TOPICS
Search

Riemann Zeta Function zeta(2)

DOWNLOAD Mathematica NotebookDownload Wolfram Notebook

The value for

zeta(2)=sum_(k=1)^infty1/(k^2)
(1)

can be found using a number of different techniques (Apostol 1983, Choe 1987, Giesy 1972, Holme 1970, Kimble 1987, Knopp and Schur 1918, Kortram 1996, Matsuoka 1961, Papadimitriou 1973, Simmons 1992, Stark 1969, 1970, Yaglom and Yaglom 1987).

zeta(2) is therefore the definite sum version of the indefinite sum

H_n^((2))=sum_(k=1)^(n)1/(k^2)
(2)
=zeta(2)-psi_1(n+1),
(3)

where H_n^((2)) is a generalized harmonic number (whose numerator is known as a Wolstenholme number) and psi_n(z) is a polygamma function.

The problem of finding this value analytically is sometimes known as the Basel problem (Derbyshire 2004, pp. 63 and 370) or Basler problem (Castellanos 1988). It was first proposed by Pietro Mengoli in 1644 (Derbyshire 2004, p. 370). The solution

zeta(2)=(pi^2)/6
(4)

was first found by Euler in 1735 (Derbyshire 2004, p. 64) or 1736 (Srivastava 2000).

Yaglom and Yaglom (1987), Holme (1970), and Papadimitriou (1973) all derive the result, pi^2/6 from de Moivre's identity or related identities.

zeta(2) is given by the series

zeta(2)=3sum_(k=1)^infty1/(k^2(2k; k)).
(5)

(Knopp 1990, pp. 266-267), probably known to Euler and rediscovered by Apéry.

Bailey (2000) and Borwein and Bailey (2003, pp. 128-129) give a collection of BBP-type formulas that include a number for zeta(2),

zeta(2)=(27)/4sum_(k=0)^(infty)1/(64^k)[(16)/((6k+1)^2)-(24)/((6k+2)^2)-8/((6k+3)^2)-6/((6k+4)^2)+1/((6k+5)^2)]
(6)
=4/9sum_(k=0)^(infty)1/(729^k)[(243)/((12k+1)^2)-(405)/((12k+2)^2)-(81)/((12k+4)^4)-(27)/((12k+5)^2)-(72)/((12k+6)^2)-9/((12k+7)^2)-9/((12k+8)^2)-5/((12k+10)^2)+1/((12k+11)^2)].
(7)

zeta(2) is given by the double series

zeta(2)=sum_(i=1)^inftysum_(j=1)^infty((i-1)!(j-1)!)/((i+j)!)
(8)

(B. Cloitre, pers. comm., Dec. 9, 2004).

One derivation for zeta(2) considers the Fourier series of f(x)=x^(2n)

f(x)=1/2a_0+sum_(m=1)^inftya_mcos(mx)+sum_(m=1)^inftyb_msin(mx),
(9)

which has coefficients given by

a_0=(2pi^(2n))/(2n+1)
(10)
a_m=(2pi^(2n))/(2n+1)_1F_2(n+1/2;1/2,n+3/2;-1/4mpi^2)
(11)
b_m=0,
(12)

where _1F_2(a;b,c;z) is a generalized hypergeometric function and (12) is true since the integrand is odd. Therefore, the Fourier series is given explicitly by

x^(2n)=(pi^(2n))/(2n+1)+sum_(m=1)^inftya_mcos(mx).
(13)

If n=1, then

a_m=(4(-1)^m)/(m^2),
(14)

so the Fourier series is

x^2=(pi^2)/3+4sum_(m=1)^infty((-1)^mcos(mx))/(m^2).
(15)

Letting x=pi gives cos(mpi)=(-1)^m, so

pi^2=(pi^2)/3+4sum_(m=1)^infty1/(m^2),
(16)

and we have

zeta(2)=sum_(m=1)^infty1/(m^2)=(pi^2)/6.
(17)

Higher values of n can be obtained by finding a_m and proceeding as above.

The value zeta(2) can also be found simply using the root linear coefficient theorem. Consider the equation sinz=0 and expand sin in a Maclaurin series

sinz=z-(z^3)/(3!)+(z^5)/(5!)+...=0
(18)
0=1-(z^2)/(3!)+(z^4)/(5!)+...
(19)
=1-w/(3!)+(w^2)/(5!)+...,
(20)

where w=z^2. But the zeros of sinz occur at z=pi, 2pi, 3pi, ..., or w=pi^2, (2pi)^2, .... Therefore, the sum of the roots equals the coefficient of the leading term

1/(pi^2)+1/((2pi)^2)+1/((3pi)^2)+...=1/(3!)=1/6,
(21)

which can be rearranged to yield

zeta(2)=(pi^2)/6.
(22)

Yet another derivation (Simmons 1992) evaluates zeta(2) using Beukers's (1979) integral

I=int_0^1int_0^1(dxdy)/(1-xy)
(23)
=int_0^1int_0^1(1+xy+x^2y^2+...)dxdy
(24)
=int_0^1[(x+1/2x^2y+1/3x^3y^2+...)]_0^1dy
(25)
=int_0^1(1+1/2y+1/3y^2+...)dy
(26)
=[y+(y^2)/(2^2)+(y^3)/(3^2)+...]_0^1
(27)
=1+1/(2^2)+1/(3^2)+...
(28)
=zeta(2).
(29)

To evaluate the integral, rotate the coordinate system by pi/4 so

x=ucostheta-vsintheta=1/2sqrt(2)(u-v)
(30)
y=usintheta+vcostheta=1/2sqrt(2)(u+v)
(31)

and

xy=1/2(u^2-v^2)
(32)
1-xy=1/2(2-u^2+v^2).
(33)

Then

I=4int_0^(sqrt(2)/2)int_0^u(dudv)/(2-u^2+v^2)+4int_(sqrt(2)/2)^(sqrt(2))int_0^(sqrt(2)-u)(dudv)/(2-u^2+v^2)
(34)
=I_1+I_2.
(35)

Now compute the integrals I_1 and I_2.

I_1=4int_0^(sqrt(2)/2)[int_0^u(dv)/(2-u^2+v^2)]du
(36)
=4int_0^(sqrt(2)/2)[1/(sqrt(2-u^2))tan^(-1)(v/(sqrt(2-u^2)))]_0^udu
(37)
=4int_0^(sqrt(2)/2)1/(sqrt(2-u^2))tan^(-1)(u/(sqrt(2-u^2)))du.
(38)

Make the substitution

u=sqrt(2)sintheta
(39)
sqrt(2-u^2)=sqrt(2)costheta
(40)
du=sqrt(2)costhetadtheta,
(41)

so

tan^(-1)(u/(sqrt(2-u^2)))=tan^(-1)((sqrt(2)sintheta)/(sqrt(2)costheta))=theta
(42)

and

I_1=4int_0^(pi/6)1/(sqrt(2)costheta)thetasqrt(2)costhetadtheta=(pi^2)/(18).
(43)

I_2 can also be computed analytically,

I_2=4int_(sqrt(2)/2)^(sqrt(2))[int_0^(sqrt(2)-u)(dv)/(2-u^2+v^2)]du
(44)
=4int_(sqrt(2)/2)^(sqrt(2))[1/(sqrt(2-u^2))tan^(-1)(v/(sqrt(2-u^2)))]_0^(sqrt(2)-u)du
(45)
=4int_(sqrt(2)/2)^(sqrt(2))1/(sqrt(2-u^2))tan^(-1)((sqrt(2)-u)/(sqrt(2-u^2)))du.
(46)

But

tan^(-1)((sqrt(2)-u)/(sqrt(2-u^2)))=tan^(-1)((sqrt(2)-sqrt(2)sintheta)/(sqrt(2)costheta))
(47)
=tan((1-sintheta)/(costheta))=tan^(-1)((costheta)/(1+sintheta))
(48)
=tan^(-1)[(sin(1/2pi-theta))/(1+cos(1/2pi-theta))]
(49)
=tan^(-1){(2sin[1/2(1/2pi-theta)]cos[1/2(1/2pi-theta)])/(2cos^2[1/2(1/2pi-theta)])}
(50)
=1/2(1/2pi-theta),
(51)

so

I_2=4int_(pi/6)^(pi/2)1/(sqrt(2)costheta)(1/4pi-1/2theta)sqrt(2)costhetadtheta
(52)
=4[1/4pitheta-1/4theta^2]_(pi/6)^(pi/2)
(53)
=4[((pi^2)/8-(pi^2)/(16))-((pi^2)/(24)-(pi^2)/(144))]=(pi^2)/9.
(54)

Combining I_1 and I_2 gives

zeta(2)=I_1+I_2=(pi^2)/(18)+(pi^2)/9=(pi^2)/6.
(55)

See also

Apéry's Constant, Hadjicostas's Formula, Riemann Zeta Function

Explore with Wolfram|Alpha

References

Apostol, T. M. "A Proof That Euler Missed: Evaluating zeta(2) the Easy Way." Math. Intel. 5, 59-60, 1983.Bailey, D. H. "A Compendium of BBP-Type Formulas for Mathematical Constants." 28 Nov 2000. http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf.Beukers, F. "A Note on the Irrationality of zeta(2) and zeta(3)." Bull. London Math. Soc. 11, 268-272, 1979.Borwein, J. and Bailey, D. Mathematics by Experiment: Plausible Reasoning in the 21st Century. Wellesley, MA: A K Peters, pp. 89-90, 2003.Castellanos, D. "The Ubiquitous Pi. Part I." Math. Mag. 61, 67-98, 1988.Choe, B. R. "An Elementary Proof of sum_(n=1)^(infty)1/(n^2)=(pi^2)/6." Amer. Math. Monthly 94, 662-663, 1987.Derbyshire, J. Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. New York: Penguin, 2004.Giesy, D. P. "Still Another Proof That sum1/k^2=pi^2/6." Math. Mag. 45, 148-149, 1972.Havil, J. Gamma: Exploring Euler's Constant. Princeton, NJ: Princeton University Press, pp. 37-40, 2003.Holme, F. "Ein enkel beregning av sum_(k=1)^(infty)1/(k^2)." Nordisk Mat. Tidskr. 18, 91-92 and 120, 1970.Kimble, G. "Euler's Other Proof." Math. Mag. 60, 282, 1987.Knopp, K. Theory and Application of Infinite Series. New York: Dover, 1990.Knopp, K. and Schur, I. "Über die Herleitug der Gleichung sum_(n=1)^(infty)1/(n^2)=(pi^2)/6." Archiv der Mathematik u. Physik 27, 174-176, 1918.Kortram, R. A. "Simple Proofs for sum_(k=1)^(infty)1/(k^2)=(pi^2)/6 and sinx=xproduct_(k=1)^(infty)(1-(x^2)/(k^2pi^2))." Math. Mag. 69, 122-125, 1996.Matsuoka, Y. "An Elementary Proof of the Formula sum_(k=1)^(infty)1/(k^2)=(pi^2)/6." Amer. Math. Monthly 68, 486-487, 1961.Papadimitriou, I. "A Simple Proof of the Formula sum_(k=1)^(infty)1/(k^2)=(pi^2)/6." Amer. Math. Monthly 80, 424-425, 1973.Simmons, G. F. "Euler's Formula sum_1^(infty)1/n^2=pi^2/6 by Double Integration." Ch. B. 24 in Calculus Gems: Brief Lives and Memorable Mathematics. New York: McGraw-Hill, 1992.Spiess, O. "Die Summe der reziproken Quadratzahlen." In Festschrift zum 60 Geburtstag von Dr. Andreas Speiser (Ed. L. V. Ahlfors et al. ). Zürich: Füssli, pp. 66-86, 1945.Srivastava, H. M. "Some Simple Algorithms for the Evaluations and Representations of the Riemann Zeta Function at Positive Integer Arguments." J. Math. Anal. Appl. 246, 331-351, 2000.Stark, E. L. "Another Proof of the Formula sum_(k=1)^(infty)1/(k^2)=(pi^2)/6." Amer. Math. Monthly 76, 552-553, 1969.Stark, E. L. "1-1/4+1/9-1/(16)+...=(pi^2)/(12)." Praxis Math. 12, 1-3, 1970.Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. Middlesex, England: Penguin Books, p. 40, 1986.Yaglom, A. M. and Yaglom, I. M. Problem 145 in Challenging Mathematical Problems with Elementary Solutions, Vol. 2. New York: Dover, 1987.

Referenced on Wolfram|Alpha

Riemann Zeta Function zeta(2)

Cite this as:

Weisstein, Eric W. "Riemann Zeta Function zeta(2)." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html

Subject classifications