arXiv:1806.07962v1 [math.CA] 15 Jun 2018 64,33C60 26A42, codes AMS 2010 ent nerl,Gushpremti ucin aoielpi f elliptic Jacobi function, hypergeometric Gauss integrals, definite in o ute netgto,icuigtesignificant desc the functions. noncomprehensive including investigation, and further brief for a id tions with integral conclude generali We of generalizations many Moll. of and and Russell-style in derived, given are is integrands, function metric is oedfiieitgaso .H .Rsel lotalw all almost Russell, L. H. W. of integrals definite some First eeaiain fRselsyeintegrals Russell-style of Generalizations ooaoSho fMines of School Colorado eateto Physics of Department e od n phrases and words Key mcoff[email protected] odn O80401 CO Golden, akW Coffey W. Mark ue6 2018 6, June Abstract USA 1 eeaiaint elliptic to generalization niiso Amdeberhan of entities e.Te list a Then zed. ito fdirec- of ription t trigono- ith unction 1. Introduction
This paper first reviews, evaluates, and extends various definite integrals of W.
H. L. Russell [6]. It then generalizes several integrals of [2] that are stated to be of
“Russell-style”. Because several of the results are expressed in terms of the Gauss
hypergeometric function 2F1 for general argument and rational parameters, they likely
have application in the calculation of hypervolumes and of the terms of Feynman
diagrams. Several directions for further research are mentioned in the last section.
2. Some integrals of Russell
There are a dozen integrals presented without proofs in the listing [6], which we
label successively with a preceding R. We evaluate the first seven of these integrals,
giving generalizations for most of them. For several, it is useful to employ the families
of integrals contained in the following.
Lemma 1. There holds for integer m 0 ≥ π π θ sin θ cos2m θdθ = , 2m +1 Z0 π 2π θ sin3 θ cos2m θdθ = , 4m2 +8m +3 Z0 π 8π θ sin5 θ cos2m θdθ = , (2m + 1)(2m + 3)(2m + 5) Z0 π 48π θ sin7 θ cos2m θdθ = , (2m + 1)(2m + 3)(2m + 5)(2m + 7) Z0 and
π 2jj!π θ sin2j+1 θ cos2m θdθ = . (2m + 1)(2m + 3)(2m + 5) (2m +2j + 1) Z0 ··· There are many relations between these integrals, which may be studied via recursion relations and otherwise.
Let
I xr sinp x cosq x dx. r,p,q ≡ Z 2 Then the latter entries of Lemma 1 follow from the first as the r = 1 special case of
1 I = (p + q)xr sinp+1 x cosq−1 x + rxr−1 sinp x cosq x r,p,q (p + q)2 r(r 1)I − rpI − − − +(q 1)(p + q)I − ] . − − r 2,p,q − r 1,p 1,q 1 − r,p,q 2 Specifically for the first entry of Lemma 1 we have the easy integration by parts,
followed by the use of a Beta function integral,
π 1 π π θ sin θ cosm θdθ = cosm+1 θdθ ( 1)m+1 m +1 − m +1 − Z0 Z0 √π Γ m +1 π = 2 [1 ( 1)m] ( 1)m+1. 2(m + 1) Γ m + 3 − − − m +1 − 2 2 The first entry of Lemma 1 is a highly specialized case of the following indefinite integral.
Lemma 2. Up to an additive constant,
cosm+1 t 1 m m t sin t cosm t dt = cos t F ; + 1; + 2; cos2 t +(m + 2)t . −(m + 1)(m + 2) 2 1 2 2 2 Z The form of this result shows that explicit evaluations are possible at least for t = 0,
π/4, π/2, and π. This follows since F (α, β; γ; 0) = 1, F (α, β; γ;1) = Γ(γ)Γ(γ 2 1 2 1 − α β)/[Γ(γ α)Γ(γ β)] in terms of the Gamma function, and F (α, β; γ;1/2) − − − 2 1 transforms to 2α F (α,γ β; γ; 1) and, equivalently, to 2β F (β,γ α; γ; 1). The 2 1 − − 2 1 − − Gauss hypergeometric function in Lemma 2 has the further property that Clausen’s
identity applies to 2F1(α, β; α + β +1/2; x). One method to develop Lemma 2 is to
apply the expansion, for example, for even exponents as we are interested herein,
m−1 2m 1 2m 2m cos x = 2m 2 cos2(m k)x + , 2 " k − m # Xk=0 wherein cos 2(m k)x = 2 cos2(m k)x 1 and 2m is the central binomial coefficient. − − − m Remark. We refrain from doing so here, but it is possible to then consider the
analogous integrals
t cos t sinm t dt, t cos2j+1 t sinm t dt, Z Z 3 and analogs of the Russel integrals which are next taken up.
R1 is given as
∞ x −z e r 1 e−(r+1)z+xe dz = 1 + r(r 1) ... . x − x − x2 − Z0 This integral may be written in terms of the Gamma Γ(x) and incomplete Γ(x, y) functions [1, 4]:
∞ ∞ n ∞ −z x e−(r+1)z+xe dz = e−(n+r+1)zdz n! 0 n=0 0 Z X Z ∞ xn 1 = n! (n + r + 1) n=0 X =( x)−(r+1)[Γ(r + 1) Γ(r +1, x)]. − − − R2 is: π θ sin θ π (1 + x) dθ = ln , 1 x2 cos2 θ 2x (1 x) Z0 − − while R3 is: π θ sin θ π = tan−1 x. 1+ x2 cos2 θ x Z0 Hence, using complex variables, these are equivalent, and R2 could just as well be
− written with the inverse function tanh 1. In turn, these integrals are subsumed by the
generalization given for R7 below. By expanding the denominator of the integrand
as geometric series and using Lemma 1 we obtain
∞ π θ sin θ π dθ = x2n θ sin θ cos2n θdθ 1 x2 cos2 θ 0 n=0 0 Z − X Z ∞ 2n x π − = = tanh 1 x. 2n +1 x n=0 X By using other parts of Lemma 1, generalizations of R2/R3 are easily obtained.
For instance: ∞ π θ sin3 θ π dθ = x2n θ sin3 θ cos2n θdθ 1 x2 cos2 θ 0 n=0 0 Z − X Z 4 ∞ 2n 2 x π x 1 − = = 1+ − tanh 1 x (2n + 1)(2n + 3) x2 x n=0 X and ∞ π θ sin5 θ π dθ = x2n θ sin5 θ cos2n θdθ 1 x2 cos2 θ Z0 − n=0 Z0 ∞ X 2n 2 2 x π 5 (x 1) − = = x2 1+ − tanh 1 x . (2n + 1)(2n + 3)(2n + 5) x4 3 − x n=0 X R4 is: π/2 3 π ecos θ cos 3θ cos(cos3 θ sin 3θ)dθ = e1/8. 2 Z0 We show the generalization
π/2 j π j ecos θ cos jθ cos(cosj θ sin jθ)dθ = e1/2 . 2 Z0 π/2 π/2 j jiθ j ecos θ cos jθ cos(cosj θ sin jθ)dθ = Re ee cos θdθ 0 0 Z ∞ Z 1 π/2 = Re cosjn θejinθdθ n! n=0 Z0 ∞X 1 1 π/2 = cosjn θejinθdθ 2 n! − n=0 Z π/2 X ∞ π 1 1 π j = = e1/2 . 2 n! 2jn 2 n=0 X The integral in the penultimate line was evaluated by way of the reciprocal Beta function expression 3.892.2 of [4] (p. 476).
R5 is stated as
π π θ sin θ√1+ x2 cos2 θdθ = [x√1+ x2 + ln(x + √1+ x2)]. 2x Z0 We demonstrate the generalization
π 1 1 3 θ sin θ(1 + x2 cos2 θ)1/kdθ = π F , ; ; x2 , 2 1 2 −k 2 − Z0
5 where 2F1 is the Gauss hypergeometric function [1, 4]. From binomial expansion
within the integrand, ∞ π 1/k π θ sin θ(1 + x2 cos2 θ)1/kdθ = x2j θ sin θ cos2j θdθ j 0 j=0 0 Z X Z ∞ 1/k x2j = π j 2j +1 j=0 X 1 1 3 = π F , ; ; x2 . 2 1 2 −k 2 − For the last step we have noted that 1/k = ( 1)j( 1/k) /j! and 1/(2j +1) = j − − j
(1/2)j/(3/2)j in terms of Pochhammer symbol (a)j = Γ(a + j)/Γ(a). Another way to
obtain this result is to write ∞ ∞ 1/k x2j 1/k ∞ π = π x2j e−(2j+1)tdt j 2j +1 j j=0 j=0 0 X X Z ∞ 1 = π e−t(1 + e−2tx2)1/kdt = π (1 + x2u2)1/kdu Z0 Z0 1 1 3 = π F , ; ; x2 . 2 1 2 −k 2 − For k = 2 we have the R5 reduction
π π 1 − θ sin θ(1 + x2 cos2 θ)1/2dθ = √1+ x2 + sinh 1 x . 2 x Z0 Here the case 1 1 3 F , ; ; x2 2 1 2 −2 2 − may be obtained from ∞ 1 1 3 (2k)! 1 − F , ; ; x2 = ( 1)k x2k = sinh 1 x 2 1 2 2 2 − − 4k(k!)2(2k + 1) x Xk=0 via the relation (1/2) = 2(j 1/2)( 1/2) . The connection of the middle display j − − − j expression to hypergeometric form is by way of the duplication formula for Pochham-
mer symbols λ λ +1 (λ) =22n . n 2 2 n n 6 Remark. The integrals of R2, R3, and R5 may be generalized to arbitrary powers of sin θ within the integrand as we show below for R6.
Entry R6 is
π 3π θ sin θ[(1 + x cos θ)1/3 + (1 x cos θ)1/3]dθ = [(1 + x)4/3 (1 x)4/3], − 4x − − Z0 whereas we first demonstrate the generalization
π πk θ sin θ[(1+x cos θ)1/k +(1 x cos θ)1/k]dθ = [(1+x)(k+1)/k (1 x)(k+1)/k]. − (k + 1)x − − Z0 Again binomial expansion and Lemma 1 are applied so that
∞ π 1/k ∞ θ sin θ[(1+x cos θ)1/k+(1 x cos θ)1/k]dθ = xj[1+( 1)j] θ sin θ cosj θdθ − j − 0 j=0 0 Z X Z ∞ 1/k ∞ =2 x2m θ sin θ cosj θdθ 2m m=0 Z0 X ∞ 1/k x2m =2π 2m (2m + 1) m=0 X πk = [(1 + x)(k+1)/k (1 x)(k+1)/k]. (k + 1)x − − The final summation follows from an easy integration of
∞ 1/k 1 x2m = [(1 x)1/k +(1+ x)1/k]. 2m 2 − m=0 X We have several further generalizations by using Lemma 1, including
π 4π 1 k 1 5 θ sin3 θ[(1 + x cos θ)1/k + (1 x cos θ)1/k]dθ = F , − ; ; x2 . − 3 2 1 −2k 2k 2 Z0 When k = 2 we have the result
π θ sin3 θ[(1 + x cos θ)1/2 + (1 x cos θ)1/2]dθ − Z0 8π = [√1 x(2 + x 8x2 +5x3)+ √1+ x( 2+ x +8x2 +5x3)]. 105x3 − − −
7 In fact, for k > 2,
π θ sin3 θ[(1 + x cos θ)1/k + (1 x cos θ)1/k]dθ − Z0 has the form 4π q p(x) p˜(x) + , 3 x3 (1 x)(2k−1)/2 (1 + x)(2k−1)/2 − where q is a rational number, p is a quintic polynomial, andp ˜(x) = p( x). More- − − over, explicitly,
k k p(x)=(x 1)4 x + , p˜(x)=(x + 1)4 x . − 2k + 1) − 2k + 1) Omitting many other special cases and examples, we present the following general
result coming from Lemma 1.
For odd positive integers p,
π 2(p+1)/2[(p 1)/2]!π 1 k 1 p +2 θ sinp θ[(1+x cos θ)1/k+(1 x cos θ)1/k]dθ = − F , − ; ; x2 . − p!! 2 1 −2k 2k 2 Z0 R7 reads
π θ sin θ π 1+ √2x + x2 π √2x dθ = ln + tan−1 , 4 4 5/2 2 3/2 2 0 1+ x cos θ 2 x 1 √2x + x ! 2 x 1 x Z − − and is also easily generalized.
We have
∞ π θ sin θ π dθ = ( 1)jxjn θ sin θ cosjn θdθ 1+ xj cosj θ − 0 n=0 0 Z X Z ∞ xjn = π ( 1)j − jn +1 n=0 X ( 1)j 1 = π − Φ xj, 1, , j j wherein Φ(z, a, b) is the Lerch zeta function [1, 4]. For j = 4 we obtain the R7
4 reduction to π 2F1(1/4, 1;5/4; x ).
8 Remarks. In regard to R8 one has available the expansions
∞ 22k−1(22k 1)B ln cos x = − 2k x2k k(2k)! Xk=1 ∞ 1 sin2k x π = , x < , −2 k | | 2 Xk=1 wherein Bn are the Bernoulli numbers. For R10 and R12 one may apply, together
with double angle formulas,
∞ 1+ p cos x ( p)k cos kx = , − 1+2p cos x + p2 Xk=0 with p exp(a cos2 θ) and x a sin θ cos θ. R10 reads → → 2 π/2 1+ ea cos θ cos(a sin θ cos θ) π 2 2 dθ = . 1+2ea cos θ cos(a sin θ cos θ)+ e2a cos θ 2(ea/2 + 1) Z0 π/2 1 2π Due to the periodicity and symmetry of the integrand, 0 (...)dθ = 4 0 (...)dθ. Very recently other integrals with trigonometric-hyperbolicR function integrandsR have been discussed in [3].
3. Generalized Russell-style integrals
In referring to entries of the list of [2], the integral identities are labelled with a preceding A. We present several generalizations. The entries of [2] are said to result from the application of an integral transformation with an integrand with a Gaussian factor and a kernel function of either 1/ cosh y or 1/ sinh y. Instead, many of these identities may be obtained with
∞ ∞ 1 1 2 e−(2j+1)κx = , 2 ( 1)je−(2j+1)κx = sinh κx − cosh κx j=0 j=0 X X and the interchange of summation and integration.
A1 is generalized to
∞ ∞ −x2/π2 g sinh gx 2 2 √π sinh x 2 2 e x e−x /π + e−g x dx = dx. cosh2 gx cosh2 x cosh gx Z0 Z0 9 In [2], this identity is stated only for the specific value of g = γ, the celebrated Euler
constant [5].
Similarly, A10 is generalized to
∞ 2 3/2 ∞ −gx2/π 3 2 2 x dx πg e π5e−π x /g + g5/2e−gx /π = dx, 0 cosh πx 2 0 cosh πx Z Z whereas in [2] g is restricted to the value of the famous Catalan constant G =
∞ (−1)k k=0 (2k+1)2 . P A3 is generalized to
∞ ∞ −r2x2/π −x2/π −r4x2/π sinh rxdx e dx x e + re 2 = 2 , 0 cosh rx 0 cosh r x Z Z whereas in [2] r = 2.
A5 is generalized to
∞ ∞ −r2x2/π 2 sinh xdx r e xe−x /π = dx. cosh2 x 2 cosh rx Z0 Z0 In [2], this identity is stated only for r = 2.
In regard to A6 we remark that, by using the substitution x 1/x, → 1 x− ln xdx ∞ x− ln xdx = . 1+ x2 1+ x2 Z0 Z1 We then give the generalization
2 2 1 x− ln xdx r ∞ e−r x /π = dx, 1+ x2 2 cosh r√πx Z0 Z0 rather than simply r = 2 as in [2].
A9 is generalized to
∞ 2 ∞ −px2 − 2 − − 2 2 x dx pπ√p e dx √πe x /p + p2√pπ 2e px /π = , 0 cosh x 2 0 cosh πx Z Z while in [2] this identity is stated just for p = 3.
10 4. Further directions
There exists many other analogous integrations, of which we mention only a few other approaches. One is q-extensions, in some cases using the q-binomial series. Since the 2F1 function (and so many of its specializations) have appeared, this suggests in
addition a consideration of finite field analogs.
Bessel functions provide an analog of trigonometric functions, and although oscil-
latory, are not periodic. A better analogy to many Russell-type integrals is provided
by elliptic functions, of which there are many choices. This opens many avenues for
investigation.
One example is to consider the set of Jacobi elliptic functions. A general idea is to consider definite integrations over an interval(s) including rational multiples of one of the two periods. We hereby provide two very specialized analogs of R5, with further x = 1. Jacobi elliptic functions are denoted sn(z, m), etc. with modulus m.
In preparation, we have the particular complete elliptic integral K of the first kind constant 1 Γ2(1/4) K = . 2 4√π Then K (1/2) 1 1 π z sn z, 1+ cn2 z, dz = , 2 2 2 Z0 s and K 2 (1/2) 1 1 Γ2(1/4) z sn z, 1+ cn2 z, dz = . 2 2 √2π Z0 s Other possibilities are to consider various analogs of Lemma 1 in anticipation of series expansions of integrands of rational form in elliptic functions. Two of many such families of definite integrals are to accordingly consider
2K(m) K(m) z sn(z, m)dn(z, m)cnk(z, m)dz, z sn(z, m)dn(z, m)cnk(z, m)dz, Z0 Z0
11 2 2 d with again 2K being a half period of cn and sn, sn z + cn z = 1, and dz cnz = snz dnz. By way of integration by parts, there are again various generalizations of − Lemma 1, and further integrals with integrands of higher powers of sn in terms of recursion relations.
We present the following examples for m = 1/2, being again the lemniscatic setting. K (1/2) 1 z sn(z, 1/2)dn(z, 1/2)cn2(z, 1/2)dz = , 3√2 Z0 K (1/2) π z sn(z, 1/2)dn(z, 1/2)cn4(z, 1/2)dz = , 20√2 Z0 and K (1/2) 1 z sn(z, 1/2)dn(z, 1/2)cn5(z, 1/2)dz = , 10√2G Z0 where G 0.83462684167... is the Gauss constant, given by ≈ √2 1 1 1 G = K = Γ2 . π √2 (2π)3/2 4
Acknowledgements
T. Amdeberhan is thanked for useful correspondence, and P. Martin for reading
the manuscript.
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