The Connection between the Basel Problem and a Special Integral ()
1. Introduction
Basel problem asks for the precise value of the progression
. It was first posed by Pietro Mengoli in 1644 and solved by Leonhard Euler in 1735 [1] . The value is known as
.
There are more general results [2] about the progression,
Let
, it becomes the
.
Moreover, using Fourier expansion of
,
we will get
(1.1)
In the end of Section 5, we give another proof of (1.1) by using the relationship of two special integrals which are introduced in Sections 3 and 4. Also, inspired by this, in Section 6, we discuss about Bernoulli numbers and Genocchi numbers. We obtain some properties of Bernoulli numbers and Bernoulli polynomials.
2. Basic Properties
The convergence of the infinite series 
is obvious. We can use various methods to prove it. Especially, when we consider Riemann-Zeta function, 
, the progression diverges when
and converges when
. Also, we can use the estimate of the partial sum of the series.
when
. Or we can use the Cauchy principle. In fact, for
,
thus
when
. Then, the progression converges.
3. Calculation of 
There are various proofs of the Basel problem and Robin Chapman wrote a survey [3] about these. Some are elementary and some will use advanced mathematics such as Fourier analysis, complex analysis or multivariable calculus. Here we review the method of Jiaqiang Mei [4] , which is rather elementary and easy to understand. There is also an elementary proof on the Wiki [1] .
Repeated use of the equation
we get
(1.2)
Note that
we may rewrite the Equation (1.2) as
where
Using the inequality
we get the estimation
Let
, we obtain the following equation
The above progression is uniformly convergent in any closed interval not containing 
and can be written as
Especially, we have
Therefore,
4. As a Special Case of Power Series
For the power series
, we calculate the domain of convergence. Since
the radius of convergence equals
. If
, the power series becomes the progression 
which is convergent. If
, then the power series becomes 
which is also convergent. Therefore, the domain of convergence is
.
Suppose
. We can do the derivation item by item in the inteval
. That is,
Multiply both sides by
,
Derivate both sides again, we get
Thus,
We obtain a second order ordinary differential equation
(1.3)
If we set
, the Equation (1.3) is converted to a first order equation,
Multiplie both side by
,
Let
, we have
Then
Using the initial conditions
, 
, we have
Then, if
,
That is
Note that
. Then,
Particularly,
Therefore, for the improper integral
, we know its value is equal to
.
5. From the Special Integral to the Basel Problem
In this section, we will calculate the special integral arised in the last section, i.e.
For
,
Thus
Let
, 
,
. Obviously, 
, where 
is Lebesgue measure.
For simplicity, we denote 
by
.
By Minkowski inequality [5] , we have
(1.4)
Then,
We will prove the equality holds in our case. First we have the following lemma.
Lemma 1. 
if and only if there is a real valued function h that is nonnegative a.e. such that when both f and g are not 0 then 
a.e.
Proof. Please refer to [6] .
Lemma 1 can be generalized to infinite summation case.
Lemma 2. 
if and only if there is a real valued function f and a series of real valued functions 
which have the same signs such that 
a.e.
Proof. First, by induction, the lemma holds for finite sum. That is
Let
. Then,
Since the measure of 
is zero, we have
Combine Equation (1.4), we complete the proof.
On the other hand, we observe that, for 
Then we get
We give a remark about the second equality of the above equation. It can be infered by Fubini theorem or Tonelli’s theorem [7] .
Infact,
where 
is the counting measure on
, and 
is the Lebesgue measure on
. Obviously, they are both 
-finite measures. And since 
is non-negative, by Tonelli’s theorem,
There are other ways to get this relationship from this special integral to
. First, recall the lemma established by James P. Lesko and Wendy D. Smith [2] .
Lemma 3. For
, 
and
, we have
(1.5)
Especially, when 
and
, (1.5) yields
By this lemma,
Then by monotone convergence theorem, it equals to
. See [3] .
Or, we can do it in this way,
6. Relationship between Two Special Integrals
We will use a result from ([8] , Exer 20).
Lemma 4. Assume that the function 
is monotone on the interval
. It need not be bounded at the points
,
; we assume however that the improper integral 
exists. Under these conditions,
Then for our case, the integral 
exists and satisfies the conditions. In fact, let
, 
then
Let
, then
Since
, we have 
for
. So 
is monotone on the interval
. Then by Lemma 4, we have
If we consider the improper integral
. Let
,
. Then
Let
, then
. Since
. Thus,
. So 
is also monotone on the interval
. By Lemma 4, we have
Next, we deduce the following equation and give another discription of (1.1),
(1.6)
Lemma 5. Let
, 
, then
.
Proof. Let
, it is easy to note that its derivate 
equals to zero and
.
By changing variables,
Then
Here, note that
. By Lemma 5,
For the integral
, we have a relevant result.
Lemma 6. Let
, 
, then
.
Proof.
then,
Hence,
The second equality holds, because
Observe that
Thus
Applying the same argument in beginning of this section, we get
and
Remark. It must be very interesting if we could calculate the integral 
not using the progression
.
Remark. Similarly, we can prove that
.
7. Bernoulli Numbers and Bernoulli Polynomials
Recall some facts of Bernoulli numbers, and for more information, please refer to [4] [9] -[12] .
The Bernoulli numbers 
are defined by the power series expansion
Then
Thus we get a recursion formula for the Bernoulli numbers, namely
We get
. From the identity
the function 
is an even function. Hence it has only even terms in its power series expansion.
Sometimes, people prefer to use 
to denote
, then
We have various ways to get the important equation.
Lemma 7.
Proof. By replacing 
by 
in the identity
we get
Then by taking the logarithmic derivative of the product expansion for the sine,
we get the expansion of
.
Comparing the coefficents of
, we get (7).
Another way is to take the logarithmic derivative of the identity [4]
which yields
Then we get (7) by comparing with
Proposition 1.
Proof.
where
It is easy to prove that 
by induction. Therefore,
Let us consider the expansion
where 
are Genocchi numbers. Then
Thus, 
and
, which infers that
. For
,
That is,
(1.7)
Note that
Taking expansion,
we have
(1.8)
This give a quick way to compute Bernoulli numbers since in (1.7) we have
.
If let 
denote
, then
Proposition 2.
Proof. By changing variables,
Note that
then
Set
, then
. By induction, we prove that
. Therefore,
Together with (1.6), we have
Remark. Since 
and
, the proposition can also be written as
Bernoulli polynomials 
are defined by the formula
The functions 
are polynomials in 
and
Similarly, we define 
by the formula
(1.9)
The functions 
are polynomials in
. In fact,
Comparing the coefficients of
, we have
(1.10)
and
. Using (1.7), for
, we have
On the other hand, by definition,
Do the addition,
Comparing the coefficients of
, we have
Let 
and summation these equations, we get
From the equation
, we infer that
.
Therefore,
If
,
If
,
Whether 
is odd or even, we always have the following trivial identity.
By differentiating on 
at both sides of (1.9), we also have
But being different from
, we have
Proposition 3. i)
.
ii)
.
iii)
.
iv)
.
Proof. i)
thus,
ii)
By comparing the coeffients of
, we get
iii)
By comparing the coeffients of
, we get
iv) by ii) and iii).
Remark. 1) Especially, we have
, since
.
2) Let 
or 1 in iv), we have
. Thus,
.
3) Let 
in iii), we will get
.
Equation (1.8) can also be deduced in the following way. Using
we obtain
(1.11)
Similarly,
This infers that
By substituting (1.11) in the above formula, we obtainde
Acknowledgements
We express our gratitude to David Harvey who point out that the numbers 
in our manuscript (here is
) are essentially the Genocchi numbers, see [13] .
The work is partially supported by National Natural Science Foundation of China (NSFC), Tianyuan fund for Mathematics, No. 11126046, and the University Science Research Project of Jiangsu Province (13KJB110029).