The Basel Problem

The Basel Problem

Abstract Because of its relation to the distribution of prime numbers, the Riemann zeta function ζ (s) is one of the most important functions in mathematics. The zeta function is defined by the following formula for any complex ∞ s number s with the real component greater than 1. ζ (s) = ∑n=1 1/n .Taking s=2, we see that ζ(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.

Keywords: Zeta Function, Pi, Basal Problem.

1. 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. The Basel problem asks for the sum of the series:

1 1 1 1 ∑∞ = + + + ⋯ (1) n=1 n2 12 22 32

휋2 It was Euler who found the exact sum to be . In this paper we discuss some of the notable proofs given by 6 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. The R.H.S. of (1) can be written as

1 1 1 1 1 3 1 ∑∞ = ∑∞ − ∑∞ = (1 − ) ∑∞ = ∑∞ (2) 푛=0 (2푛+1)2 n=1 n2 n=1 (2n)2 4 n=1 n2 4 n=1 n2

Proof 1:

Let us first take a proof that is due to Calabi, Beukers and Kock1 and also mentioned in Robin Chapman14. They first noticed the following equality:

1 1 1 푑푥푑푦 ∑∞ = ∫ ∫ (3) 푘=0 (2푘+1)2 0 0 1−푥2푦2

1−푦2 1−푥2 After that they made the substitution 푢 = 푡푎푛−1푥√ 푎푛푑 푣 = 푡푎푛−1푦√ . We now get the transformed 1−푥2 1−푦2 sin 푢 sin 푣 variables after substitution as 푥 = 푎푛푑 푦 = . Now notice the following important equation which cos 푣 cos 푢 connects 휁(2) and the left hand side of the equation (3).

3 1 1 1 휁(2) = ∑∞ − ∑∞ = ∑∞ (4) 4 푛=1 푛2 푚=1 2푚2 푘=0 (2푘+1)2

The Right hand side of the equation (3) can be rewritten after the substitution as

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1 1 푑푥푑푦 푑푢푑푣 휋 ∫ ∫ =∫ ∫ 퐽 ℎ푒푟푒 퐴 = {(푢, 푣): 푢 > 0, 푣 > 0, 푢 + 푣 < } (5) 0 0 1−푥2푦2 퐴 1−푢2푣2 2

Since the 퐽 the Jacobian matrix will be

2 휕(푥,푦) 푐표푠 푢/ 푐표푠 푣 푠푖푛 푢 푠푖푛 푣/ 푐표푠 푣 퐽 = = det | | = 1 − 푢2푣2 (6) 휕(푢,푣) 푠푖푛 푢 푠푖푛 푣/ 푐표푠2푢 푐표푠 푣/ 푐표푠 푢 1 휋 휋 휋2 휋2 Now as the area of 퐴 = ( ) ∗ ( ) ∗ ( ) = . We get from equation (1),(2) and (3) that 휁(2) = . 2 2 2 8 6

Proof 2:

We now derive a proof that comes from a note by Boo Rim Choe2 in the American Mathematical Monthly in 1987 and also mentioned in Robin Chapman14. We start by taking a note of the power series expansion of the inverse sine function.

1.3…(2푛−1) 푥2푛+1 푠푖푛−1(푥) = ∑∞ 푓표푟 |푥| ≤ 1 (7) 푛=0 2.4…(2푛) 2푛+1

Now let us put 푡 = 푠푖푛−1(푥). After the substitution we get

1.3…(2푛−1) sin (푡)2푛+1 푡 = ∑∞ 푓표푟 |푡| ≤ 휋/2 (8) 푛=0 2.4…(2푛) 2푛+1

2푛+1 To proceed further we first need to find the integration from 0 푡표 휋/2 of sin (푡) or 퐼2푛+1 = 휋/2 2푛+1 ∫0 (sin (푥)) 푑푥, we need to use the recursion formula which can be expressed as:

휋/2 2푛−1 퐼2푛+1 = 퐼2푛−1 − ∫0 [(sin (푥)) cos (푥)] 푐표푠푥. 푑푥 (9)

Now using integration by parts now we can derive that

2푛+1 퐼 = 퐼 (10a) 2푛 2푛+1 2푛−1

Therefore by recursion we can write that

2푛 2푛.2(푛−1) 2.4…(2푛) 퐼 = 퐼 = 퐼 = (10b) 2푛+1 2푛+1 2푛−1 (2푛+1)(2푛−1) 2푛−3 3.5…(2푛+1)

We therefore get that

휋/2 2.4…(2푛) 퐼 = ∫ (sin (푥))2푛+1 푑푥 = (11) 2푛+1 0 3.5…(2푛+1)

Now if we use equation (11) and (8) we get

휋2 휋⁄ 1 =∫ 2 푡. 푑푡 = ∑∞ (12) 8 0 푛=0 (2푛+1)2

휋2 The right hand side of the equality (12) is (3/4)휁(2). Therefore we get that 휁(2) = . 6

Proof 3:

We now take a proof from an article in the Mathematical Intelligencer by Apostol3 in 1983 and also mentioned in Robin Chapman14. We first start noting the following equation:

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1 1 1 = ∫ ∫ 푥푗−1푦푗−1푑푥푑푦 (13) 푗2 0 0

It is also to note that we get by the monotone convergence theorem

1 1 1 1 1 푑푥푑푦 ∑∞ = ∫ ∫ (∑∞ (푥푦)(푛−1)). 푑푥푑푦 = ∫ ∫ (14) 푘=1 푘2 0 0 푘=1 0 0 1−푥푦

Let us now introduce change of variables by changing the variables 푢 = (푥 + 푦)/2) 푎푛푑 푣 = (푥 − 푦)/2)). After the substitution of the variables we get the relationship between changed variables and actual variables as 푥 = 푢 – 푣 and = 푢 + 푣 . Now notice that the Left hand side of the equation (14) is 휁(2). We therefore can rewrite equation (14) in terms of the changed variables as:

푑푢푑푣 휁(2) = 2 ∬ (15) 푆 1−푢2+푣2

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 1 1 1 half of the square which can be itself be divided into two triangles with coordinates ((0, 0), ( , ) , ( , 0)) and 2 2 2 1 1 1 (( , ) , ( , 0) , (1,0)) respectively. We can therefore express 휁(2) as 2 2 2

1 푢 푑푢푑푣 1 1−푢 푑푢푑푣 ( ) 2 1 휁 2 = 4 ∫0 ∫0 2 2 + 4 ∫ ∫0 2 2 1−푢 +푣 2 1−푢 +푣

1 1 푢 1 1 1−푢 = 4 ∫2 푡푎푛−1 ( ) . 푑푢 + 4 ∫ 푡푎푛−1 ( ) . 푑푢 (16) 0 √1−푢2 √1−푢2 1/2 √1−푢2 √1−푢2

푢 Now we notice that 푡푎푛−1 ( ) = 푠푖푛−1(푢) . Let us this identity by again changing the variables by √1−푢2 1−푢 substituting 휃 = 푡푎푛−1( ). After substitution we get √1−푢2

2 푡푎푛2휃 = (1 − 푢)/(1 + 푢) and 푠푒푐2휃 = . (17) 1+푢

1 휋 1 It follows that 푢 = 2푐표푠2휃 − 1 = 푐표푠2휃 and therefore we get that 휃 = ( ) 푐표푠−1푢 = − 푠푖푛−1푢. We can 2 4 2 hence rewrite equation (16) in terms of the changed variables as

1 −1 −1 2 2 푠푖푛 푢 1 1 휋 푠푖푛 푢 −1 2 1/2 −1 −1 2 1 휋 휋 휁(2) = 4 2 푑푢 + 4 1 ( − ) . 푑푢 = [2(푠푖푛 푢) ] + [휋푠푖푛 푢 − (푠푖푛 푢) ] = + − ∫0 2 ∫ 2 0 1/2 √1−푢 2 √1−푢 4 2 18 2 휋2 휋2 휋2 휋2 − + = . (18) 4 6 36 6

Proof 4:

We now present some textbook proofs, found in many books on Fourier analysis to find the value of 휁(2). In this proof we are going to discuss next we will use the 퐿2 completeness of the trigonometric functions.

Suppose that 푯 is a Hilbert space with inner product. Let (푒푛) be an ortho-normal basis of H such that

〈푒푚, 푒푛〉 = 1 푖푓 푚 = 푛

0 푖푓 푚 ≠ 푛 (19)

Then we know that by Parseval's identity, we can write for every 푔 ∈ 푯,

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2 2 〈푔, 푔〉 = ||푔|| = ∑푛|〈푔, 푒푛〉| (20)

Let the Hilbert Space 푯 be 퐿2[0,1]. Let us now apply this theorem 푔(푥) = 푥. The Left hand side of the 1 1 1 equation (20) will be 〈푔, 푔〉 = and 〈푔, 푒 〉 = and 〈푔, 푒 〉 = for 푛 ≠ 0. Therefore we can re express 3 0 2 푛 2휋푖푛 equation (20) as:

1 1 1 = + ∑ (21) 3 4 풏∈풁,풏≠ퟎ 4휋2푛2

Which therefore gives us 휁(2) = 휋2/6.

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 푓(0) = 푓(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 푓 satisfies a Holder condition. • Fourier series converges everywhere if 푓 is of bounded variation. • Fourier series converges uniformly if 푓 is continuous and its Fourier coefficients are absolutely summable.

Then the Fourier series of 푓 converges to 푓 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 푓(푥) = 푥(1 − 푥) which then gives us the following equation:

1 cos(2휋푛푥) 푥(1 − 푥) = − ∑∞ (22) 6 푛=1 휋2푛2

If we put x=0 in the LHS of the equation (22) we get (2) = 휋2/6. We can also get the identity if we put x=1/2 which gives us:

휋2 (−1)푛 = − ∑∞ (23) 12 푛=1 푛2

Now as we know that

∞ 2 ζ (2) = ∑n=1 1/n (24)

Now if we subtract the equation (23) from the equation (24). We get

휋2 1 1 ζ (2) − = 2 ∗ (∑∞ ) = ∗ ∑∞ 1/n2 = ζ (2)/2 (25) 12 n=1 (2n)2 2 n=1

휋2 We therefore get ζ (2)/2 = which gives us 휁(2) = 휋2/6. 12

This is mentioned in Robin Chapman14.

Proof 6:

Let us now derive the original proof given by Euler. We use the infinite product

푥2 sin(휋푥) = 휋푥 ∏∞ (1 − ) (26a) 푛=1 푛2

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The Basel Problem for the sine function.

푥2 푥2 푥2 sin 휋푥 = 휋푥(1 − 푥2) (1 − ) (1 − ) (1 − ) … … (26b) 4 9 16

Now the RHS can be written as

1 1 1 1 1 1 푅퐻푆 = 휋푥 + 휋푥3 (1 + + + + ⋯ ) + 휋푥5 ( + +. . + +. . ) +.. (27a) 4 9 16 1.4 1.9 4.9

Whereas the LHS can be expanded in power series as:

(휋푥)3 (휋푥)5 퐿퐻푆 = 휋푥 − + (27b) 3! 5!

Now if we compare the coefficients of 푥3 in the MacLaurin series of both the LHS and RHS sides we get ζ(2) = 휋2/6.

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:

∞ 2 푓(푡) = ∑푛=1 cos 푛푡 /푛 (28)

First of all notice that the function is uniformly convergent on the real line. Now we know that sin 푛푡 can also be written as (푒푖푛푡 − 푒−푖푛푡)/2푖 . Therefore we can write as follows:

푒푖푡−푒푖(푁+1)푡 푒−푖푡−푒−푖(푁+1)푡 ∑푁 sin 푛푡 = ∑푁 (푒푖푛푡 − 푒−푖푛푡)/2푖 = − 푛=1 푛=1 2푖(1−푒푖푡) 2푖(1−푒−푖푡)

푒푖푡−푒푖(푁+1)푡 1−푒−푖푁푡 = − (29) 2푖(1−푒푖푡) 2푖(1−푒푖푡)

2 1 And so this sum is bounded above in absolute value by = . Hence these sums are uniformly |1−푒푖푡| sin 푡/2 sin 푛푡 bounded and by Dirichlet’s test the sum ∑푁 is uniformly convergent. It therefore follows that for t ∈ 푛=1 푛 (0, 2π) the following equation holds

푛푡 푒푖푛푡 푡−휋 푓′(푡) = − ∑∞ sin = −퐼푚 (∑∞ ) = 퐼푚(log(1 − 푒푖푡)) = 푎푟푔 (1 − 푒푖푡) = (30) 푛=1 푛 1 푛 2

휋 휋 푡−휋 휋2 Now notice that 푓(휋) − 푓(0) = ∫ 푓′(푡)푑푡 = ∫ 푑푡 = − (31) 0 0 2 4

(−1)푛 But 푓(0) = 휁(2) and 푓(휋) = ∑∞ = −휁(2)/2. Hence 휁(2) = 휋2/6. 푛=1 푛2

This is mentioned in Robin Chapman14.

Proof 8:

Let us discuss about another textbook proof, found in many books on complex analysis. We use the calculus of residues. Let 푔(푥) = 휋푧−2푐표푡 휋푧. It can be easily seen that 푔 has poles at precisely the integers where the value of 푐표푡 휋푧 becomes infinite.

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Now notice that at the pole at zero 푔(푥) has residue −휋2 /3, and that at a non-zero integer has residue 1/푛2.

Let N be a natural number and let 퐶푁 be the square contour with vertices (±1 ± 푖)(푁 + 1/2). Now we know that by the calculus of residues the following identity holds:

2 −휋 N 1 1 + 2 ∑n=1 = ∫ 푔(푧). 푑푧 = 퐼푛 (32) 3 n2 2휋푖 퐶푁

Now if 휋푧 = 푥 + 푖푦 the following equation holds:

푐표푠2푥+푠푖푛ℎ2푥 | cot πz |2 = (33) 푠푖푛2푥+푠푖푛ℎ2푥

Now if z lies on the vertical edges of 퐶푁 then the following equation holds

푠푖푛ℎ2푥 | cot πz |2 = < 1 (34) 1+푠푖푛ℎ2푥 and if z lies on the horizontal edges of 퐶푁 then the following equation holds

1 1+sinh2π(N+ ) 2 2 2 1 2 | cot πz | ≤ 1 = 푐표푡ℎ π(N + ) ≤ 푐표푡ℎ (π/2) (35) sinh2π(N+ ) 2 2

휋퐾 Hence | 푐표푡 휋푧| ≤ 퐾 = 푐표푡ℎ (휋/2) on 퐶푁 , and so |푓(푧)| ≤ 1 on 퐶푁 . This estimate shows that the (푁+ )2 2 following equation holds:

1 8휋퐾(푁+ ) 2 |퐼푛| ≤ 1 (36) 2휋(푁+ )2 2

2 Notice that as 퐼푛 → 0 as 푁 → ∞. Therefore we get ζ(2) = 휋 /6.

This is mentioned in Robin Chapman14.

Proof 9:

Let us discuss another proof that can be found as an exercise in Apostol4 and also mentioned in Robin Chapman14.We first start by taking the note of two important results

If 0 < 푥 < 휋/2 then the following 2 inequality holds:

• 푠푖푛 푥 < 푥 < 푡푎푛 푥 and so • 푐표푡2푥 < 푥−2 < 1 + 푐표푡2푥

We can therefore write that If n and N are natural numbers with 1 ≤ 푛 ≤ 푁, then

nπ (2n+1)2 nπ cot2 < < 1 + cot2 (37) 2N + 1 n2π2 2N + 1

Using the first inequality of equation (37) we get

π2 nπ 1 푁휋2 휋2 nπ ∑푁 cot2 < ∑푁 < + ∑N cot2 (38) (2n+1)2 푛=1 2N + 1 푛=1 푛2 (2n+1)2 (2n+1)2 푛=1 2N + 1

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2 N 2 nπ 휋 퐴푛 2 Now if 퐴푛 = ∑푛=1 cot , we can prove the identity ζ(2) = by simply showing that lim = . Now If 2N + 1 6 푁→∞ 푁2 3 푛휋 1 ≤ 푛 ≤ 푁 and 휃 = , then 푠푖푛(2푁 + 1)휃 = 0 . 2푁 + 1

Now by De-moivre’s theorem we know that 푠푖푛(2푁 + 1)휃 is the imaginary part of (cos θ + isin θ)2푁+1, and so

푠푖푛(2푁 + 1)휃 1 = ∑푁 (−1)푘퐶2푁+1푐표푠2(푁−푘)휃 푠푖푛2푘+1휃 = ∑푁 (−1)푘퐶2푁+1푐표푡2(푁−푘)휃 (39) 푠푖푛2푁+1휃 푠푖푛2푁+1휃 푘=0 2푁−푘 푘=0 2푁−푘

This can be written in the form

푁 2푁+1 푁−1 푓(푥) = (2푁 + 1)푥 − 퐶3 푥 +· · ·. (40)

2 푁 푘 2푁+1 2(푁−푘) Where 푓(푐표푡 휃) = ∑푘=0(−1) 퐶2푁−푘 푐표푡 휃.

2 Hence the roots of 푓(푥) = 0 are 푐표푡 (푛휋/(2푁 + 1)) where 1 ≤ 푛 ≤ 푁 and so 퐴푛 = 푁(2푁 − 1)/3. Thus 퐴 2 푛 → , as required. 푁2 3

Proof 10:

This comes from a note by Kortram5 and also mentioned in Robin Chapman14. Given an odd integer n = 2m +

1 a very important identity is that 푠푖푛 푛푥 = 퐹푛(푠푖푛 푥) where 퐹푛 is a polynomial of degree 푛. An important 퐹푛(푦) property to notice is that the zeros of 퐹푛(푦) are the values 푠푖푛(푗휋/푛) (−푚 ≤ 푗 ≤ 푚) and lim ( ) = 푛. 푦→0 푦 We can then say that

2 푚 푦 퐹푛(푦) = 푛푦 ∏푗=1(1 − 푗휋 ) (41) 푠푖푛2( ) 푛 and therefore by our definition the following equality holds

2 푚 푠푖푛 푥 sin 푛푥 = 푛푠푖푛푥 ∏푗=1(1 − 푗휋 ) (42) 푠푖푛2( ) 푛

Now let us compare the coefficients of 푥3 in the MacLaurin expansion of both sides LHS and RHS. Now the MacLaurin expansion of sin 푛푥 gives us:

푛3푥3 푛5푥5 sin 푛푥 = 푛푥 − + +.. (43) 3! 5!

3 푛 푚 1 And the coefficients of 푥 in RHS will be – − 푛 ∑푗=1 푗휋 . Therefore we can write by equation (42) and 6 푠푖푛2( ) 푛 3 푛 푛 푚 1 (43) that− =– − 푛 ∑푗=1 푗휋 .We can therefore say that using the earlier equations (43) the following 6 6 푠푖푛2( ) 푛 identity holds:

1 1 푚 1 = − ∑푗=1 푗휋 (44) 6푛2 6 푛2푠푖푛2( ) 푛

Let us fix an integer 푀 and let 푚 > 푀 without any loss of generality. Then we can say that

1 푚 1 1 푀 1 + ∑푗=푀+1 푗휋 = − ∑푗=1 푗휋 (45) 6푛2 푛2푠푖푛2( ) 6 푛2푠푖푛2( ) 푛 푛

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2 휋 and using the inequality 푠푖푛 푥 > 푥 for 0 < 푥 < , we get 휋 2

1 푀 1 1 푚 1 0 < − ∑푗=1 푗휋 < + ∑푗=푀+1 (46) 6 푛2푠푖푛2( ) 6푛2 4푗2 푛

Now notice that if we let n towards infinity we get

1 1 1 0 < − ∑푀 < ∑∞ (47) 6 푗=1 휋2푗2 푗=푀+1 4푗2

1 1 1 휋2 Hence we get easily get ∑∞ = → ∑∞ = as wanted. 푗=1 휋2푗2 6 푗=1 푗2 6

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 푓 be an entire function and let { 푎푛} be the nonzero zeros of 푓. Suppose f has a zero at 푧 = 0 of order 푚 ≥ 0 (where order 0 means 푓(0) ≠ 0). Then there exist a function 푔 and a sequence of integers {푝푛} such that the following equation holds:

∞ 푚 푓(푧) = 푧 exp (푔(푧)) ∏ 퐸푝푛 (푧/푎푛) 푛=1

Where 퐸푛(푦) = (1 − 푦) 푖푓 푛 = 0;

푦2 푦푛 퐸 (푦) = (1 − 푦)exp (푦 + + ⋯ + ) 푖푓 푛 = 1,2,3. . ; 푛 2 푛

This. Let us assume 푓 = 푠푖푛(휋푥), the sequence 푝푛 = 1 and the function 푔(푧) = 푙표푔(휋푧). Therefore we can write as follows:

푡 푡 푡 푡 sin 푡 = 푡 (1 − ) (1 + ) (1 − ) (1 + ) (48) 휋 휋 2휋 2휋

Since 푠푖푛 푡 have roots precisely at 푡 ∈ 푍. Now if we substitute 푡 = 휋푦 we get the following equation:

푦 푦 푦2 푦2 sin(휋푦) = 휋푦(1 − 푦)(1 + 푦) (1 − ) (1 + ) … = 휋푦(1 − 푦2) (1 − ) (1 − ) .. (49) 2 2 4 9

Taking logarithm of both sides of the equality

ln(sin(휋푦)) = 푙푛휋 + 푙푛푦 + ln(1 − 푦2) + [푙푛(4 − 푦2) − 푙푛4] + ⋯ (50)

Now if we differentiate both sides of the equality with the variable 푦 we get the following equation:

1 1 2푦 2푦 2푦 휋 cos(휋푦) ∗ ( ) = − − − − ⋯ (51) sin(휋푦) 푦 1−푦2 4−푦2 9−푦2

This gives us

1 1 1 1 1 1 + + + + ⋯ . = − 휋 cos(휋푦) ∗ ( ) (52) 푦 1−푦2 4−푦2 9−푦2 2푦2 2ysin(휋푦)

Now we put 푦 = −푖푥, we can write the above equation (52) as follows:

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The Basel Problem

1 1 1 1 1 + + + ⋯ = − + 휋 cos(−푖휋푥) ∗ ( ) (53) 1+푥2 4+푥2 9+푥2 2푥2 2ixsin(−푖휋푥)

Use Euler’s Formula we can write that

1 (푒푖푧+푒−푖푧) 2푖푧 푐표푠(푧) 2 푖(푒 +1) = 1 = (54) 푠푖푛(푧) (푒푖푧−푒−푖푧) 푒2푖푧−1 2

휋푐표푠(−푖휋푥) 휋 푖(푒2휋푥+1) 휋 푒2휋푥+1 휋 휋 = . = . = + (55) 2푖푥푠푖푛(−푖휋푥) 2푖푥 푒2휋푥−1 2푥 푒2휋푥−1 2푥 푥(푒2휋푥−1)

Now if we use equation (54) and (55) then we can write as follows:

1 1 1 1 1 1 휋 휋 + + + ⋯ = − + 휋 cos(−푖휋푥) ∗ ( ) = − + + = 1+푥2 4+푥2 9+푥2 2푥2 2ixsin(−푖휋푥) 2푥2 2푥 푥(푒2휋푥−1) 푒2휋푥(휋푥−1)+휋푥+1 (56) 2푥2(푒2휋푥−1)

1 0 LHS of the equation can be easily be transformed into ∑∞ by substituting푥 = 0. Notice that the form is in 푗=1 푗2 0 the Right hand side of the equation (56) .Therefore we can use L’hospital’s rule which yields ζ(2) = 휋2/6.

Proof 12:

This is proof is due to Luigi Pace, Dept of Econ & Stats at Udine, Italy6 .It was inspired by a 2003 note that 2 solves the problem using a double integral on 푅+ via Fubini’s Theorem; a result that gives conditions under which it is possible to compute a double integral by using an iterated integral.

We know define two variables 푋1, 푋2 ∶ 푅 → 푅+ random variables having a probability density function:

휌푋푖: 푅+ → [0, 1] ]. Let us now define another variable 푌 = 푋1/푋2 . We claim that Probability density function ∞ for 푌 is 휌 = 푡 휌 (푡휇)휌 (푡). 푑푡. To prove that joint density: 푌 ∫0 푋1 푋2

∞ 푏푡2 ∞ 푏 Pr(푎 ≤ 푌 ≤ 푏) = ∫ ∫ 휌푋 (푡1)휌푋 (푡2). 푑푡1 푑푡2 = ∫ ∫ 푡2휌푋 (푡2푢)휌푋 (푡2). 푑푢 푑푡2 = 0 푎푡2 1 2 0 푎 1 2 푏 ∞ 푡 휌 (푡 푢)휌 (푡 ). 푑푢 푑푡 (57) ∫푎 ∫0 2 푋1 2 푋2 2 2

2 For the proof let us assign half-Cauchy distribution to 푋 , 푋 independently: 휌 (푡) = . Now we shall use 1 2 푋푖 휋(1+푡)2 this in the formula for 휌푌(푢) obtained above:

4 ∞ 푡 1 2 1+푡2푢2 4 ln 푢 휌 (푢) = ∫ . 푑푡 = [ln ( )]∞ = (58) 푌 휋2 0 1+푡2푢2 1+푡2 휋2(푢2−1) 1+푡2 0 휋2 푢2−1

Integrating 휌푌(푢) from 0 to 1 we get 푃푟(0 ≤ 푌 ≤ 1). Therefore we can say that

1 1 4 ln 푢 푃푟(0 ≤ 푌 ≤ 1) = ∫ 휌 (푢). 푑푢 = ∫ . 푑푢 (59) 0 푌 0 휋2 푢2−1

But note that 푃푟(0 ≤ 푌 ≤ 1) = ½ , Therefore we can say using equations (59) that

1 ln 푢 휋2 ∫ . 푑푢 = . (60) 0 푢2−1 8

1 Let us simplify the integral by using = 1 + 푢2 + 푢4 + ⋯. We can therefore get as follows: 1−푢2

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휋2 1 ln 푢 1 ln푢 1 3 = − ∫ . 푑푢 = ∑∞ ∫ . 푑푢 = ∑∞ = ζ(2) (61) 8 0 1−푢2 푛=0 0 푢2푛 푛=0 (2푛+1)2 4

Which gives us as required ζ(2) = 휋2/6.

Proof 13:

This proof is due to Matsuoka7 and also mentioned in Robin Chapman14.Consider the two integrals

휋 휋 2 2푛 2 2 2푛 퐼푛 = ∫0 푐표푠 푥. 푑푥 푎푛푑 퐽푛 = ∫0 푥 푐표푠 푥. 푑푥 (62)

Now as we discussed earlier

1.3.5….(2푛−1) 휋 (2푛)!휋 퐼 = = (63) 푛 2.4.6…2푛 2 4푛푛!22

If n > 0 then we can say that if we do integration by parts gives

휋 휋 휋 휋 2푛 2 2 2푛−1 2 2푛−1 2 2 2( 2푛 퐼푛 = [푥푐표푠 푥]0 + 2푛 ∫0 푥푠푖푛푥푐표푠 푥 = 푛[푥 푠푖푛푥푐표푠 푥]0 − 푛 ∫0 푥 푐표푠 푥 − (2푛 − 2 2푛−2 2 1)푠푖푛 푥푐표푠 푥). 푑푥 = 푛(2푛 − 1)퐽푛−1 − 2푛 퐽푛 (64) we can therefore write using equation (64) and (63) that

(2푛)!휋 = 푛(2푛 − 1)퐽 − 2푛2퐽 (65) 4푛푛!22 푛−1 푛

Therefore we can rewrite expression (65) as:

휋 4푛−1[(푛−1)!]2 (4푛푛!2) = 퐽 − 퐽 (66a) 4푛2 (2푛−2)! 푛−1 (2푛)! 푛

Now summing up both the sides of the equation (66) we get:

휋 1 4푛−1[(푛−1)!]2 (4푛푛!2) 42−1[(2−1)!]2 (411!2) (4푛푁!2) ∑푁 = ∑푁 퐽 − ∑푁 퐽 = 퐽 + [ 퐽 − 퐽 ] +. . − 퐽 4 푛=1 푛2 푛=1 (2(푛−1))! 푛−1 푛=1 (2푛)! 푛 0 (2(2−1))! 2−1 (2)! 1 (2푁)! 푁 (66b)

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

휋 1 (4푛푁!2) ∑푁 = 퐽 − 퐽 (67) 4 푛=1 푛2 0 (2푁)! 푁

푛 2 3 (4 푁! ) Since 퐽0 = 휋 /24 it sufficient to show that lim = 0 to prove our identity. 푁→∞ (2푁)!

휋 휋 Notice that the inequality 푥 < 푠푖푛 푥 for 0 < 푥 < gives 2 2

2 휋 휋 휋 휋 휋2 휋2 휋2 .퐼푁 퐽 < ∫2 푠푖푛2푥. 푐표푠2푛푥. 푥. 푑푥 = (∫2 푐표푠2푛푥. 푑푥 − ∫2 푐표푠2푛+2푥. 푑푥 ) = (퐼 − 퐼 ) = 8 (68) 푁 4 0 4 0 0 4 푁 푁+1 푁+1

Thus we can write

(4푛푁!2) 휋3 0 < 퐽 < (69) (2푁)! 푁 16(푁+1)

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Hence proved that ζ(2) = 휋2/6.

Proof 14:

This proof is due to Stark8 and also mentioned in Robin Chapman14. 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

1 퐹 (푥) = ∑푛−1 퐷 (푥) (70) 푛 푛 푘=0 푘

푘 푖푠푥 th Where 퐷푘(푥) = ∑푠=−푘 푒 which is also known as the 푘 order Dirichlet Kernel. It can also be expressed as

|푗| 퐹 (푥) = ∑ (1 − )푒푖푗푥 = ∑푛 (n − |j|)eijx (71) 푛 |푗|≤푛−1 푛 푗=−푛

Using equation (70) and (71) we get

nx sin 2 2 n ikx n ( x ) = ∑ (n − |k|)e = n + 2 ∑ (n − k) cos kx (72) sin k=−n k=1 2

Hence we can write

nx sin 2 2 k 2 흅 2 2 nπ n π nπ n 1−(−1) nπ 푰 = ∫ x( x ) . dx = + 2 ∑ (n − k) ∫ xcos kx. dx = − 2 ∑ (n − k) = − 풏 ퟎ sin 2 k=1 0 2 k=1 k2 2 2 1 1 4n ∑ + 4 ∑ (73) 1≤k≤n,k is a odd number k2 1≤k≤n,k is a odd number k

If we assume that 푛 = 2푁 with 푁 an integer then by equation (73) we can say that

휋 푥 푥 휋2 1 푙표𝑔푁 ∫ (sin 푁푥/ sin )2 . 푑푥 = − ∑푁−1 + 푂( ) (74) 0 8푁 2 8 푟=0 (2푟+1)2 푁

Now for 0 < x < π we can say that

푠푖푛 푥/2 > 푥/휋 (75)

Therefore we can write the following equation:

휋 푥 푥 휋2 휋 푑푥 휋2 푁휋 푑푦 ∫ (sin 푁푥/ sin )2 . 푑푥 < ∫ 푠푖푛2푁푥 = ∫ 푠푖푛2푦 = 푂(푙표푔푁/푁) (76) 0 8푁 2 8푁 0 푥 8푁 0 푦

Taking limits as 푁 → ∞ gives us the following equation:

휋2 1 3 = ∑∞ = ζ(2) (77) 8 푟=0 (2푟+1)2 4

Hence proved that ζ(2) = 휋2/6.

Proof 15:

This is an exercise in Borwein & Borwein’s Pi and the AGM9 and also mentioned in Robin Chapman14. We carefully square Gregory’s formula which states that

휋 1 1 1 (−1)푛 = 1 − + − + ⋯ . = ∑∞ (78) 4 3 5 7 푛=0 2푛+1

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(−1)푛 Let us now denote 푎 = ∑푛=푁 . Now notice that the sum of the positive powers will be same as sum of 푛 푛=−푁 2푛+1 the negative powers. Since

(−1)푛 (−1)푛 (−1)푙+1 (−1)푙 (−1)푙 ∑−1 = ∑∞ = ∑∞ = − ∑∞ = ∑∞ (79) 푛=−∞ 2푛+1 푛=1 −2푛+1 푙=0 −2(푙+1)+1 푙=0 −2푙−1 푙=0 2푙+1

푛 푛 푛=∞ (−1) ∞ (−1) 푛=푁 1 Therefore lim 푎푛 = ∑푛=−∞ = 2 ∗ ∑푛=0 = 휋/2 . Let us denote 푏푛 = ∑푛=−푁 . Now to prove the 푁→∞ 2푛+1 2푛+1 (2푛+1)2 휋2 2 identity we require that lim 푏푛 = , so we have to show that lim (푎푛 − 푏푛) = 0. We now use the 푁→∞ 4 푁→∞ following identity:

1 1 1 1 = ( − ) 푖푓 푚 ≠ 푛 (80) (2푛+1)(2푚+1) 2(푚−푛) 2푛+1 2푚+1

Now if we square 푎푛 we get

1 1 푎 2 = ∑푛=푁 + ∑푁 ∑푁 (81) 푛 푛=−푁 (2푛+1)2 푛=−푁 푚=−푁 (2푛+1)(2푚+1)

(−1)푚+푛 1 1 (−1)푁푐 and therefore we can say that (푎 2 − 푏 ) = ∑푁 ∑푁 ∗ ( − ) = ∑푁 푛,푁 (82) 푛 푛 푛=−푁 푚=−푁 2(푚−푛) 2푛+1 2푚+1 푛=−푁 2푛+1

Here a point to note is that all the terms with zero denominators are omitted which comes when 푚 = 푛 ,

푐푛,푁 referred above in the equation (82) is defined as

(−1)푚 푐 = ∑푁 (83) 푛,푁 푛=−푁 (푚−푛)

(−1)푗 It is easy to see that 푐 = −푐 and so 푐 = 0. If n > 0 then 푐 = ∑푁+푛 and so |푐 | ≤ 1/(푁 − −푛,푁 푛,푁 0,푁 푛,푁 푗=푁−푛+1 푗 푛,푁 푛 + 1). Thus we can write that

1 1 1 2 1 1 2 (푎 2 − 푏 ) = ∑푁 ( − ) = ∑푁 ( + ) + ∑푁 ( + 푛 푛 푛=1 (2푁−1)(푁−푛+1) (2푁+1)(푁−푛+1) 푛=1 2푁+1 2푛−1 (푁−푛+1) 푛=1 2푁+3 2푛+1 1 1 ) ≤ (2 + 4 log(2푁 + 1) + 2 + 2log (푁 + 1)) (84) (푁−푛+1) 2푁+1

2 And so a (푎푛 − 푏푛)→ 0 as 푁 → ∞ as required.

Proof 16:

This is an exercise in Hua’s textbook on number theory10 and also mentioned in Robin Chapman14. This depends on the formula for the number of representations of a positive integer as a sum of four squares. Let 푟(푛) be the number of quadruples (푥, 푦, 푧, 푡) of integers such that = 푥2 + 푦2 + 푧2 + 푡2 . Trivially 푟(0) = 1 and it is well known that

푟(푛) = 8 ∑푚|푛,푚 푖푠 푛표푡 푑푖푣푖푠푖푏푙푒 푏푦 4 푚 (85)

푁 for n > 0. Let 푅(푁) = ∑푛=0 푟(푛) . Now notice that 푅(푁) is asymptotic to the volume of the 4D ball of radius √푁, i.e., R(N) ∼ (휋2/2). 푁2 . But we know that

N R(N) = 1 + 8 ∑N ∑ m = 1 + 8 ∑ m ⌊ ⌋ = 1 + 8(θ(n) − n=1 푚|푛,푚 푖푠 푛표푡 푑푖푣푖푠푖푏푙푒 푏푦 4 푚≤푁,푚 푖푠 푛표푡 푑푖푣푖푠푖푏푙푒 푏푦 4 m 4θ(N/4)) (86)

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x 1 x x 2 Where θ(x) = ∑ m ⌊ ⌋. But we know that θ(x) = ∑ 푚 = ∑ ∑⌊푥/푟⌋ 푚 = . ∑ (⌊ ⌋ + ⌊ ⌋ ) m≤x m 푚푟≤푥 푟≤푥 푚=1 2 r≤x r r

Now θ(x) can be further written as:

x 2 x2 1 θ(x) = ∑ (( ) + O(x/r)) = (ζ(2) + O ( )) + O(xlogx ) (87) r≤x r 2 x

푁2 as x → ∞. Hence it can be derived that 푅(푁)~ (휋2/2). 푁2~4ζ(2)(푁2 − ) .Hence proved that ζ(2) = 휋2/6. 4

Proof 17:

Let us now discuss about a proof given by Ivan11 which is almost similar to that of James D. Harper’s12 simple proof. They used the Fubini theorem for integrals and McLaurin’s series expansion for 푡푎푛ℎ−1.

To start with let’s first state this basic identity as follows:

1 1+푦 푦2푛+1 log ( ) = ∑∞ 푓표푟 |푦| < 1 (88) 2 1−푦 푛=0 2푛+1

Another equality that we shall use is:

1 1 1 1 1 1 ∫ ∫ . 푑푦. 푑푥 = ∫ ∫ . 푑푥. 푑푦 (89) −1 −1 1+2푥푦+푦2 −1 −1 1+2푥푦+푦2

Now note that

푥+푦 arctan 2 1 1 1 1 √1−푥2 1 1 휋 휋 ∫ ∫ . 푑푦. 푑푥 = ∫ [ ]−1. 푑푥 = ∫ 푑푥 = (90) −1 −1 1+2푥푦+푦2 −1 √1−푥2 −1 2√1−푥2 2

The right hand side of the equation can written as follows:

1+푦 1 1 1 1 log (1+2푥푦+푦2) 1 푙표𝑔 1 푦2푛+1 1 ∫ ∫ . 푑푥. 푑푦 = ∫ [ ]1 . 푑푦 = ∫ 1−푦 . 푑푦 = 2 ∫ ∑∞ = 4 ∑∞ = 4 ∗ −1 −1 1+2푥푦+푦2 −1 2푦 −1 −1 푦 −1 푛=0 2푛+1 푛=0 (2푛+1)2 3 ( ζ(2)) = 3ζ(2) (91) 4

Therefore it follows that ζ(2) = 휋2/6.

Proof 18:

This is a proof due to Josef Hofbauer13 .We start off with the simple identity:

1 1 1 1 1 1 1 1 = 푥 푥 = [ 푥 + 푥 ] = [ 푥 + 휋+푥 ] (92) 푠푖푛2푥 4 푠푖푛2( )푐표푠2( ) 4 푠푖푛2( ) 푐표푠2( ) 4 푠푖푛2( ) 푠푖푛2( ) 2 2 2 2 2 2

1 Now if we put 푥 = 휋/2. The 퐿퐻푆 = = 1 . And the RHS will be 푠푖푛2휋/2

1 1 1 1 1 1 1 1 RHS = ( ) [ 휋 + 3휋 ] = ( ) [ 휋 + 3휋 + 5휋 + 7휋 ] (93) 4 푠푖푛2( ) 푠푖푛2( ) 16 푠푖푛2( ) 푠푖푛2( ) 푠푖푛2( ) 푠푖푛2( ) 4 4 8 8 8 8

Now notice that the expression can be expanded into the form mentioned below:

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1 2n−1 1 2 2n−1−1 1 RHS = [∑k=0 2푘+1 ] = [∑k=0 2푘+1 ] (94) 4n 푠푖푛2( )휋 4n 푠푖푛2( )휋 2n+1 2n+1

Taking limit 푛 → ∞ and using lim 푁 푠푖푛(푥/푁) = 푥 for 푁 = 2푛 and 푥 = (2푘 + 1)휋/2 yields gives us the 푁→∞ following equation:

8 1 = ∑∞ 1/(2푘 + 1)2 (95) 휋2 푘=0

This as was done earlier many times leads to ζ(2) = 휋2/6.

2. Conclusion None of these proofs is original; most are well known, though some of them might not be very familiar. Many of the proofs mentioned in the paper were discovered in a survey article by Robin Chapman14. Another Proof of the Basel Problem is recently found by author15-16. In this paper we discuss some of the notable proofs given by mathematicians to the basal problem. Most of the methods used in this paper are very well known whereas some can be found as proofs of problems present in textbooks.

3. References 1. F. Beukers; E. Calabi; J.A.C. Kolk, Sums of generalized harmonic series and volumes, Nieuw Arch. Wisk. No.11(1993) pp. 217- 224. 2 2. Boo Rim Choe; An Elementary Proof of ∑∞ 1 = 휋 ,The American Mathematical Monthly, Vol. 94, No. 7. (1987), pp. 662-663. n=1 n2 6 3. T. Apostol, A proof that Euler missed: Evaluating ζ(2) the easy way, Math. Intelligencer 5 (1983) pp.59–60 4. T. Apostol ,Mathematical Analysis (Addison-Wesley, 1974). 2 5. R. A. Kortram, "Simple proof for ∑∞ 1 = 휋 and 푠푖푛(푥) = 푥훱푘(1 − 푥2/푘2휋2)" Mathematics Magazine, 69, . (1996),pp. 122- k=1 k2 6 125 2 6. L. Pace, Probabilistically Proving that ζ(2) = 휋 , Amer. Math Monthly, 118 (2011) 641-643 6 2 7. Y. Matsuoka, An elementary proof of the formula ∑ 1 = 휋 , Amer. Math. Monthly 68 (1961) 485– 487 k2 6 2 8. E. L. Stark, "Another Proof of the Formula ∑ 1 = 휋 " The. American Mathematical Monthly, 76, 552-553 (1969). k2 6 9. J. M. Borwein and P. B. Borwein, ”Pi and the AGM” , Canadian Mathematical Society Series of Monographs and Advanced Texts, no. 4, John Wiley & Sons, New York, 1998. 10. L.-K. Hua, “Introduction to Number Theory”, Springer,(1982) 11. Mircea Ivan ,“A simple solution to Basel problem”, General Mathematics Vol. 16, No. 4 (2008), pp. 111–113 2 12. James D. Harper, ”A simple proof of 1 + 1/22 + 1/32 + ··· = 휋 ”, The American Mathematical Monthly 109(6) (Jun. - Jul., 6 2003) 540–541. 2 13. Josef Hofbauer, “A simple proof of 1 + 1/22 + 1/32 + ··· = 휋 and related identities”, The American Mathematical Monthly 6 109(2) (Feb., 2002) 196–200. 14. Robin Chapman, Evaluating ζ(2), Preprint, http: //www.maths.ex.ac.uk/∼rjc/etc/zeta2.pdf 15. Ghosh, S. (2021, February 24). Another Proof of Basel Problem. https://doi.org/10.13140/RG.2.2.29833.06249/2 16. Ghosh, Sourangshu. (2020). Solution to the Basel Problem using Calculus of Residues. 10.13140/RG.2.2.29833.06249/2.

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