Integrals evaluated in terms of Catalan s constant. Graham Jameson and Nick Lord (Math. Gazette, March 2017)

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1 Integrals evaluated in terms of Catalan s constant Graham Jameson and Nick Lord (Math. Gazette, March 7) Catalan s constant, named after E. C. Catalan (84 894) and usually denoted by G, is defined by G = n= It is, of course, a close relative of n= ( ) n (n + ) = (n + ) = 3 π ζ() = 4 8. The numerical value is G It is not known whether G is irrational: this remains a stubbornly unsolved problem. The best hope for a solution might appear to be the method of Beukers [] to prove the irrationality of ζ() directly from the series, but it is not clear how to adapt this method to G. A remarkable assortment of seemingly very different definite integrals equate to G, or are evaluated in terms of G. A compilation of no fewer than eighty integral and series representations for G, with proofs, is given by Bradley []. Another compilation, based on Mathematica, is [3]. Here we will present a selection of some of the simpler integrals and double integrals that are evaluated in terms of G, including a few that are actually not to be found in [] or [3]. The most basic one is tan x x obtained at once by termwise integration of the series dx = G, () tan x x = ( ) n xn n +. n= Termwise integration (for those who care) is easily justified. Write s n (x) = n ( ) n xr r +. r= Since the series is alternating, with terms decreasing in magnitude, we have tan x/x = s n (x) + r n (x), where r n (x) x n+ /(n + 3), so that r n(x) dx as n. We now embark on a round trip of integrals derived directly from (). Most of them can be seen in [4], but we repeat them here for completeness. First, the substitution x = tan

2 gives π/4 G = tan sec d = d. () sin cos Since tan + cot = /(sin cos ), () can be rewritten (tan + cot ) d = G. (3) The substitution = φ gives one of the most important equivalent forms: π/4 sin d = 4φ π/4 sin φ dφ = φ dφ = G. (4) sin φ cos φ We mention in passing that (4) can be rewritten in terms of the gamma function. Recall that by Euler s reflection formula, Γ( + x)γ( x) = πx/(sin πx). Substituting πx = in (4), we deduce: / Γ( + x)γ( x) dx = G π. The substitution x = e t in () gives at once G = Next, we integrate by parts in (), obtaining G = tan (e t ) dt. (5) [ ] tan log x x log x + x dx. The first term is zero, since tan x log x x log x as x +, hence log x dx = G. (6) + x Now substituting x = /y, we deduce log x dx = G. (7) + x Substituting x = e y in (7), we obtain hence G = y + e y ey dy = y cosh y dy, x dx = G. (8) cosh x

3 Now substituting x = tan in (6), we obtain G = log tan sec sec d = Since ( + cos )/( cos ) = cot, we can deduce log + cos d = cos log tan d. (9) log tan d = 4 log tan φ dφ = 4G. () This, in turn, leads to an interesting series representation. Given the series log + x x = x n+ n + and the well-known identity we deduce Write G = n= cos n+ d = (n) (n )(n + ) = n= sin n+ d = n= (n) (n + ), () n (n!) (n)!(n + ) = Using () again (but none of the earlier integrals), we now establish By the binomial series, Hence As above, sinh x = so by a neat cancellation of a n, a n = n= n ( n n ) (n + ). () sinh (sin ) d = G. (3) (n ) (n) ( + x ) = ( ) n a / n x n. x n= ( + t ) dt = ( ) n a n / n + xn+. sin n+ d = sinh (sin ) d = n= n= a n (n + ), n= ( ) n (n + ) = G. We remark that the same method, applied to sin (sin ) (= ) delivers the series π/ (n + ) = d = π 8. 3

4 Integrals of logsine type and applications A further application of (9) is to integrals of the logsine type, of which there are a rich variety. We start by stating the most basic such integral, which does not involve G: log sin d = log cos d = π log. (4) Our proof follows [5, p. 46]). The substitution = π φ shows that the two integrals are equal: denote them both by I. Also, the substitution = π φ gives π log sin d = I. π/ Hence I = π log sin d. Now substituting = φ and using sin φ = sin φ cos φ, we have hence (4). I = = = 4I + π log, log sin φ dφ (log sin φ + log cos φ + log ) dφ Note that the equality of the two integrals in (4) (without knowing their value) implies that log tan d =. Also, (4) can be restated neatly as follows: log( sin ) d = log( cos ) d =. The integral (4) has numerous equivalent forms. For example, first substituting x = sin and then integrating by parts, we find sin x x dx = cot d = log sin d = π log. (5) Further equivalents, and a survey of Euler s work in this area, are given in [6]. Catalan s constant enters the scene when we integrate from to π 4 instead of π. Let I S = log sin d, I C = log cos d. Substituting = π φ, we have I C = log sin d. So by (4), I π/4 S + I C = log sin d = π log. Meanwhile by (9), I S I C = log tan d = G. So we conclude Again there is a neat restatement: I S = G π 4 log, I C = G π log. (6) 4 log( sin ) d = G, log( cos ) d = G. 4

5 However, (6) will be more useful in the ensuing applications. Alternatively, (4) and (6) can be derived from the series log( sin ) = cos n (e.g. see [7]); however, justification of termwise conver- log( cos ) = ( ) n n= n gence is more delicate in this case. n= cos n, n The substitution x = tan delivers the following reformulation of (6) in terms of the log function: log( + x ) + x dx = log sec d = log cos d = π log G. (7) One can easily check that by substituting x = /y and combining with (7), one obtains log( + x ) dx = G + π log. (8) + x Now integrate by parts in (6): we find log cos d = [ ] π/4 log cos + = π 8 log + tan d, sin cos d so that tan d = G π log, (9) 8 an identity not given in [] or [3]. cot d = G + π log. 8 Next, write I = Similarly, or by (9) combined with (3), we have log( + sin ) d = By (6) and the identity + cos = cos, we obtain I = log( + cos ) d. (log + log cos ) d = π log + 4 log cos φ dφ = π log + G π log = G π log, () 5

6 another integral not given in [] or [3]. Combining () and (4), we have the pleasingly simple result log( + cosec ) d = and of course the same applies with cosec replaced by sec. Writing ( + sin )( sin ) = cos, we deduce further log( sin ) d = ( ) log( + sin ) log sin d = G, () log cos d log( + sin ) d = G π log. () where hence Integrating by parts in (), we have log( + sin ) d = J = [ ] π/ log( + sin ) J = π log J, Furthermore, the substitution x = sin gives cos + sin d, J = π log G. (3) sin x + x dx = J. (4) A further deduction from (6) is derived using the identity cos +sin = cos( π 4 ): log(cos + sin ) d = ( log + log cos( π 4 ) ) d = π 4 log + log cos φ dφ π/4 = π 4 log + G π log = G π log. (5) 4 Alternatively, (5) can be deduced from () and the identity (cos + sin ) = + sin. By (5) and (4), log( + tan ) d = ( ) log(cos + sin ) log cos d = G π 4 log + π log = G + π log, (6) 4 6

7 and hence also, with the substitution x = tan, which is proved by a rather longer method in []. log( + x) dx = G + π log, (7) + x 4 Of course, tan can be replaced by cot in (6). With () and (), this means that we have obtained the values of log[ + f()] d, where f() can be any one of the six trigonometric functions. Yet further integrals can be derived by integrating by parts in (5) and (6) (rather better with cot ): we leave it to the reader to explore this.. Double integrals An obvious double-integral representation of G, not involving any trigonometric or logarithmic functions, follows at once from the geometric series. For x, y [, ), we have /( + x y ) = n= ( )n x n y n. Since termwise integration gives x n y n dx dy = + x y dx dy = n= (n + ), ( ) n = G. (8) (n + ) Termwise integration is justified as in (). (Alternatively, one stage of integration immediately equates the double integral to ()). Substituting x = e u and y = e v, we deduce G = e u e v du dv = + e u e v du dv. (9) cosh(u + v) (This is really a pair of successive single-variable substitutions, not a full-blooded twovariable one.) Next, we establish a much less transparent double-integral representation: d dφ = G. (3) + cos cos φ This, again, is not in [] or [3]. It was given in [4], possibly its first appearance. Here we present a proof based on (). We actually show + cos cos φ d dφ = G, (3) 7

8 from which (3) follows by substituting = and φ = φ. Write Now J() = + cos cos φ dφ. + cos cos φ = (cos + sin )(cos φ + sin φ) + (cos sin )(cos φ sin φ) = (cos cos φ + sin sin φ). (3) The substitution t = tan φ gives Applied with a = cos and b = sin, this gives a cos φ + b sin φ dφ = ab tan b a. J() = sin cos. By (), it follows that J() d = G, which is (3). Alternatively, substitute x = tan and y = tan φ in (8). Again, (3) is obtained after applying (3). (This is the method of [4].) We now establish two further double integrals: dx dy = G, (33) x y dx dy = 4G. (34) (x + y)( x) / ( y) / Result (34) is identity (44) in [], achieved with some effort; we present a somewhat simpler proof. The substitution x = u, y = v reduces (34) to (33). Denote the integral in (33) by I. Since the integrand satisfies f(y, x) = f(x, y), the contributions with x y and y x are the same, hence Transform to polar coordinates: I = I = x sec dy dx. x y r dr d. r Now sec r [ ] sec r dr = log( r ) = log sec 8

9 and so sec = tan I = = tan tan, (log tan log tan ) d. As seen in (4), log tan d = π/ log tan φ dφ =. By (9), we deduce that I = G. The discussion site [8] gives the following variant of (33): 4 x y dx dy = 3 G. The method is similar, but the writer resorts to Mathematica for the final stage, the integral log[4/(4 sec )] d; with a bit of effort, this can be deduced from our results above using the identity 4/(4 sec ) = sin cos /(sin 3). Applications to elliptic integrals The complete elliptic integral of the first kind is defined, for t <, by K(t) = ( t sin d. ) / The value of K(t) is given explicitly by a theorem of Gauss: K(t) = π M(, t ), where M(a, b) is the arithmetic-geometric mean of a and b, and t = ( t ) /. Two quite different proofs of this theorem can be seen in [9] and []. However, without any reference to Gauss s theorem, we can apply (4) to evaluate K(t) dt. Indeed, reversing the implied double integral, we have K(t) dt = F () d, where F () = Substituting t sin = sin φ, we have Hence, by (4), F () = ( t sin dt. ) / cos φ cos φ sin dφ = sin. K(t) dt = G. (35) Of course, this really amounts to another double-integral representation of G. 9

10 The complete elliptic integral of the second kind is E(t) = ( t sin ) / d. In the same way, we have E(t) dt = G() d, where G() = = = = ( t sin ) / dt cos φ cos φ sin dφ ( + cos φ) dφ sin sin + cos, hence E(t) dt = G +. (36) A generalisation: the function Ti (x) Going back to the original integral in (), we can define a function of x by considering the integral on [, x]. The more-or-less standard notation is Ti (x) = x tan t t Clearly, G = Ti (). Integrating termwise as in (), we have for x, n= dt. Ti (x) = x x3 3 + x5 5 = ( ) n x n+ (n + ). (37) This can be compared with dilogarithm function Li (x) = n= xn /n. We remark that G appears in the evaluation Li (i) = ζ() + ig. 8 We can use (37) to calculate values for x, for example Ti ( ).487. Since tan( π ) = / tan for < < π, we have tan x + tan x = π for x >. This translates into a corresponding functional equation for Ti (x): ( ) Ti (x) Ti = π log x. (38) x

11 It is sufficient to prove this for x. Write Ti (x) Ti (/x) = F (x). Then F (x) = x /x tan t t dt = = x /x x /x u tan u du ( π u) u tan = π log x F (x). du So for x >, we have the following series expansion in powers of /x: Ti (x) = π log x + x 3 x x 5, showing very clearly the nature of Ti (x) for large x. Of course, the procedures that gave equivalent expressions for G can be also applied to Ti (x). For example, the substitution t = tan leads to Ti (x) = tan x sin d. Integration by parts and then again the substitution t = tan gives Ti (x) = tan x log x = tan x log x x log t + t dt tan x log tan d. As with the function Li (x), not many other explicit values of Ti (x) are known. One, proved by ingenious methods in [] (formula (33)) is: Ti ( 3) = 3 G π log( + 3). References. F. Beukers, A note on the irrationality of ζ() and ζ(3), Bull. London Math. Soc. (979), David M. Bradley, Representations of Catalan s constant (), at 3. Victor Adamchik, 33 Representations for Catalan s constant, at adamchik/articles/catalan 4. Nick Lord, Intriguing integrals: an Euler-inspired odyssey, Math. Gazette 9 (7), George Boros and Victor H. Moll, Irresistible Integrals, Cambridge Univ. Press (4). 6. Nick Lord, Variations on a theme - Euler and the logsine integral, Math. Gazette 96 (),

12 7. J. A. Scott, In praise of Catalan s constant, Math. Gazette 86 (), G. J. O. Jameson, Elliptic integrals, the arithmetic-geometric mean and the Brent-Salamin algorithm for π, at jameson. Nick Lord, Evaluating integrals using polar areas, Math. Gazette 96 (), GRAHAM JAMESON Dept. of Mathematics and Statistics, Lancaster University, Lancaster LA 4YF, UK g.jameson@lancaster.ac.uk NICK LORD Tonbridge School, Tonbridge, Kent TN9 JP njl@tonbridge-school.org 3 May 6