November 10, 2016 Some integrals related to the Basel problem
Khristo N. Boyadzhiev Department of Mathematics and Statistics, Ohio Northern University, Ada, OH 45810, USA [email protected]
Abstract. We evaluate several arctangent and logarithmic integrals depending on a parameter. This provides a closed form summation of certain series and also integral and series representation of some classical constants.
Key words and phrases, Basel problem, arctan integrals, dilogarithm, trilogarithm, Catalan constant.
2010 Mathematics Subject Classification. Primary 11M06; 30B30; 40A25
1. Introduction
The famous Basel problem posed by Pietro Mengoli in 1644 and solved by Euler in 1735 asked for a closed form evaluation of the series
11 1 ... 2322
(see [7], [9], [10]). Euler proved that
11 2 (1) 1 ... . 222 3 6
In the meantime, trying to evaluate this series, Leibniz discovered the representation
1
1 log(1t ) 1 1 dt 1 ... , 22 0 t 23 but was unable to find the numerical value of the integral (see comments in [7]). How to relate this integral to 2 /6 is discussed in [7] and [10]; it is shown that by using the complex logarithm one can solve the Basel problem. There exist, however, other integrals which can be used to quickly prove (1) without complex numbers. Possibly, the best example is the integral
1 arcsin x 1 (2) dxLi ( ) Li ( ) 2 22 0 1 x 2 for | | 1 . Here
n Li ( ) 2 2 n1 n is the dilogarithm [12]. Setting 1 in (2), the left hand side becomes
2 1 2 1 arcsin x 280 while the RHS is
1 3 3 1 1 Li (1) Li ( 1) Li (1) (1 ...) 22 2 4 2 4 222 3 and (1) follows immediately. This proof was recently published by Habib Bin Muzaffar in [13]. It is possibly one of the best solutions to the Basel problem. Euler also used the arcsin function in one of his proofs. Euler’s approach is explained on pp. 85-86 in [15].
In this short paper we follow the idea from [13] and consider some other integrals that can be associated to the Basel problem, either solving it, or leading to similar results. In the process we evaluate a number of integrals from the tables [8] and [14].
We start with three arctangent integrals in Section 2, and then we also discuss two logarithmic integrals. Among other things, in section 2 we find the curious representation (equation (8) below)
n1 3 1 1 1 1 ( 1) 192 1 ... 1 ... . n1 n3 2 n 1 3 2 n 1 4
2
In Section 3 we focus on several integral and series representation of some classical constants (see equations (15) and (17) below). In particular, we list two integral representations of (3) and evaluate one special arctangent integral (see (16) and (19) below).
2. Special integrals with arctangents and logarithms
At the end of [13] it was mentioned that instead of arcsin(x ) in (2) one could use arctan(x ) . However, as we shall see here, integrals with arctangents are not so simple.
Two natural candidates for the described method are clear
arctanxx1 arctan (3) d x, and dx . 22 0011xx
When 1, these integrals evaluate to a multiple of 2 and in order to prove (1) we need to evaluate them also as a multiple of the series Li2 (1) . Here is how the fist one works.
Proposition 1. For any 01 ,
arctanx 1 (4) 2dx log log Li ( ) Li ( ) . 2 22 0 11x
It easy to see (by using limits) that the RHS extends to 0 and 1. The function
1 log log becomes zero for 0 and 1. With 1 we compute immediately 1
2 3 (arctanx )2 Li (1) 0 422
2 and hence Li (1) . 6 2
Proof. Let J() be the LHS in (4). By differentiation (for 01 )
x1 2 x 2 2 x J( ) 2 dx dx 2 2 2 2 2 2 2 00(1x )(1 x ) 1 1 x 1 x
1 1x2 2log 2log 2 2 2 . 1 1 x 0 1
3
Thus, since J() is defined for 0
2logt J() dt . 2 0 1t
Integrating by parts we find
1 log(1 tt ) log(1 ) J( ) log log dt 1 0 tt
1 log log Li ( ) Li ( ) . 1 22
This integral was recently evaluated in [2]. It is missing from the popular table [8], but appears in [14] as entry 2.7.4 (12). However, it appears there in a different form
arctanx 2 1 1 2dx log2 (1 ) Li Li (1 ) 2 22 0 1x 3 2 1 which is less helpful for proving (1) .
Now we look at the second integral in (3). Although it does not lead directly to the proof of (1), it provides the closed form evaluation of one interesting series.
Proposition 2. For every | | 1 we have
1 arctanx 21n (5) 2dx log 2 H , 2 n 0 1xnn0 2 1 where
1 1 ( 1)n1 H1 ... , n 1,2,...; H 0 , n 23 n 0 are the skew-harmonic numbers (see [4]).
In particular, for 1 we find from (5)
2 1 (6) log 2 Hn . 16n0 2n 1
Note that Hn are the partial sums in the expansion of log 2 and the series in (6) is alternating.
Proof. Using the Taylor series for arctan(x ) we write
4
11arctanx ( 1)n2 n 1 x 2 n 1 dx dx 22 001xn0 2 n 1 1 x
( 1)n 2 n 11 x 2 n 1 dx . 2 n0 2nx 10 1
Entry 3.241(1) in [8] says that
1 x21n 1 dx( n 1) , 2 0 12 x where ()x is the incomplete beta function (see 8.370, pp. 947-948 in [8] and also [5]). According to equation 8.375 (2) in [8] we have
n (nH 1) ( 1) log 2 n .
Putting all these pieces together we arrive at (5). The proof is completed.
It would be interesting to see what happens when in Proposition 2 we replace arctanx by (arctanx )2 , which leads to 3 on the LHS.
With the notation
1 1 1 h1 ... ( n 1,2...), h 0 n 3 5 2n 1 0 we have
Proposition 3. For every | | 1,
1 (arctanx )21 h 1 ( 1)n (7) dx n 1 ... 2n . 2 0 1xn1 n 3 2 n 1 4
In particular, when 1, we have the curious representation
n1 3 hn 1 ( 1) (8) 192 1 ... . n1 nn3 2 1 4
Proof. Using the Taylor series for |x | 1,
x3 x 5( 1)nn 1 x 2 1 arctanxx ... ... , 3 5 2n 1
5
it easy to compute the expansion
n1 2( 1) 2nn 2 (arctanx ) hn x . n1 n
Therefore, according to entry 3.241(1) in [8]
11(arctan xx )2 ( 1)nn 1 2 dx h 2n d x . 22 n 0011xn1 n x
n1 ( 1)2n 1 1 hnn , n1 n 22 where, as above, ()x is the incomplete beta function. According to the representation ( 1)k ()z k0 zk
(see 8.372 (1) in [8]) we compute
n 1 1 1 1 1n 1 1 ( 1) n ... ( 1) 1 ... , 2 2212325n n n 435 21 n and this finishes the proof.
Next we present two logarithmic integrals which can naturally be associated to the arctangent integrals in (3). Using the same technique, differentiation on a parameter, they can be evaluated to somewhat similar outcomes.
Proposition 4. For every 11 ,
log(1 x ) (9) dxlog log(1 ) Li ( ) . 2 0 xx(1 )
When 1 this is the integral
log(1 x ) 2 dx , 0 xx(1 ) 6 which is equivalent to entry 4.295.18 in [8] and is also a particular case of entry 2.6.10.52 in [14].
Proof. Setting h() to be the LHS we have
6
1 1 1 x h( ) d x log 0 (1x )(1 x ) 1 1 x 0
1 1 log log . 11
From here, integrating by parts,
logtt log(1 ) h( ) dt log log(1 ) dt 001tt
log log(1 ) Li2 ( ).
Done!
Proposition 5. For every 11 ,
1 log(1x ) 1 1 (10) dx Li Li . 22 0 xx(1 ) 2 2
When 1, this is entry 4.291.12 in [8] and entry 2.6.10.8 in [14].
1 log(1 x ) 1 2 1 dx Li log2 2 . 2 0 xx(1 ) 2 12 2
Proof. Setting g() to be the LHS we have
1 1 1 1 1 x 1 2 g( ) d x log log . 0 (1x )(1 x ) 1 1 x 0 1 1
At the same time we notice that
n d 1 1 1 1 1 1 Li2 log , dn2 1 n1 2 1 2 so the conclusion is
d 1 g( ) Li2 . d 2
Therefore, for some constant C we have
7
1 gC( ) Li2 . 2
1 With 0 we compute C Li2 and the proof is finished. 2
We can evaluate the integral in (10) also in terms of a power series in . Using the Taylor series for log(1x ) we compute
11log(1x ) ( 1)n n1 x n d x dx 00x(1 x )n0 n 1 1 x
( 1)n n111 x n ( 1) n n d x ( n 1) nn00n10 1 x n 1
(log 2 H ) n n1 . n0 n 1
Comparing this to (10) we come to the following result.
Corollary 6. For every 11 ,
11 (log 2 Hn ) n1 Li22 Li . 2 2 n0 n 1
In Particular, with 1,
2 (log 2 Hn ) 11 2 Li2 log 2 . n0 n 1 2 12 2
(See also [4].)
3. Two integral representations for (3) and a special arctangent integral
n For | | 1, let Lip ( ) p be the polylogarithm [12] n1 n
Lema 7. For every | | 1 and p 0,
8
1 11 x (11) (logx )pp log d x ( 1)1 ( p 1) Li ( ) Li ( ) pp22 0 xx1
1 1 (12) (logx )pp log(1 x ) d x ( 1)1 ( p 1)Li ( ) . p2 0 x
Equation (12) is entry 2.6.19.6 in [14].
Proof. With xe t the first integral becomes
( 1)p1 t p log(1 e t ) log(1 e t ) dt 0
nnn1 p11 p nt p ( 1) p nt ( 1)t e dt ( 1) t e dt nn11nn00
nn1 ( 1)n1 1 ( 1)p1 (p 1) pp11 nn11n n n n
p1 (1) (p 1)Li pp22 ()Li ( )
The same substitution in the second integral provides
n p11 p t p p nt ( 1)t log(1 e ) dt ( 1) t e dt 00n1 n
n 1 ( 1)pp11 (pp 1) ( 1) ( 1)Li ( ) p1 p2 n1 nn and the lemma is proved.
Corollary 8. We have the representations
811 dt (13) (3) arctant arctan , 7 0 tt
1 1 dt (14) (3) log(1 t )log 1 , 0 tt
where the first integral is equivalent to
9
1 (arctant )2 7 (15) dt G (3) . 0 t 28
The most remarkable integral in (15) brings together three important constants, , the Catalan constant G , and (3) . This result is known; see entry 8 in the list [1] and also p. 18 in [6].
Proof. The starting point is equation (4), where in the integral we make the substitution tx to bring it to the form
arctant 1 2dt log log Li ( ) Li ( ) . 22 22 0 t 1
Here we divide both sides by and integrate for from 0 to 1 ,
1d 11 1 1 1 2 arctant dt log log d Li ( ) Li ( ) d . 22 22 0 0t 0 1 0
Evaluating these integrals (using the above lemma for the second one) we come to the equation
17dt arctant arctan Li (1) Li ( 1) (3) . 33 0 tt 4
1 We shall transform now this integral. First we split it this way: , and then in the last 0 0 1 1 one we make the substitution x to get t
11dt1 dt arctantt arctan 2 arctan arctan . 00t t t t
This proves the first representation above, equation (13) ,
1 17dt arctant arctan (3) . 0 tt 8
Next we use the identity (t 0)
1 arctan arctan t , t 2 and the well-known fact that
10
1 arctan t dt G , 0 t to prove (15).
For the second representation (14) we use (9) in the form (with tx )
log(1 t ) dt log log(1 ) Li ( ) . 2 0 tt()
Dividing by and integrating for from 0 to 1 we write
log(1 td )1 1 1 1 Li ( ) dt log log(1 ) d 2 d , 0tt 0 0 0 that is,
log(1 t ) 1 log 1dt Li (1) Li (1) 2 (3). 33 0 tt
In the same way as above we transform this integral to
log(1tt ) 1 1 log(1 ) 1 log 1 dt 2 log 1 dt , 00t t t t which yields (14). The proof of the corollary is finished.
We end with an extension of equation (15) to power series.
Proposition 9. For any | | 1,
1 (arctan x )2 1 h (16) dx( 1)nn12n . 2 0 xn2 n1
In particular, for 1,
h 7 ( 1)n1 n G (3) (17) 2 . n1 n 4
(The numbers hn were defined right before Proposition 3.)
The series (17) is entry (59) in [6].
11
Proof. By expanding (arctanx )2 in power series as in the proof of Proposition 3,
11(arctan x )2 ( 1)nn 1 1 ( 1) 1 dx h2n x 2 n 1 d x h 2 n . nn 2 00xnn11 n2 n
With 1 the assertion (17) follows from equation (15).
Remark 10. The series and the integral in (16) can be evaluated explicitly in a closed form by using a result of Ramanujan. Namely, Ramanujan proved that for 0 1, ,
hn 1 1 1 1 1 (18) 22n log log Li Li log 2 22 n1 n 2 1 1 1 1 1 1 7 Li33 Li (3) 1 1 4
(see [3], p. 255). We shall use the principle branch of the logarithm. The above equation can be extended by analytic continuation in the disc | | 1 where the RHS is defined. In particular, we can replace by i in order to obtain an alternating series. To simplify the RHS we use the dilogarithm identity
1 1 1 2 Li2 Li 2 log log Li 2 ( ) Li 2 ( ) 1 1 1 4
1i and also we use the formulas log(ii ) log and log 2i arctan . The result is. 2 1 i
1 2 hn (arctanx ) 1 i i 1 (19) ( 1)nn12 2dx Li Li 2 33 n1 n0 x11 i i
2 2 7 2i arctan( ) Li22 ( i ) Li ( i ) 2 i log( ) arctan( ) (3) . 4 2 4
Remark 11. Propositions 3 and 9 admit natural extensions when replacing (arctanx )2 by (arctanx )p , for any integer p 2 . In this case we use the expansion
(20) (arctanx )pn A ( n , p ) x , n1 where A( n , p ) 0 for np and for np
12
3n p n p p!(,)n n 1 s k p A( n , p ) ( 1)22 ( 1) 2k p1 2!kp k 1 k or,
3n p n p p! n A(,) n p (1) 22 (1) 2(,)(,)k L n k s k p p1 , n!2 kp
(see [11], Table 3). Here s(,) k p are the Stirling numbers of the first kind and n 1 n! L(,)nk are he Lah numbers. Thus we have k 1 k!
11pn (arctanx ) nn1 (21) dx Anp(,)(,) xdx Anp ; 00xnnn11
11(arctan xx ) pn d x A(,) n p n d x , 22 0011xxn1 that is,
1 (arctan xn )p 1 1 (22) d x A(,) n p n 2 0 1 x 2n1 2
(see [8], entry 3.241 (1). When 1 we find from here
pp1 2 1 n 1 (23) (p 1)2 A ( n , p ) . n1 2
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