aa r X i v : . [ m a t h . G M ] N ov LOG-TRIGONOMETRIC INTEGRALS AND ELLIPTIC FUNCTIONS
MARTIN NICHOLSON
Abstract.
A class of log-trigonometric integrals are evaluated in terms of elliptic functions. From this, by usingthe elliptic integral singular values, one can obtain closed form evaluations of integrals such as π/ ż ln (cid:18) cosh x √ x ) √ (cid:19) dx = π √ − π (cid:16) √ (cid:17) + 13 π
24 ln 2 . Introduction
The following integrals were calculated in [7] π/ ż ln (cid:2) x + ln (2 e − a cos x ) (cid:3) dx = π ln ae b − , (1) π/ ż ln (cid:2) x + ln (2 e − a cos x ) (cid:3) cos 2 x dx = π (cid:18) − a − e b + 1 e b − (cid:19) , (2) π/ ż x sin 2 xx + ln (2 e − a cos x ) dx = π (cid:18) a + e b − e b ( e b − (cid:19) , (3) π/ ż − π/ (cid:0) e ix (cid:1) γ ix − a + ln (2 cos x ) dx = − πa + π e ( γ +1) a e a − H (ln 2 − a ) , (4)where a ∈ R , b = min { a, ln 2 } , and H is the unit step function. These are log-trigonometric integrals ofthe type whose study was initiated in the series of papers [4, 6, 1, 2, 3]. The author of the paper [7] alsonoted that integral (1) can be used to obtain integrals that can be evaluated in terms of logarithm ofDedekind eta function. However the resulting integrals contained special functions of complex argument.In this paper we modify this approach to obtain integrals of elementary functions of real argument thatare evaluated in terms of infinite products or Lambert series. These infinite products and Lambert seriescan be expressed in terms of elliptic integrals and allow one to obtain closed form evaluation of certainlog-trigonometric integrals at particular values of the parameter.In the following we will use standard notations from the theory of elliptic functions. Let k be modulus, k ′ = √ − k the complementary modulus, and define the complete elliptic integrals of the first andsecond kind with modulus kK = K ( k ) = π/ ż dϕ p − k sin ϕ , E = E ( k ) = π/ ż q − k sin ϕ dϕ, respectively. Let K ′ = K ( k ′ ) and define α according to α = K ′ /K . These notations will be usedthrought the paper. So that whenever a parameter α is encountered in a formula it is assumed that thecorresponding values of k , k ′ , K , K ′ , E contained in the same formula are determined from the formulasabove.The outline of the paper is as follows. In the next section 2 we formulate and prove main theoremsof this paper. These theorems are derived from the integrals (1-4). However these integrals can be MARTIN NICHOLSON generalized, as shown in Appendix A for the case γ = 0 of integral (4). Section 3 is devoted to the studyof these integrals that are consequences of the integral in Appendix A. All integrals in sections 2 and3 are expressed in terms of elliptic integrals. This means that elliptic integral singular values provideclosed form evaluations of log-trigonometric integrals derived in sections 2, 3 for particular values of theparameter α . Corresponding illustrative examples are considered in section 4. Last section 5 is devotedto discussion of the results presented in this paper.2. Main theorems
Theorem 1. If k , k ′ , and K are defined in terms of α as described in the introduction, then (5) π/ ż ln (cid:18) cosh xα − cos ln(2 cos x ) α (cid:19) dx = − π α − π kk ′ K α π , α > ln 22 π ;(6) π/ ż ln (cid:18) cosh xα + cos ln(2 cos x ) α (cid:19) dx = π α
24 + π √ kk ′ , α > ln 2 π . Proof.
For r > ln 2 and n ∈ N , one has from equation (1) π/ ż ln ( x r n + (cid:20) ln(2 cos x ) rn − (cid:21) ) dx = 0 , π/ ż ln ( x r n + (cid:20) ln(2 cos x ) rn + 1 (cid:21) ) dx = − π ln(1 − e − rn ) . Now we take the sum of these equations from n = 1 to infinity. Using the formula ( x r n + (cid:20) ln(2 cos x ) rn − (cid:21) ) ( x r n + (cid:20) ln(2 cos x ) rn + 1 (cid:21) ) = 1 + 2 x − log (2 cos x ) r n + (cid:2) x + log (2 cos x ) (cid:3) r n , and ∞ Y n =1 ( x − log (2 cos x ) r n + (cid:2) x + log (2 cos x ) (cid:3) r n ) = r π cosh πxr − cos π ln(2 cos x ) r x + log (2 cos x ) , yields π/ ż ln ( r π cosh πxr − cos π ln(2 cos x ) r x + log (2 cos x ) ) dx = − π ln ∞ Y n =0 (1 − e − rn ) . Interchanging the order of infinite product and integral can be justified using Fubini’s theorem. Using π/ ż ln (cid:2) x + ln (2 cos x ) (cid:3) dx = 0 , and replacing r by 2 πα one gets(7) π/ ż ln (cid:18) cosh xα − cos ln(2 cos x ) α (cid:19) dx = − π α ) − π ln ∞ Y n =1 (1 − e − παn ) , α > ln 22 π ; OG-TRIGONOMETRIC INTEGRALS AND ELLIPTIC FUNCTIONS 3 (8) π/ ż ln (cid:18) cosh xα + cos ln(2 cos x ) α (cid:19) dx = π π ln ∞ Y n =1 (1 + e − παn ) , α > ln 2 π . Equation (8) is easily deduced from (7) by taking its linear combinations. The infinite products in theseformulas are calculated in [8], ch.21, ex.10: ∞ Y n =1 (1 − e − παn ) = e πα (cid:18) kk ′ K π (cid:19) / , ∞ Y n =1 (1 + e − παn ) = e πα √ k k ′ ! / . This completes the proof. (cid:3)
Theorem 2. (9) π/ ż cosh x α cos ln(2 cos x )2 α cosh xα + cos ln(2 cos x ) α dx = π α + 2) − α K, α > ln 2 π .
Proof.
The case γ = 0 of (4) gives for r > ln 2 π/ ż − π/ ix + r (2 n + 1) + ln (2 cos x ) dx = πr (2 n + 1) , n = − , − , − , ... π/ ż − π/ ix + r (2 n + 1) + ln (2 cos x ) dx = πr (2 n + 1) − π e r (2 n +1) − , n = 0 , , , ... Summing these equations with the help of ∞ X n = −∞ ( − n ix + r (2 n + 1) + log(2 cos x ) = πr cosh πx r cos π ln(2 cos x )2 r cosh πxr + cos π ln(2 cos x ) r , and replacing r by πα , one is immediately lead to π/ ż cosh x α cos ln(2 cos x )2 α cosh xα + cos ln(2 cos x ) α dx = π − πα ∞ X n =0 ( − n e πα (2 n +1) − , α > ln 2 π . The interchanging of summation and integration is easily justified by Fubini’s theorem. The infinite seriesin this expression is calculated in [8], ch.22.6: ∞ X n =0 ( − n e πα (2 n +1) − K π − . (cid:3) Theorem 3. (10) π/ ż sin 2 x sinh xα cosh xα − cos ln(2 cos x ) α dx = 13 πα
48 + π α + πα πα + α π (cid:18) E − − k K (cid:19) K, α > ln 22 π ;(11) π/ ż sin 2 x sinh xα cosh xα + cos ln(2 cos x ) α dx = π α + πα πα + α π ( E − K ) K, α > ln 2 π .
MARTIN NICHOLSON
Proof.
It is convenient to rewrite formula (3) as(12) π/ ż − π/ sin 2 xix + ln (2 cos x ) − a dx = πi (cid:18) a + e b − e b ( e b − (cid:19) . Thus for r > ln 2 and n ∈ N π/ ż − π/ sin 2 xix + ln (2 cos x ) − rn dx = πi r n , π/ ż − π/ sin 2 xix + ln (2 cos x ) + rn dx = πi (cid:18) r n + e − rn − e − rn (1 − e − rn ) (cid:19) . From this it follows with the help of the summation1 ix + log(2 cos x ) + 2 ∞ X n =1 ix + log(2 cos x )[ ix + log(2 cos x )] − r n = πr sin π ln(2 cos x ) r − i sinh πxr cosh πxr − cos π ln(2 cos x ) r and integration ([7], eq. (47))(13) π/ ż − π/ sin 2 xix + ln (2 cos x ) dx = 13 π , that π/ ż sin 2 x sinh πxr cosh πxr − cos π ln(2 cos x ) r dx = π r + 13 r
48 + r e r − − r ∞ X n =0 rn . After change of parameter r = 2 πα , this identity takes the form π/ ż sin 2 x sinh xα cosh xα − cos ln(2 cos x ) α dx = π α + 13 πα
24 + πα e πα − − πα ∞ X n =1 παn , α > ln 22 π . One can easily deduce taking linear combinations of the previous identity that π/ ż sin 2 x sinh xα cosh xα + cos ln(2 cos x ) α dx = π α + πα πα − πα ∞ X n =0 πα (2 n +1)2 , α > ln 2 π . The infinite series in these formulas were calculated in [5] in terms of elliptic integrals(14) ∞ X n =1 παn = 16 − π KE + 2(2 − k )3 π K , (15) ∞ X n =0 πα (2 n +1)2 = 2 π K ( K − E ) . The proof of (10) and (11) is complete. (cid:3)
OG-TRIGONOMETRIC INTEGRALS AND ELLIPTIC FUNCTIONS 5
Theorem 4. (16) π/ ż cos 2 x sin ln(2 cos x ) α cosh xα − cos ln(2 cos x ) α dx = 11 πα − π α + πα πα − α π (cid:18) E − − k K (cid:19) K, α > ln 22 π ;(17) π/ ż cos 2 x sin ln(2 cos x ) α cosh xα + cos ln(2 cos x ) α dx = π α − πα πα + α π ( E − K ) K, α > ln 2 π .
Proof.
These identities are proved in a similar manner to the proof of the previous theorem. The startinglog trigonometric integral is(18) π/ ż − π/ cos 2 xix + ln (2 cos x ) − a dx = πe a H (ln 2 − a ) − π (cid:18) a + e b − e b ( e b − (cid:19) , where a ∈ R , b = min { a, ln 2 } , which is a consequence of (3) and the γ = 0 and γ = 1 cases of (4). Fromthis, it follows that π/ ż cos 2 x sin π ln(2 cos x ) r cosh πxr − cos π ln(2 cos x ) r dx = 11 r − π r + r e r −
1) + r ∞ X n =1 rn , r > ln 2 . Thus(19) π/ ż cos 2 x sin ln(2 cos x ) α cosh xα − cos ln(2 cos x ) α dx = 11 πα − π α + πα e πα −
1) + πα ∞ X n =1 παn , α > ln 22 π ;(20) π/ ż cos 2 x sin ln(2 cos x ) α cosh xα + cos ln(2 cos x ) α dx = π α − πα πα − πα ∞ X n =0 πα (2 n +1)2 , α > ln 2 π . (20) is a consequence of (19). The infinite series in these formulas are the same as in the previous theorem,eqs. (14) and (15). (cid:3) Some other integrals
Theorem 5. (21) π ż sinh x − πα cosh x − πα − cos x ) α dx = π coth π α − πα √ − − αK (cid:0) √ k (cid:1) , α > ln 2 π ;(22) π ż sinh x − πα cosh x − πα + cos x ) α dx = π tanh π α − αkK (cid:0) p /k (cid:1) , α > ln 4 π . Proof.
The case θ = π/ π ż dxi ( x − π/ − a + ln (2 sin x ) = 4 ππi − a + π − e πi/ − a H (ln 2 − a ) , a ∈ R . MARTIN NICHOLSON
Replacing a with παn/
2, where α > ln 2 π and n ∈ Z , and summing all these equations one can get π ż sinh x − πα cosh x − πα − cos x ) α dx = π coth π α − πα √ − − πα ∞ X n =1 √ παn − , and as a consequence π ż sinh x − πα cosh x − πα + cos x ) α dx = π tanh π α − πα ∞ X n =0 √ πα (2 n +1)4 − , α > ln 4 π . The series in these equations is calculated in terms of elliptic integrals in the appendix B. (cid:3)
Another case is θ = π . It is interesting in that it has a weaker restriction on the parameter α thanother formulas obtained above. In terms of the base of the elliptic functions this restriction is q < Theorem 6. If α > , then (23) π ż sinh π − x α cosh π − x α + cos x ) α dx = αkK √ (cid:0) iK ′ , k (cid:1) − π tanh π α . Proof.
The proof is similar to the proof of the previous theorem. We get π/ ż − π/ sinh (cid:0) xα + πα (cid:1) dx cosh (cid:0) xα + πα (cid:1) + cos x ) α = π tanh π α − πα √ ∞ X n =0
12 cosh πα (2 n +1)3 − , α > , and the series in this formula was calculated in the Appendix B. (cid:3) Theorem 7. If α > , then (24) π ż arctan ( tanh π − x α cot x )2 α ) cos x dx = π √
34 sinh πα − π α tanh π α + √ kK (cid:0) iK ′ , k (cid:1) . Proof.
Only an outline of the proof is given here. We start from the generalization of (18)(25) π/ ż − π/ cos 2 x dxi ( x + θ ) − a + ln (2 cos x ) = − π iθ − a ) + π e a − iθ + e a − iθ (1 − e a − iθ ) ! H [ln(2 cos θ ) − a ] , where − π < θ < π , a ∈ R , and a = ln(2 cos θ ). Its proof is similar to the one considered in AppendixA with slight modifications. First we set θ = π/ θ ) = 0, thenreplace a with (2 n + 1) r where r >
0, and take the sum from n = − N − n = N ( N ∈ N ). It is easilychecked that integration and the limit lim N →∞ can be interchanged in this case by Fubini’s theorem.Hence π r π/ ż − π/ cos 2 x tan π ( ix + iθ + ln(2 cos x ))2 r dx = ∞ X n =0 π ( iθ − r (2 n + 1)) − π ∞ X n =0 e − (2 n +1) r − iθ + e − (2 n +1) r − iθ (cid:0) − e − (2 n +1) r − iθ (cid:1) ! . OG-TRIGONOMETRIC INTEGRALS AND ELLIPTIC FUNCTIONS 7
On the RHS of this equation, we first integrate wrt to θ and then take real part, while on the lhs it iseasier first to take real part and then to integrate wrt to θ by using the value of the elementary integral ż sin πyr cosh π ( θ + x ) r + cos πyr dθ = 2 rπ arctan (cid:26) tanh π ( θ + x )2 r tan πy r (cid:27) . The result is π/ ż − π/ arctan (cid:26) tanh π ( π + 3 x )6 r tan π ln(2 cos x )2 r (cid:27) cos 2 x dx = π r tanh π r − π √
38 sinh r − π √ ∞ X n =0
12 cosh(2 n + 1) r − . The sum on the RHS is calculated in Appendix B. (cid:3)
It is clear that a lot of other log-trigonometric evaluations in terms of elliptic integrals can be obtainedin this way. 4.
Examples
Using elliptic integral singular values one can find closed form evaluations of log-trigonometric integralsconsidered above at certain values of the parameter α , e.g. π/ ż ln (cid:18) cosh x √ x ) √ (cid:19) dx = π √ − π (cid:16) √ (cid:17) + 13 π
24 ln 2 , (26) π/ ż cosh x cos ln(2 cos x )4 cosh x + cos ln(2 cos x )2 dx = π − (cid:0) √ (cid:1) Γ (cid:0) (cid:1) √ π , (27) π ż sinh π − x √ cosh π − x √ + cos x ) √ dx = Γ (cid:0) (cid:1) / π − π tanh π √ . (28)Equation (6) in Theorem 1 is particularly interesting because in this case k and k ′ are algebraic. That’swhy evaluation of the integral (26) does not contain any gamma functions.5. Discussion
To better understand the integrals calculated in the previous sections, let’s consider the integral (9)from Theorem 2 and transform it into another form I ( α ) = π/ ż cosh x α cos ln(2 cos x )2 α cosh xα + cos ln(2 cos x ) α dx = 14 π/ ż − π/ dx cos (cid:16) ix α + ln(2 cos x )2 α (cid:17) = 14 π/ ż − π/ dx cos ln(1+ e ix )2 α . Here, making the substitution z = ln(1 + e ix ) as explained in the Appendix A, one obtains(29) I ( α ) = 18 i ż C dz (1 − e − z ) cos z α . MARTIN NICHOLSON
Contour C is the same as in Appendix A. This integral can be evaluated directly by means of residuetheorem in terms of a Lambert series.Thus, what we essentially did in this paper was to rewrite certain contour integrals as integrals of realvalued functions over the real interval.There are other integrals that can be evaluated in terms of Lambert series, e.g.(30) π/ ż sin ln(2 cos x ) α cosh xα − cos ln(2 cos x ) α dx = πα − πα ∞ X n =1 e παn − , (31) π/ ż sin ln(2 cos x ) α cosh xα + cos ln(2 cos x ) α dx = πα ∞ X n =0 e πα (2 n +1) − . However these Lambert series can not be expressed in terms of elliptic integrals.For θ = π/ a disappears altogether: ż π xx + ln (2 e a sin x ) dx = 2 π π + 4 a , ż π ln(2 e a sin x ) x + ln (2 e a sin x ) dx = 4 πaπ + 4 a , a ∈ R . These formulas lead to the analogs of (31) ż π sin ln(2 sin x ) α cosh xα + cos ln(2 sin x ) α dx = 0 , ż π sinh xα cosh xα + cos ln(2 sin x ) α dx = π tanh π α , α ∈ R , that obviously does not contain any Lambert series.Another type of integrals can be obtained from log-trigonometric integrals (1-4) when | Im a | ≥ π , e.g. π/ ż sinh ln(2 cos x ) α cosh ln(2 cos x ) α − cos xα = πα , α ∈ R . These integrals also does not contain any Lambert series.
Appendix A. Generalization of equation (4) with γ = 0Here we consider a generalization of (3) when β = 0. Namely, it will be proved that(32) π/ ż − π/ dxi ( x + θ ) − a + ln (2 cos x ) = πiθ − a + π − e iθ − a H [ln(2 cos θ ) − a ] , where − π < θ < π , a ∈ R , and a = ln(2 cos θ ).To do this first rewrite this integral as a contour integral. Let(33) z = ix + ln(2 cos x ) , − π < x < π . When x varies from − π to π the complex variable z traverses the path C given by the parametric equationRe z = ln(2 cos x ), Im z = x . This path is plotted in the figure below (the blue line). OG-TRIGONOMETRIC INTEGRALS AND ELLIPTIC FUNCTIONS 9 − − − − Im z = θ Re z =ln(2 cos θ ) Re z Im z As can be seen, this path extends from −∞ below the line Im z = 0, passes through the point (ln 2 , −∞ above the line Im z = 0. It is easy to solve eq. (33) for x in terms of z : x = i ln( e z − −∞ . Thus π/ ż − π/ dxi ( x + θ ) − a + ln (2 cos x ) = 12 i ż C dz ( z + iθ − a )(1 − e − z ) . The integral in the z -domain does not have any branching points, therefore there is no need to includeany branch cuts in the picture above.There can be at most two poles of the function z + iθ − a )(1 − e − z ) inside the contour of integration. Thepole z = 0 is always inside the contour, while the pole z = a − iθ can be either inside or outside ofthe contour depending on the values of a and θ . Let’s fix the value of − π < θ < π and plot the lineIm z = θ in the picture above (the dashed horizontal line). It has only one point of intersection withpath C which is (ln(2 cos θ ) , θ ). Thus if a < ln(2 cos θ ), then the pole z = a − iθ is inside the contour C .If a > ln(2 cos θ ) then the pole z = a − iθ is outside the contour C . Now one can easily apply the residuetheorem to complete the proof of (32). Appendix B. Calculating certain sums with hyperbolic functions
Let S ( α ) = ∞ X n = −∞ √ παn + 1 ,S ( α ) = ∞ X n = −∞ √ παn − , then due to 2 cosh x − x one obtains S ( α ) − S ( α ) = 2 ∞ X n = −∞ παn ,S ( α ) + S ( α ) = 2 √ ∞ X n = −∞ cosh παn cosh παn . Well known formulas from the theory of elliptic functions [8] state that ∞ X n = −∞ παn = 2 Kπ , ∞ X n = −∞ cosh παn cosh παn = 2 Kπ √ k. One can deduce from this by trivial algebra that S ( K ′ /K ) = ∞ X n = −∞ √ παn − Kπ (cid:0) √ k (cid:1) . Similarly(34) ∞ X n =0 √ πα (2 n +1)4 − kKπ (cid:0) p /k (cid:1) . To calculate the sums we encountered in the proofs of thereoms 6 and 7 note that 2 cosh 2 x − cosh 3 x cosh x .Then it is easy to deduce that(35) ∞ X n =0
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