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Some Integrals Related to the Basel Problem

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Some Integrals Related to the Basel Problem 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] 112 nn 1 2 (arctan xx ) ( 1) 2n dx h 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 , 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 Li log , 1 2 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 .
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