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 + ) = 3 + 5.... (n + ) = 3 π ζ() = 4 8. The numerical value is G.959656. 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
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
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 =.4.... (n).3.... (n )(n + ) = n= sin n+ d = n=.4.... (n).3.... (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).3.... (n )..4.... (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
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
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
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
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
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
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
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
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 3 + 5 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), 68 7.. David M. Bradley, Representations of Catalan s constant (), at http://germain.umemat.maine.edu/faculty/bradley/papers/c.ps 3. Victor Adamchik, 33 Representations for Catalan s constant, at www.cs.cmu.edu/ adamchik/articles/catalan 4. Nick Lord, Intriguing integrals: an Euler-inspired odyssey, Math. Gazette 9 (7), 45 47. 5. 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 (), 45 458.
7. J. A. Scott, In praise of Catalan s constant, Math. Gazette 86 (), 3. 8. http://mathoverflow.net/questions/8635/conjectured-integral-for-catalans-constant/ 87 9. G. J. O. Jameson, Elliptic integrals, the arithmetic-geometric mean and the Brent-Salamin algorithm for π, at www.maths.lancs.ac.uk/ jameson. Nick Lord, Evaluating integrals using polar areas, Math. Gazette 96 (), 89 96. GRAHAM JAMESON Dept. of Mathematics and Statistics, Lancaster University, Lancaster LA 4YF, UK e-mail: g.jameson@lancaster.ac.uk NICK LORD Tonbridge School, Tonbridge, Kent TN9 JP e-mail: njl@tonbridge-school.org 3 May 6