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The Basel Problem

Because of its relation to the distribution of prime numbers, the Riemann zeta function {\zeta} (s) is one of the most important functions in mathematics. The zeta function is defined by the following formula for any complex number s with the real component greater than 1.Taking s=2, we see that {\zeta}(2) is equal to the sum of the squares of reciprocals of all positive integers. This leads to the famous problem by Basel in mathematical analysis with important relevance to number theory, solved by Leonhard Euler in 1734. In this paper, we discuss some of the notable proofs given by mathematicians to the basal problem. Most of the theorems are very well known whereas some can be found as proofs of problems present in textbooks.We also give one new proof using the theory of calculus of residues.


Introduction
The Basel problem was first introduced in 1644 by Pietro Mengoli, Italian mathematician and clergyman (1626-1686) who is known for his Known (nowadays) for work in infinite series. The Problem remained open for 90 years, until by solving it, Euler gave his first proof in 1734. It was Euler who found the exact sum to be 2 6 . He would eventually propose three separate solutions to the problem during his lifetime for ζ(2). Bernhard Riemann took up these ideas years later in his seminal 1859 paper, "On the Number of Primes Less Than a Given Magnitude," in which he described his zeta function and demonstrated its fundamental properties. In this paper we discuss about different solutions that researchers have since found about evaluating the value of ζ(2) drawing from diverse areas as complicated analysis, calculus, probability, and the theory of Hilbert space. To start with let's define what is Riemmann ζ function.

Proof 1:
Let us first take a proof that is due to Calabi, Beukers and Kock 1 . They first noticed the following equality: (2) = ∑ The Right hand side of the equation (1) can be rewritten after the substitution as Since the the Jacobian matrix will be Now as the area of = ( 1 2 ) * ( 2 ) * ( 2 ) = 2 8 . We get from equation (1), (2) and (3) that (2) = 2 6 .

Proof 2:
We now derive a proof that comes from a note by Boo Rim Choe 2 in the American Mathematical Monthly in 1987. We start by taking a note of the power series expansion of the inverse sine function.

Proof 3:
We now take a proof from an article in the Mathematical Intelligencer by Apostol 3 in 1983. We first start noting the following equation: It is also to note that we get by the monotone convergence theorem Let us now introduce change of variables by changing the variables = ( + )/2) = ( − )/2)). After the substitution of the variables we get the relationship between changed variables and actual variables as =and = + . Now notice that the Left hand side of the equation (12) is (2). We therefore can rewrite equation (12) in terms of the changed variables as: Here S is the square with vertices (0, 0), (1/2, −1/2), (1, 0) and (1/2, 1/2). Now because of the symmetry present (symmetrical about the line y=0). We can find the value by only doubling the integration of the upper half of the square which can be itself be divided into two triangles with coordinates It follows that = 2 2 − 1 = 2 and therefore we get that = (  (18)

Proof 4:
We now present some textbook proofs, found in many books on Fourier analysis to find the value of (2). In this proof we are going to discuss next we will use the 2 completeness of the trigonometric functions. Suppose that is a Hilbert space with inner product. Let ( ) be an ortho-normal basis of H such that Then we know that by Parseval's identity, we can write for every ∈ , Which therefore gives us (2) = 2 /6.

Proof 5:
Let us now see another similar way of proving the identity by fourier transform. We first take a function f which is continuous in [ 0, 1 ] and (0) = (1). Let us first explore the criteria for point-wise convergence of a periodic function f. The criteria for point-wise convergence of a periodic function f are as follows: • Fourier series converges uniformly if satisfies a Holder condition.
• Fourier series converges everywhere if is of bounded variation.
• Fourier series converges uniformly if is continuous and its Fourier coefficients are absolutely summable.
Then the Fourier series of converges to pointwise. As our function holds true in both the circumstances we can say that the Fourier series of f converges to f point wise. Let us apply this to the function with these properties let's say ( ) = (1 − ) which then gives us the following equation: If we put x=0 in the LHS of the equation (20) we get (2) = 2 /6. We can also get the identity if we put x=1/2 which gives us: Now as we know that Now if we subtract the equation (21) from the equation (22). We get We therefore get ζ (2)/2 = 2 12 which gives us (2) = 2 /6.

Proof 6:
Let us now derive the original proof given by Euler. We use the infinite product for the sine function.
Now the RHS can be written as Whereas the LHS can be expanded in power series as: Now if we compare the coefficients of 3 in the MacLaurin series of both the LHS and RHS sides we get ζ(2) = 2 /6.

Proof 7:
The two earlier proofs using Fourier transform can be proved in a similar way but without using Fourier transform in the following manner. Consider the series: First of all notice that the function is uniformly convergent on the real line. Now we know that sin can also be written as ( − − )/2 . Therefore we can write as follows: And so this sum is bounded above in absolute value by is uniformly convergent. It therefore follows that for t ∈ (0, 2π) the following equation holds = − (2)/2. Hence (2) = 2 /6.

Proof 8:
Let us discuss about another textbook proof, found in many books on complex analysis. We use the calculus of residues. Let ( ) = −2 . It can be easily seen that has poles at precisely the integers where the value of becomes infinite.
Now notice that at the pole at zero ( ) has residue − 2 /3, and that at a non-zero integer has residue 1/ 2 . Let N be a natural number and let be the square contour with vertices (±1 ± )( + 1/2). Now we know that by the calculus of residues the following identity holds: Notice that as → 0 as → ∞. Therefore we get ζ(2) = 2 /6.

Proof 10:
This comes from a note by Kortram 5 . Given an odd integer n = 2m + 1 a very important identity is that = ( ) where is a polynomial of degree . An important property to notice is that the zeros of ( ) are the values ( / ) (− ≤ ≤ ) and lim →0 ( ( ) ) = . We can then say that and therefore by our definition the following equality holds And the coefficients of 3 in RHS will be -6 − ∑ 1 2 ( ) =1 . Therefore we can write by equation (42) and (43)  .We can therefore say that using the earlier equations (43) the following identity holds:

Proof 11:
This proof again comes from Euler which uses the Weierstrass Factorization Theorem which is a direct generalization of the Fundamental Theorem of Algebra. We first state that theorem.
Let be an entire function and let { } be the nonzero zeros of . Suppose f has a zero at = 0 of order ≥ 0 (where order 0 means (0) ≠ 0). Then there exist a function and a sequence of integers { } such that the following equation holds: This. Let us assume = ( ), the sequence = 1 and the function ( ) = ( ). Therefore we can write as follows: Since have roots precisely at ∈ . Now if we substitute = we get the following equation: Now we put = − , we can write the above equation (52) Now if we use equation (54) and (55) (56) .Therefore we can use L'hospital's rule which yields ζ(2) = 2 /6.

Proof 12:
This is proof is due to Luigi Pace, Dept of Econ & Stats at Udine, Italy 6 .It was inspired by a 2003 note that solves the problem using a double integral on + 2 via Fubini's Theorem; a result that gives conditions under which it is possible to compute a double integral by using an iterated integral.
If n > 0 then we can say that if we do integration by parts gives we can therefore write using equation (64) and (63) that (2 )! 4 ! 2 2 = (2 − 1) −1 − 2 2 (65) Therefore we can rewrite expression (65) as: All the middle terms present in the sequence gives a value a zero, and we are left with only the first term and the last term. Hence we can write that Since 0 = 3 /24 it sufficient to show that lim →∞ (4 ! 2 ) (2 )! = 0 to prove our identity.

Proof 14:
This proof is due to Stark 8 . Consider a very important property of identity of the Fejer kernel. Before that we first give the definition of Fejer kernel. Fejér kernel is defined as Where ( ) = ∑ =− which is also known as the th order Dirichlet Kernel. It can also be expressed as Using equation (70) and (71) Now for 0 < x < π we can say that Therefore we can write the following equation: Taking limits as → ∞ gives us the following equation: Hence proved that ζ(2) = 2 /6.

Proof 15:
This is an exercise in Borwein & Borwein's Pi and the AGM 9 . We carefully square Gregory's formula which states that Let us now denote = ∑ . Now notice that the sum of the positive powers will be same as sum of the negative powers. Since Here a point to note is that all the terms with zero denominators are omitted which comes when = , , referred above in the equation (82) And so a ( 2 − )→ 0 as → ∞ as required.

Proof 16:
This is an exercise in Hua's textbook on number theory 10 . This depends on the formula for the number of representations of a positive integer as a sum of four squares. Let ( ) be the number of quadruples ( , , , ) of integers such that = 2 + 2 + 2 + 2 . Trivially (0) = 1 and it is well known that for n > 0. Let ( ) = ∑ ( ) =0 . Now notice that ( ) is asymptotic to the volume of the 4D ball of radius √ , i.e., R(N) ∼ ( 2 /2). 2 . But we know that Where Now θ(x) can be further written as: ) .Hence proved that ζ(2) = 2 /6.

Proof 17:
Let us now discuss about a proof given by Ivan 11 which is almost similar to that of James D. Harper's 12 simple proof. They used the Fubini theorem for integrals and McLaurin's series expansion for ℎ −1 .
To start with let's first state this basic identity as follows: Another equality that we shall use is: Therefore it follows that ζ(2) = 2 /6.

Proof 18:
This is a proof due to Josef Hofbauer 13 .We start off with the simple identity: Taking limit → ∞ and using lim →∞ ( / ) = for = 2 and = (2 + 1) /2 yields gives us the following equation: This as was done earlier many times leads to ζ(2) = 2 /6.

New Proof
We first take a note of the Taylor series expansion of (1 + ) which is as follows: (1 + ) = − Now if we divide the expression by we get Integrating both sides from -1 to 1 we get the following equality: . . Let us solve this integral by using complex variables. For our convenience we change the variable of integration as z. Therefore our only task remaining is to compute the integral Here the direction of the integration in the arc is anticlockwise. Now by using the calculus of residues we know that Since the analytic function has no poles inside the semicircle ( ) = , 0 ≤ ≤ , we can then write using equation (6) and (7) that Here the direction of the integration in the is clockwise. Now if we put = , we can rewrite the analytic function ( ) as  ( 2 )). = 2 + = ln 2 − ln 2 = 0 (112) Therefore from equation (11) and (17) Now from equation (4)