A sequence of elementary integrals related to integrals studied by Glaisher that contain trigonometric and hyperbolic functions
aa r X i v : . [ m a t h . G M ] A ug A SEQUENCE OF ELEMENTARY INTEGRALS RELATED TO INTEGRALSSTUDIED BY GLAISHER THAT CONTAIN TRIGONOMETRIC ANDHYPERBOLIC FUNCTIONS
MARTIN NICHOLSON
Abstract.
We generalize several integrals studied by Glaisher. These ideas are then applied to obtainan analog of an integral due to Ismail and Valent. Introduction
The following integral Z ∞ sin x sinh( x/a )cos(2 x ) + cosh (2 x/a ) dxx = tan − a (1.1)can be deduced as a particular case of entry . . from [2]. The case a = 1 of this integral can be foundin an old paper by Glaisher [1]. We are going to generalize the above integral as Theorem 1.
Let n be an odd integer. Then Z sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − ( t/a ) (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − ( t/a ) (cid:1) dtt √ − t p t /a = tan − a . (1.2)When n is large, then the main contribution to the integral 1.2 comes from a small neighbourhood around t = 0 and the integral reduces to 1.1.Another integral by Glaisher reads (equation in [1]) Z ∞ cos x cosh x cos(2 x ) + cosh (2 x ) x dx = 0 . It would be generalized as
Theorem 2.
Let n be an even integer. Then Z cos( n sin − t ) cosh (cid:0) n sinh − t (cid:1) cos(2 n sin − t ) + cosh (cid:0) n sinh − t (cid:1) tdt √ − t = 0 . (1.3)Unfortunately there doesn’t seem to be any nice parametric extensions similar to that in Theorem 1.A particularly interesting integral is this one due to Ismail and Valent [4] Z ∞−∞ dt cos( K √ t ) + cosh (cid:0) K ′ √ t (cid:1) = 1 , where K = K ( k ) and K ′ = K ( √ − k ) are elliptic integrals of the first kind. Berndt [3] gives ageneralization of this formula and as an intermediate result proves that (see Corollary . ) Z ∞ x k +1 dx cos x + cosh x = ( − k π k +2 k +1 ∞ X j =0 ( − j (2 j + 1) k +1 cos π (2 j +1)2 n . (1.4)The next theorem gives an elementary analog of this formula: Theorem 3.
Let k and n be a positive integers such that k < (cid:4) n (cid:5) . Then Z t k cos(2 n sin − √ t ) + cosh (cid:0) n sinh − √ t (cid:1) dt √ − t = π ( − k k +1 n n/ X j =1 ( − j − tan π (2 j − n cosh (cid:16) n sinh − tan π (2 j − n (cid:17) sin π (2 j − n cos π (2 j − n ! k . (1.5) Date : August 11, 2020. Proofs of these theorems are given in the subsequent sections 2, 3, and 4. In the last section 5, it willbe explained that the form of the integral in Theorem 3 is not arbitrary. Its form has been chosen toreflect a certain kind of symmetry satisfied also by integrals in Theorems 1 and 2.2.
Proof of Theorem 1
We break the proof into a series of lemmas.
Lemma 4.
Let n be an odd integer. Then we have the partial fractions expansion sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − ( t/a ) (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − ( t/a ) (cid:1) nt = n X j =1 i ( − j − sin π (2 j − n · (cid:16) a cos π (2 j − n + i (cid:17) (cid:16) a + i cos π (2 j − n (cid:17) t (cid:16) a − ia cos π (2 j − n (cid:17) − a sin π (2 j − n . (2.1) Proof.
When n is an odd integer, the expressions n sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − ( t/a ) (cid:1) /t and cos (cid:0) n sin − t (cid:1) +cosh (cid:0) n sinh − ( t/a ) (cid:1) are polynomials in t of degrees n − and n , respectively: sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − ( t/a ) (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − ( t/a ) (cid:1) nt = P n − ( t ) Q n ( t ) . Let’s find the n roots of the denominator polynomial Q n ( x ) . Q n ( x ) can be written as Q n ( x ) = cos (cid:0) n sin − √ x + in sinh − ( √ x/a ) (cid:1) cos (cid:0) n sin − √ x − in sinh − ( √ x/a ) (cid:1) , and thus its roots can be found from the equations sin − √ x ± i sinh − ( √ x/a ) = π (2 j − n , j = 1 , , ..., n, or equivalently from the equations √ x r xa ± i √ xa √ − x = sin π (2 j − n , j = 1 , , ..., n, One can get rid of the radical expressions to come to a quadratic equation wrt x : x (cid:18) (1 − a ) + 4 a cos π (2 j − n (cid:19) + 2 xa (1 − a ) sin π (2 j − n + sin π (2 j − n = 0 , j = 1 , , ..., n. One can easily deduce from this that the n roots of the denominator polynomial are thus x j = (cid:18) a − ia cos π (2 j − n (cid:19) − a sin π (2 j − n , j = 1 , , ..., n. (2.2)Now we can find the partial fractions expansion P n − ( t ) Q n ( t ) = n X j =1 P n − ( x j ) Q ′ n ( x j ) 1 t − x j . (2.3)A simple calculation shows that Q ′ n ( x j ) P n − ( x j ) = r x j a + x j cosh (cid:0) n sinh − ( √ x j /a ) (cid:1) sin (cid:0) n sin − √ x j (cid:1) − r x j − x j cos (cid:0) n sin − √ x j (cid:1) sinh (cid:0) n sinh − ( √ x j /a ) (cid:1) . The equation cos (cid:0) n sin − √ x j (cid:1) + cosh (cid:0) n sinh − ( √ x j /a ) (cid:1) = 0 implies cosh (cid:0) n sinh − ( √ x j /a ) (cid:1) = µ j sin (cid:0) n sin − √ x j (cid:1) , cos (cid:0) n sin − √ x j (cid:1) = iν j sinh (cid:0) n sinh − ( √ x j /a ) (cid:1) where µ j = ± , ν j = ± . To determine the signs µ j , ν j , one can consider the limiting case a >> . Wehave √ x j = sin π (2 j − n − ia sin π (2 j − n cos π (2 j − n + O ( a − ) . This means sin − √ x j = π (2 j − n − ia sin π (2 j − n + O ( a − ) . From this it follows that µ j = ν j = ( − j − and thus Q ′ n ( x j ) P n − ( x j ) = ( − j − (cid:18)r x j a + x j − i r x j − x j (cid:19) = i ( − j sin π (2 j − n (cid:16) a − ia cos π (2 j − n (cid:17)(cid:16) a cos π (2 j − n + i (cid:17) (cid:16) a + i cos π (2 j − n (cid:17) . Substituting this into 2.3 we get the desired result. (cid:3)
Lemma 5. Z t (cid:16) a − ia cos π (2 j − n (cid:17) − a sin π (2 j − n t dt √ − t p t /a (2.4) = tan − a + i tanh − cos π (2 j − n i (cid:16) a cos π (2 j − n + i (cid:17) (cid:16) a + i cos π (2 j − n (cid:17) . (2.5) Proof.
Composition of two substitutions t = 1 − (1 + 1 /a ) sin φ , ( < φ < tan − a ) and tan φ = s ,( < s < a ) reduces this integral to an integral of a rational function. (cid:3) Lemma 6.
For n odd, one has n X j =1 ( − j − sin π (2 j − n = n. Proof.
Put t = 1 , a = i in Lemma 4. (cid:3) From the three lemmas above it follows immediately that Z sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − ( t/a ) (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − ( t/a ) (cid:1) dtt √ − t p t /a = tan − a i n n X j =1 ( − j − sin π (2 j − n tanh − cos π (2 j − n . To finish the proof, note that the sum in this formula is because (since n is odd) j -th and ( n + 1 − j ) -thterms cancel each other out. 3. Proof of Theorem 2
Lemma 7.
Let n be an even integer. Then cos( n sin − t ) cosh (cid:0) n sinh − ( t/a ) (cid:1) cos(2 n sin − t ) + cosh (cid:0) n sinh − ( t/a ) (cid:1) = ( − n/ a n a n + n X j =1 ( − j a sin π (2 j − n n (cid:16) a − ia cos π (2 j − n (cid:17) · (cid:16) a cos π (2 j − n + i (cid:17) (cid:16) a + i cos π (2 j − n (cid:17) t (cid:16) a − ia cos π (2 j − n (cid:17) − a sin π (2 j − n . Proof.
When n is even, the functions cos( n sin − t ) and cosh (cid:0) n sinh − ( t/a ) (cid:1) are a polynomials in t ofdegree n/ . This means we can write cos( n sin − t ) cosh (cid:0) n sinh − ( t/a ) (cid:1) cos(2 n sin − t ) + cosh (cid:0) n sinh − ( t/a ) (cid:1) = C + R n − ( t ) Q n ( t ) where R n − is a polynomial of order n − and Q n was defined in the proof of the Lemma 4. Q n ( x ) has n roots given by 2.2.To find the constant C consider the limit t → + ∞ assuming that a > . In this case sin − t = π − i ln(2 t ) + O ( t − ) , sinh − ( t/a ) = ln(2 t/a ) + O ( t − ) , and we get C = ( − n/ a n a n . Since the order of the polynomial R n − is smaller than the order of the polynomial Q n we can write thepartial fractions expansion R n − ( t ) Q n ( t ) = n X j =1 R n − ( x j ) Q ′ n ( x j ) 1 t − x j . A calculation similar to that in Lemma 4 shows that Q ′ n ( x j ) R n − ( x j ) = 2 nx j r x j a + x j sinh (cid:0) n sinh − ( √ x j /a ) (cid:1) cos (cid:0) n sin − √ x j (cid:1) − r x j − x j sin (cid:0) n sin − √ x j (cid:1) cosh (cid:0) n sinh − ( √ x j /a ) (cid:1) ! = ( − j − nx j (cid:18)r x j a + x j − i r x j − x j (cid:19) = 2 n ( − j (cid:16) a − ia cos π (2 j − n (cid:17) a sin π (2 j − n (cid:16) a cos π (2 j − n + i (cid:17) (cid:16) a + i cos π (2 j − n (cid:17) . This completes the proof of the lemma. (cid:3)
Using Lemmas 5 and 7 we find Z cos (cid:0) n sin − t (cid:1) cosh (cid:0) n cosh − ( t/a ) (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − ( t/a ) (cid:1) t dt √ − t p t /a = ( − n/ a n +1 a n tan − (1 /a ) + a n X j =1 ( − j tanh − cos π (2 j − n − i tan − a n (cid:16) a − ia cos π (2 j − n (cid:17) sin π (2 j − n , and in particular when a = 1 Z cos( n sin − t ) cosh (cid:0) n sinh − t (cid:1) cos(2 n sin − t ) + cosh (cid:0) n sinh − t (cid:1) tdt √ − t = π
16 ( − n/ − n X j =1 ( − j π + 4 i tanh − cos π (2 j − n n cot π (2 j − n = π n ( − n/ n − n X j =1 ( − j cot π (2 j − n . To calculate the sum in this expression we use Lemma 7 with t = 1 and a → ∞ to get n X j =1 ( − j cot π (2 j − n = ( − n/ n. This completes the proof of the theorem.4.
Proof of Theorem 3
Here we restrict the consideration to the case a = 1 . Lemma 8.
The following partial fractions expansion holds for positive integers k and n such that k < (cid:4) n (cid:5) t k cos (cid:0) n sin − √ t (cid:1) + cosh (cid:0) n sinh − √ t (cid:1) = ( − k k n n/ X j =1 t + sin π (2 j − n cos π (2 j − n ( − j − tan π (2 j − n cosh (cid:16) n sinh − tan π (2 j − n (cid:17) π (2 j − n cot π (2 j − n sin π (2 j − n cos π (2 j − n ! k . Proof.
From consideration of the limit t → + ∞ once can see (similarly to that in Lemma 7) that theleading coefficient of the polynomial Q n ( t ) = cos (cid:0) n sin − √ t (cid:1) + cosh (cid:0) n sinh − √ t (cid:1) is n − (1 + ( − n ) and thus that Q n ( t ) is an even polynomial of degree (cid:4) n (cid:5) . Its roots are (see 2.2) x j = − i sin π (2 j − n π (2 j − n , y j = i sin π (2 j − n π (2 j − n , j = 1 , , ..., j n k . For further calculations, we will need explicit values of sin − √ x j and sinh − √ x j , where the principalbranches of the multivalued functions are implied. First, one can write sin − √ x j = ξ j − iη j , sinh − √ x j = ϕ j − iψ j , with ξ j , η j , ϕ j , ψ j > . Further, from elementary identities − t = cos(2 sin − √ t ) and t =cosh(2 sinh − √ t ) one can see that cos(2 ξ j ) cosh(2 η j ) = cosh(2 ϕ j ) cos(2 ψ j ) = 1 , sin(2 ξ j ) sinh(2 η j ) = sinh(2 ϕ j ) sin(2 ψ j ) = sin π (2 j − n cos π (2 j − n . These equations can be easily solved to yield ξ j = ψ j = π (2 j − n , η j = ϕ j = 12 sinh − tan π (2 j − n . Thus sin − √ x j = π (2 j − n − i − tan π (2 j − n , sinh − √ x j = π (2 j − ni + 12 sinh − tan π (2 j − n . Similarly sin − √ y j = π (2 j − n + i − tan π (2 j − n , sinh − √ y j = − π (2 j − ni + 12 sinh − tan π (2 j − n . For k < (cid:4) n (cid:5) we have the partial fractions expansion t k Q n ( t ) = n/ X j =1 x kj Q ′ n ( x j ) 1 t − x j + y kj Q ′ n ( y j ) 1 t − y j ! . Calculations using the formulas above yield Q ′ n ( x j ) = − Q ′ n ( y j ) = 4 ni ( − j cos π (2 j − n sin π (2 j − n (cid:16) π (2 j − n (cid:17) . Now substitute this into the formula above. (cid:3)
Using Lemma 8 and the following consequence of Lemma 5 Z t + sin π (2 j − n cos π (2 j − n dt √ − t = π π (2 j − n π (2 j − n , one can easily complete the proof of Theorem 3.5. Discussion
Let’s introduce the notation α z = 2 n sinh − sin πz n , (5.1)where we assume the principal branches of the multivalued functions of complex variable. With thisdefinition one can rewrite the integral in Theorem 1 with a = 1 as Z sin (cid:0) n sin − t (cid:1) sinh (cid:0) n sinh − t (cid:1) cos (cid:0) n sin − t (cid:1) + cosh (cid:0) n sinh − t (cid:1) dtt √ − t = πn Z n sin πx sinh α x cos πx + cosh α x dx sinh α x n . As we will now show, the last integral has an interesting symmetry.When y is real, then α iy is purely imaginary. Let us define y ∗ by the equation α iy ∗ = πin, and consider the integral over an interval on the imaginary axes J = Z iy ∗ sin πz sinh α z cos πz + cosh α z dz sinh α z n = Z y ∗ sinh πy sin α iy i cos πy + cos( α iy /i ) dy sin α iy in . Making change of variables α iy = πis in 5.1 we get sin πs n = sinh πy n , which implies that πy = α s . Since cos πsn + cosh πyn = 2 , it is easy to show that dy sin πsn = ds sinh α s n . Thus the integral under consideration becomes J = Z n sin πs sinh α s cos πs + cosh α s ds sinh α s n . To recap what we have just showed: Z iy ∗ sin πz sinh α z cos πz + cosh α z dz sinh α z n = Z n sin πs sinh α s cos πs + cosh α s ds sinh α s n , y ∗ = 2 nπ ln(1 + √ . The integral in theorem 1 has been chosen to have the same kind of symmetry: Z t k cos(2 n sin − √ t ) + cosh (cid:0) n sinh − √ t (cid:1) dt √ − t = πn Z n (cid:0) sin πx n (cid:1) k +2 cos πx + cosh α x dx sinh α x n , Z iy ∗ (cid:0) sin πz n (cid:1) k +2 cos πz + cosh α z dz sinh α z n = Z n (cid:0) sin πx n (cid:1) k +2 cos πx + cosh α x dx sinh α x n , y ∗ = 2 nπ ln(1 + √ . We mention without proof an alternative representation for the sum in Theorem with k = 0 : n/ X j =1 ( − j − tan π (2 j − n cosh (cid:16) n sinh − tan π (2 j − n (cid:17) = n X y =1 coth (cid:16) n sinh − sin π (2 y − n (cid:17) coth (cid:16) sinh − sin π (2 y − n (cid:17) − n References [1] J.W.L. Glaisher,
On the summation by definite integrals of geometric series of the second and higher order , Quart. J.Math. Oxford Ser. 11, 328 (1871).[2] I.S. Gradshteyn, and I.M. Ryzhik,
Table of Integrals, Series, and Products , 6th ed., Academic Press, Boston (2000).[3] B.C. Berndt,
Integrals associated with Ramanujan and elliptic functions , The Ramanujan Journal, 41, 369 (2016).[4] M.E.H. Ismail and G. Valent,