5.
The aim from here on is to leverage off of our proposal allowing us to approxi-mate ∞ X 1 1
ns to a significant level of accuracy (i.e. Within f (1)) and to ascertain
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: nζ0(s) which we do by evaluating the derivative w.r.t n and multiplying the resulting expression by n:
n−x(−x)[Cos(−yln(n))+iSin(−yln(n))]+n1−x(−y
n )[−Sin(−yln(n))+iCos(−yln(n))] (8) To make use of Proposition(part II), we need to integrate the terms respectively so as to find
Z ∞
1
nζ0(s)dn
Let u = −yln(n). We than have that: du = −yn dn and e−u = ny; e−uy = n
and e−u(−x)y = n−x. Finally given the previous; the expression for the real
portion of the integrand follows as: x y Z eu(x−1)y Cos(u)du − Z e−u(1−x)y Sin(u)du (9) Substituting (x−1)y with a x y Z eauCos(u)du − Z eauSin(u)du (10)
in the imaginary case : x y Z eauSin(u)du + Z eauCos(u)du (11)
Using integral tables to integrate the above general form2, we have for the real part: x y[ eu(x−1)y ((x−1)y )2+ 1]( (x − 1) y Cos(u)+Sin(u))−[ eu(x−1)y ((x−1)y )2+ 1]( (x − 1) y Sin(u)−Cos(u)) (12) In the imaginary case:
−x y[ 1 ((x−1)y )2+ 1](1) − [ 1 ((x−1)y )2+ 1]( (x − 1) y ) (15) −[ 1 ((x−1)y )2+ 1]( x y( (x − 1) y ) + 1) (16) − y 2 (x − 1)2+ y2(( (2x − 1) y )) (17) Real: − y 2 (x − 1)2+ y2( x(x − 1) + y2 y2 ) (18) Imaginary: − y(2x − 1) (x − 1)2+ y2 (19) Real: −x(x − 1) + y 2 (x − 1)2+ y2 (20) − x + y2 x−1 (x − 1) +x−1y2 (21) f0(x) g0(x) = − −2x + 1 2(x − 1) (22)
At x = 12 the sum evaluates to: 14−y 2 1 4+y2
, which is less than one (under obvious 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.
