Logsine integrals. Notes by G.J.O. Jameson. log sin θ dθ = π log 2,

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1 Logsine integrals Notes by G.J.O. Jameson The basic logsine integrals are: log sin θ dθ = log( sin θ) dθ = log cos θ dθ = π log, () log( cos θ) dθ =. () The equivalence of () and () is obvious. To prove () (following [BM, p. 6]), denote the first integral by I. The substitutions θ = π φ and θ = π φ give Hence I = log cos θ dθ = π/ log sin θ dθ = I. log sin θ dθ. Now substituting θ = φ and using sin φ = sin φ cos φ, we have I = = = I + π log, log sin φ dφ hence (). Of course, we have also established (log sin φ + log cos φ + log ) dφ log sin θ dθ = π log, log( sin θ) dθ =. (3) Also (without knowing the value of I), it is clear that log tan θ dθ =. () An apparently instant alternative proof of () is by termwise integration of the series log sin θ = n= cos nθ (5) n (valid for θ kπ), which can be derived from the series log( e iθ ) = n= n einθ together with e iθ = sin θ. However, justification of termwise integration is not entirely trivial.

2 Net, we state two results obtained by considering π θ. First, we show θ log( sin θ) dθ =, θ log sin θ dθ = π log. (6) Again it is clear that the two statements are equivalent. Denote the first integral by I (we will repeatedly re-use this notation). Substituting θ = π φ and applying (3), we see that I = (π φ) log( sin φ) dφ = I, hence I =. Secondly, observe that since cos(π θ) = cos θ, we have log( + cos θ) dθ = log( cos θ) dθ. Denote this integral by J. Taking the average of the two epressions, we have J = log( cos θ) dθ = Later, we will show how to evaluate (6) and (7) on the interval [, π ]. log sin θ dθ = π log. (7) We now list a number of integrals derived from () by substitution or integration by parts. Substituting = sin θ in the following integral, we obtain: log d = ( ) / Similarly, the substitution = tan θ gives log( + ) + d = log sec θ dθ = log sin θ dθ = π log. (8) log cos θ dθ = π log. (9) Now integrate by parts in (). Since d log sin θ = cot θ, we obtain dθ log sin θ dθ = [ ] π/ θ log sin θ θ cot θ dθ = θ cot θ dθ, hence θ cot θ dθ = π log. () The substitution = sin θ now gives sin d = θ sin θ cos θ dθ = π log. () Integrating by parts again in () and using d dθ cot θ = / sin θ, we obtain π [ log = θ cot θ ] π/ + θ sin θ dθ.

3 The first term is zero, hence θ sin dθ = π log. () θ Integrals evaluated using Catalan s constant Catalan s constant, named after E. C. Catalan (8 89) and usually denoted by G, is defined by G = n= ( ) n (n + ) = The numerical value is G It is not known whether G is irrational: this remains a stubbornly unsolved problem. By termwise integration of the series tan we obtain at once = ( ) n n n +, n= tan Termwise integration (for those who care) is easily justified. Write s n () = d = G, (3) n ( ) n r r +. r= Since the series is alternating, with terms decreasing in magnitude, we have tan / = s n () + r n (), where r n () n+ /(n + 3), so that r n() d as n. Now integrating by parts in (3), we obtain G = [ ] tan log log + d. The first term is zero, since tan log log as +, hence log d = G. () + Now substituting = tan θ in (), we obtain the version relevant to logsine integrals: G = log tan θ sec θ sec θ dθ = log tan θ dθ. (5) The substitution = tan θ in (3), or integration by parts in (5), gives θ dθ = G, sin θ cos θ θ dθ = G. (6) sin θ 3

4 Numerous further integrals can be epressed in terms of G: see [Br] and [JL]. Returning to logsine integrals, let I S = log sin θ dθ, I C = log cos θ dθ. Substituting θ = π φ, we have I C = log sin θ dθ. So by (), I π/ S + I C = log sin θ dθ = π log. Meanwhile by (5), I S I C = log tan θ dθ = G. So we conclude Again there is a neat restatement: I S = G π log, I C = G π log. (7) log( sin θ) dθ = G, log( cos θ) dθ = G. However, (7) will be more useful in the ensuing applications. As with (), a number of integrals can be derived from it by substitution or parts. Corresponding to (9), the substitution = tan θ gives log( + ) + d = With (9), this gives log sec θ dθ = log( + ) d = π log + G. + Corresponding to (), integration by parts in (7) gives log cos θ dθ = [ ] π/ θ log cos θ + log cos θ dθ = π log G. (8) = π 8 log + θ tan θ dθ, θ sin θ cos θ dθ so that θ tan θ dθ = G π log, (9) 8 Similarly, we deduce θ cot θ dθ = G + π log. Corresponding to (), with cos θ instead of sin θ, 8 θ sec θ dθ = [ ] π/ θ tan θ θ tan θ dθ = π 6 G + π log. () The substitution = tan θ gives (tan ) d = θ sec θ dθ, so we can state also (tan ) d = π 6 G + π log. ()

5 Readers who are so inclined will easily be able to formulate the statements analogous to (8), (9) and (). Net, write I = log( + sin θ) dθ = By (7) and the identity + cos θ = cos θ, we obtain I = log( + cos θ) dθ. (log + log cos θ) dθ = π log + log cos φ dφ Combining () and (), we have the pleasingly simple result log( + cosec θ) dθ = = π log + G π log = G π log. () and of course the same applies with cosec θ replaced by sec θ. Writing ( + sin θ)( sin θ) = cos θ, we deduce further log( sin θ) dθ = ( ) log( + sin θ) log sin θ dθ = G, (3) log cos θ dθ log( + sin θ) dθ = G π log. () where hence Integrating by parts, we have log( + sin θ) dθ = J = [ ] π/ θ log( + sin θ) J = π log J, Furthermore, the substitution = sin θ gives θ cos θ + sin θ dθ, J = π log G. (5) sin + d = J. (6) 5

6 A further deduction from (7) is derived using the identity cos θ+sin θ = cos(θ π ): log(cos θ + sin θ) dθ = ( log + log cos(θ π ) ) dθ = π log + log cos φ dφ π/ = π log + G π log = G π log. (7) Alternatively, (7) can be derived from (), using the identity (cos θ + sin θ) = + sin θ. By (7) and (), log( + tan θ) dθ = and hence also, with the substitution = tan θ, which is proved by a longer method in [Br]. ( ) log(cos θ + sin θ) log cos θ dθ = G π log + π log = G + π log, (8) log( + ) d = G + π log, (9) + Of course, tan θ can be replaced by cot θ in (8). With () and (3), this means that we have obtained the values of log[ + f(θ)] dθ for all si trigonometric functions. Yet further integrals can be derived by integrating by parts in (7) and (8) (rather better with cot θ): we leave it to the reader to eplore this.. Integrals involving ζ(3) To obtain a companion to (6) for the interval [, π ] we allow termwise integration of the series (5); this method follows [Br, formula (35)]. Write Now λ(3) = n= (n + ) 3 = 7 8 ζ(3). [ θ ] π/ θ cos nθ dθ = n sin nθ [ cos nθ ] π/ = + n 6 sin nθ dθ n

7 = (cos nπ ) n { (n even), = (n odd). n Assuming termwise integration valid in (5), we deduce θ log( sin θ) dθ = (n + ) = λ(3). (3) 3 n= Substituting θ = π φ and applying (), we deduce further ( π ) θ log( cos θ) dθ = φ log( sin φ) dφ = λ(3). (3) Combining (3) and (3), we have Also, we deduce θ log sin θ dθ = π λ(3) log, 8 θ log tan θ dθ = λ(3). (3) θ log cos θ dθ = π λ(3) log. (33) 8 We mention two derived integrals, corresponding to () and (). Integrating by parts in (33), we obtain θ log sin θ dθ = in which the first term is zero, hence The substitution = sin θ now gives (sin ) d = [ ] π/ θ log sin θ θ cot θ dθ = π θ sin θ θ cot θ dθ, log λ(3). (3) cos θ dθ = π log λ(3). (35) Net, we evaluate θ log tan θdθ. Recall from () and (7) that log( sin θ)dθ = π/ G. Using this and (3), we have θ log tan θ dθ = = = θ log( sin θ) dθ θ log( sin θ) dθ θ log( sin θ) dθ π π/ π/ θ log( cos θ) dθ ( π φ ) log( sin φ) dφ log( sin θ) dθ = λ(3) πg. (36) 7

8 Since d log tan θ = /(sin θ cos θ), we can deduce dθ θ [ ] π/ sin θ cos θ dθ = θ log tan θ θ log tan θ dθ = πg λ(3). (37) Hence also θ sin θ dθ = φ dφ = πg λ(3). (38) sin φ cos φ Further, the substitution = tan θ gives (tan ) d = θ θ tan θ sec θ dθ = sin θ cos θ dθ = πg λ(3). (39) Although θ log( sin θ) dθ appeared in the reasoning for (36), we did not need to know its value. In fact, by a rather more elaborate consideration of the terms arising from the series (5), one can establish Details are given, for eample, in [Ar]. θ log( sin θ) dθ = πg + 5 λ(3), () 8 6 θ log( cos θ) dθ = πg 3 λ(3). () 8 6 An interesting survey of Euler s results on logsine-type integrals is given in [Lo]. References [Ar] James Arathoon, Series representations of Apéry s constant (and Catalan s, constant), notes, from james.arathoon@symde.net [BM] [Br] [JL] [Lo] George Boros and Victor H. Moll, Irresistible Integrals, Cambridge Univ. Press (). David M. Bradley, Representations of Catalan s constant (), at Graham Jameson and Nick Lord, Integrals evaluated in terms of Catalan s constant, Math. Gazette (7), Nick Lord, Variations on a theme - Euler and the logsine integral, Math. Gazette 96 (), updated 3 May 8 8