True value of an integral in Gradshteyn and Ryzhik's table
TTRUE VALUE OF AN INTEGRAL IN GRADSHTEYN ANDRYZHIK’S TABLE.
JUAN ARIAS DE REYNA
Abstract.
Victor Moll pointed out that entry 3.248.5 in the sixth editionof Gradshteyn and Ryzhik tables of integrals was incorrect. He asked someyears ago what was the true value of this integral. I evaluate it in terms oftwo elliptic integrals. The evaluation is standard but involved, using realand complex analysis.
Contents
1. Introduction. 12. Representation as a conditionally convergent double series. 23. Transforming the series into a triple integral. 44. Another simple integral. 85. Simple transformations of the last integral. 106. Integrand with only one square root. 107. Expression as a logarithmic integral. 118. Reduction to two elliptic integrals. 139. Reduction to normal form (modulus k = 2 − √ / √ Introduction.
Victor Moll in several places ([3], [2, p. 184], [1], [7], [8, p. 176–177]) tell thestory of how he started the project of proving each formula in Gradshteyn andRyzhik table of integrals. For example we quote from [8]Given the large number of entries in [5], we have not yet de-veloped an order in which to check them. Once in a while anentry catches our eye. This was the case with entry . . in Date : January 30, 2018.2010
Mathematics Subject Classification. a r X i v : . [ m a t h . C A ] J a n JUAN ARIAS DE REYNA the sixth edition of the table by Gradshteyn and Ryzhik. Thepresence of the double square root in the appealing integral (cid:90) ∞ dx (1 + x ) / (cid:2) ϕ ( x ) + (cid:112) ϕ ( x ) (cid:3) / = π √ , (26)with ϕ ( x ) = 1 + 4 x x ) , remind us of (16). Unfortunately (26) is incorrect . The numer-ical value of the left-hand side is approximately 0 . . . . , it was taken out. There is no entry . . in [6].At the present time, we are still reconciling this formula.There are two natural questions about entry . . . What is the truevalue of this integral? and, there is a variation of the integrand that integratesto π/ √
6? Our purpose here is to compute an exact value for this integralin terms of elliptic integrals at well defined arguments. We still have no goodanswer for the second question.In all the paper we call I the value of the integral. Our corrected entry is (cid:90) ∞ dx (1 + x ) / [ ϕ ( x ) + (cid:112) ϕ ( x )] / = √ − √ π/ , k, − / ) − √ F ( α, − / ) , where k = 2 − √ α = arcsin √ k and F ( ϕ, k ) and Π( ϕ, n, k ) are the ellipticintegral of the first and third kind respectively.We have used here the notations in Gradshteyn and Ryzhik tables, so that F ( α, − / ) = (cid:90) √ k dx ∆( x ) , Π( π/ , k, − / ) = (cid:90) dx (1 − kx )∆( x ) , where ∆( x ) = (cid:113) (1 − x )(1 − x ).2. Representation as a conditionally convergent double series.
Recall that I is the value of the integral. First we need to transform it alittle and then we obtain the series representation. Proposition 2.1.
We have I = (cid:90) dy (cid:104) ( y − y ) + (cid:0) ( y − y ) (cid:1) / (cid:105) / . (2.1) RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 3
Proof.
We have ϕ (1 /y ) = ϕ ( y ), so that the change of variables y = x − leavesthe integral almost equal I = (cid:90) ∞ y dy (1 + y ) / (cid:20) y y ) + (cid:16) y y ) (cid:17) / (cid:21) / . After this the change x = 1 + y yields I = 12 (cid:90) ∞ dxx / (cid:104) x − x + (cid:0) x − x (cid:1) / (cid:105) / . Finally we change again by means of xy = 1 to get (2.1). (cid:3) We have
Proposition 2.2. I = ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) ∞ (cid:88) k =0 ( − k k ! Γ( n +12 + k )Γ( n +12 ) 2 k k k (2 k )!(2 k )!(4 k + 1)! . (2.2) The series is not absolutely convergent, so the order of the sums is importanthere.Proof.
Let h ( y ) = 1 + ( y − y ), (2.1) can be written as I = (cid:90) dyh ( y ) / (1 + h ( y ) − / ) / . For 0 < y < h ( y ) >
1, therefore we may expand (1 + h ( y ) − / ) − / in power series I = (cid:90) dyh ( y ) / ∞ (cid:88) n =0 (cid:18) − / n (cid:19) h ( y ) − n/ . The binomial series (cid:80) ∞ n =0 (cid:0) − / n (cid:1) x n converges at x = 1, therefore this seriesconverges uniformly on [0 , y ∈ [0 ,
1] we have 1 (cid:54) h ( y ) (cid:54) /
3, so thatthe above series converges uniformly and can be integrated term by term. I = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) (cid:90) h ( y ) − ( n +1) / dy. For y ∈ [0 ,
1] we have 0 (cid:54) ( y − y ) (cid:54) , so that again we can expand h ( y ) − ( n +1) / applying Newton binomial series I = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) (cid:90) ∞ (cid:88) k =0 (cid:18) − ( n + 1) / k (cid:19) k k y k (1 − y ) k dy. JUAN ARIAS DE REYNA
Again this series converges uniformly and can be integrated term by term I = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) ∞ (cid:88) k =0 (cid:18) − ( n + 1) / k (cid:19) k k (cid:90) y k (1 − y ) k dy. The integral can be reduced to a binomial ([6, 3.251.1]) so that (cid:90) y k (1 − y ) k dy = 12 B (cid:16) k + 12 , k + 1 (cid:17) = 2 k (2 k )! (2 k )!(4 k + 1)! . Applying also that (cid:18) − / n (cid:19) = ( − n n (cid:18) nn (cid:19) , (cid:18) − ( n + 1) / k (cid:19) = ( − k k ! Γ( n +12 + k )Γ( n +12 ) , yields I = ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) ∞ (cid:88) k =0 ( − k k ! Γ( n +12 + k )Γ( n +12 ) 4 k k k (2 k )! (2 k )!(4 k + 1)! . (cid:3) Transforming the series into a triple integral.
We will need a Lemma in the proof of Proposition 3.2.
Lemma 3.1.
For any t > we have U ( t ) := ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k t k k = (cid:90) e − u (1 − u ) t du. (3.1) There is a constant C such that for t > we have < U ( t ) (cid:54) Ct − / .Proof. Notice that (2 k )!(2 k )!(4 k + 1)! = (cid:90) ( u (1 − u )) k du, so that ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k t k k = ∞ (cid:88) k =0 ( − k k ! 2 k (cid:90) ( u (1 − u )) k du k t k k . It is easy to justify that we may here interchange the order of sum and integralso that ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k t k k = (cid:90) e − u (1 − u ) t du. RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 5
For 0 < u < / − u ) > /
4, so that U ( t ) (cid:54) (cid:90) / e − u t du (cid:54) (cid:90) ∞ e − u t/ du = √ π √ t . (cid:3) Proposition 3.2.
We have I = (cid:90) ∞ (cid:16) ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) t n − Γ( n +12 ) (cid:17)(cid:16) ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k t k k (cid:17) e − t dt. (3.2) Proof.
Let U ( t ) be the series considered in Lemma 3.1, by this Lemma wehave 0 (cid:54) U ( t ) (cid:54) t >
0. We will apply the dominated convergencetheorem to prove thatlim N →∞ (cid:90) ∞ (cid:16) N (cid:88) n =0 ( − n n (cid:18) nn (cid:19) t n − Γ( n +12 ) (cid:17) U ( t ) e − t dt = (cid:90) ∞ (cid:16) ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) t n − Γ( n +12 ) (cid:17) U ( t ) e − t dt Assuming this, the right hand side of equation (3.2) is equal to ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) n +12 ) (cid:90) ∞ t n − U ( t ) e − t dt = ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) a n Γ( n +12 ) , (3.3)where a n = (cid:90) ∞ ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k t n − + k k e − t dt. We may integrate term by term, since in this case the integral of the absolutevalue of the terms of the series have a finite sum. a n = ∞ (cid:88) k =0 ( − k k ! 2 k (2 k )!(2 k )!(4 k + 1)! 4 k Γ( n +12 + k )3 k . Inserting this value in (3.3) we get that the right hand side of (3.2) is equalto the right hand side of (2.2), and therefore to our integral I .It remains to justify the application of the dominated convergence theoremabove. Consider the series (cid:80) ∞ n =1 ( − n f ( n ) with f ( n ) := 12 n (cid:18) nn (cid:19) t n − Γ( n +12 ) . JUAN ARIAS DE REYNA
The sequence f ( n ) for n (cid:62) n (cid:54) t − n (cid:62) t . To see it we use Gautschi’s inequalities [10, 5.6.4]. For n (cid:54) t − n (cid:62) f ( n ) f ( n + 1) = 2 n + 22 n + 1 Γ( n + 1)Γ( n + ) 1 √ t (cid:54) (cid:16) n + 22 n + 1 (cid:17)(cid:16) n (cid:17) √ t (cid:54) (cid:16) n t (cid:17) / (cid:16) n + 22 n + 1 (cid:17)(cid:16) n (cid:17) / (cid:54) (cid:16) n t (cid:17) / (cid:16) n (cid:17) (cid:54) . The last inequality is true because n t (cid:16) n (cid:17) = n t + 2 t + 2 nt (cid:54) − t + 2 t + 2 t (cid:54) n (cid:62) t we have f ( n ) f ( n + 1) = 2 n + 22 n + 1 Γ( n + 1)Γ( n + ) 1 √ t (cid:62) (cid:16) n + 22 n + 1 (cid:17)(cid:16) n (cid:17) √ t (cid:62) | (cid:80) Mn = N ( − n a n | (cid:54) max N (cid:54) n (cid:54) M | a n | . It follows thatsup M (cid:12)(cid:12)(cid:12) M (cid:88) n =0 ( − n f ( n ) (cid:12)(cid:12)(cid:12) (cid:54)
11 sup n (cid:62) f ( n )By Stirling formula we have12 n (cid:18) nn (cid:19) t n − Γ( n +12 ) ∼ t n − √ π Γ( n + 1) ∼ t n − √ π √ πn (cid:16) en (cid:17) n/ . Since the two functions are of the same order, the maximum of one is boundedby a constant for the maximum of the other. For t (cid:29) n = 2 t and is equal to e t πt For t < the sequence f ( n ) is decreasing and the maximum is f (0) > f (1).Therefore for t < / f (0) = √ πt . Applying that 0 < U ( t ) < < t < (cid:54) ct − / for t > (cid:12)(cid:12)(cid:12)(cid:16) N (cid:88) n =0 ( − n n (cid:18) nn (cid:19) t n − Γ( n +12 ) (cid:17) U ( t ) e − t (cid:12)(cid:12)(cid:12) (cid:54) (cid:40) c √ t for 0 < t < , ct / for t > , for some constant c . Therefore we have a uniform bound by an integrablefunction. (cid:3) RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 7
Let Ω be the region equal to the complex plane with a cut along the negativereal axis. Let us denote by √ z the analytic function on Ω defined as theprincipal value, taking | arg( z ) | < π . On Ω the function z + √ z is nevernegative, for example when Im( z ) > √ z ) >
0. Therefore (cid:112) z + √ z is a well defined and analytic function on Ω. We will use this notationeverywhere in the paper. Proposition 3.3.
Let δ > , for any t > we have πi (cid:90) H e tz dz (cid:112) z + √ z = ∞ (cid:88) n =0 ( − n n (cid:18) nn (cid:19) t n − Γ( n +12 ) , (3.4) where H is Hankel’s contour: the boundary of the region containing the pointswith a distance < δ to the negative real axis.Proof. By Cauchy’s Theorem the integral do not depend on δ >
0. Take δ = 2,for example, then for z in the path H we have (cid:112) z + √ z = √ z (cid:112) / √ z ,where all the square roots are principal values. Since | / √ z | (cid:54) − / we have1 (cid:112) z + √ z = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) (1 / √ z ) n +1 = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) z − n +12 , where z s := e s log z with log z meaning the principal logarithm in Ω.The series converges uniformly for z in H , therefore we may integrate termby term and 12 πi (cid:90) H e tz dz (cid:112) z + √ z = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) πi (cid:90) H z − n +12 e tz dz. Since we assume t > w = tz run through another Hankel contour when z run through H and the integral do not depend on which Hankel’s contour weintegrate, therefore12 πi (cid:90) H e tz dz (cid:112) z + √ z = ∞ (cid:88) n =0 (cid:18) − / n (cid:19) t n − πi (cid:90) H w − n +12 e w dw. Recalling the integral representation of the Γ function [10, ]1Γ( s ) = 12 πi (cid:90) H z − s e z dz we get (3.4). (cid:3) For the next Proposition we pick a particular parametrization of H . Wefix the value of δ = , and fix a concrete parametrization of H , namely γ ( ξ ) JUAN ARIAS DE REYNA defined as follows. Take γ ( ξ ) = ( ξ + 1 − i ) for ξ (cid:54) − γ ( ξ ) = e πiξ/ for − < ξ < γ ( ξ ) = (1 − ξ + i ) for ξ >
1. Then any integral (cid:90) H f ( z ) dz = (cid:90) ∞−∞ f ( γ ( ξ )) γ (cid:48) ( ξ ) dξ is an ordinary Lebesgue integral in R . Notice also that | γ (cid:48) ( ξ ) | (cid:54) π/ ξ ∈ R . Proposition 3.4.
We have I = 12 πi (cid:90) ∞ (cid:90) H (cid:90) e tz (cid:112) z + √ z e − u (1 − u ) t e − t du dz dt (3.5) where the integral is an absolutely convergent triple integral.Proof. Substituting (3.1) and (3.4) into (3.2) we have I = (cid:90) ∞ (cid:16) πi (cid:90) H e tz dz (cid:112) z + √ z (cid:17)(cid:16)(cid:90) e − u (1 − u ) t du (cid:17) e − t dt Therefore we only need to show that the integral is absolutely convergent. Wedivide the integral in ξ in three intervals. For | ξ | (cid:54) (cid:12)(cid:12)(cid:12) e tz (cid:112) z + √ z e − u (1 − u ) t e − t γ (cid:48) ( ξ ) (cid:12)(cid:12)(cid:12) (cid:28) e t cos πξ/ e − t (cid:54) e − t/ ;and e − t/ have a finite integral for ( ξ, u, t ) ∈ [ − , × [0 , × [0 , ∞ ). For ξ > (cid:12)(cid:12)(cid:12) e tz (cid:112) z + √ z e − u (1 − u ) t e − t γ (cid:48) ( ξ ) (cid:12)(cid:12)(cid:12) (cid:28) ξ − / e t (1 − ξ ) e − t (cid:54) ξ − / e − tξ/ − t/ ;which have a finite integral on the set of ( ξ, u, t ) ∈ [1 , + ∞ ) × [0 , × [0 , ∞ ).Notice that (cid:90) ∞ ξ − / e − tξ/ e − t/ dξ (cid:54) (cid:90) ∞ ξ − / e − tξ/ e − t/ dξ = e − t/ (cid:112) π/t. Which is integrable for t ∈ (0 , + ∞ ). The case ξ < − (cid:3) Another simple integral.
Proposition 4.1.
We have I = (cid:90) dx (cid:114) x (1 − x ) + (cid:113) x (1 − x ) . (4.1) RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 9 R Γ R C Proof.
Applying Fubini’s Theorem in (3.5) we may integrate first with respectto t , and we obtain I = (cid:90) (cid:16) πi (cid:90) H dz (cid:112) z + √ z − z + u (1 − u ) ) (cid:17) du. Applying Cauchy’s residue theorem the integral (cid:90) Γ R dz (cid:112) z + √ z (1 − z + u (1 − u ) ) = (cid:90) C dz (cid:112) z + √ z (1 − z + u (1 − u ) ) . where C is a circle with center at 1 + u (1 − u ) ) and radius 0 < r < andΓ R is the contour in the figure formed with part of the circle of radius R > H . When R → + ∞ the integral along the portionof the circumference of radius R tends to 0. Therefore we obtain (cid:90) H dz (cid:112) z + √ z (1 − z + u (1 − u ) ) = (cid:90) C dz (cid:112) z + √ z ( z − − u (1 − u ) ) . Notice that there is a change of sign because Γ R contains a portion of H inreverse sense.The integral in C is equal to the residue so that we get I = (cid:90) du (cid:114) u (1 − u ) + (cid:113) u (1 − u ) du. (cid:3) Simple transformations of the last integral.
Proposition 5.1. I = (cid:90) dx √ − x (cid:114) x + (cid:113) x . (5.1) Proof.
In the integral (3.5) the integrand depends only of x (1 − x ) so that theintegral is equal to I = 2 (cid:90) / dx (cid:114) x (1 − x ) + (cid:113) x (1 − x ) dx. Changing variables with x = sin θI = 2 (cid:90) π/ θ cos θ dθ (cid:114) sin θ cos θ + (cid:113) sin θ cos θ . This can be expressed in terms of the double angle I = (cid:90) π/ θ dθ (cid:114) sin θ + (cid:113) sin θ . The change of variables x = 1 − cos θ transform this in (5.1). (cid:3) Integrand with only one square root.
Proposition 6.1. I = √ (cid:112) − √ (cid:90) −√ (cid:16) − x + x x (1 − x )(2 − x ) (cid:17) / dx √ − x . (6.1) Proof.
The inner root in (5.1) can be rationalized. To this end we considerthe hyperbola x + 3 = y . Using the rational point ( x, y ) = (1 ,
2) we get therationalization x = m − m + 11 − m , y = 2 m − m − − m . There are two intervals of m that maps into 0 < x <
1. They are [0 , − √ , √ √ − x = (cid:115) − m m (2 − m ) , x / (cid:16) m ( m − − m ) (cid:17) . RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 11
We want (cid:112) x / < m < −√ m ( m − − m ) > x / (cid:112) x / (cid:16) m ( m − − m ) (cid:17) + 2 √ m ( m − − m ) . This can be simplified to2(2 − √ − m + m )( m − − √ − m ) . Finally we have dx = − − m + m )(1 − m ) dm. When m runs through the interval [0 , − √ x run from 1 to 0. Therefore I = (cid:90) −√ (cid:115) − m m (2 − m ) √ − m ) (cid:113) − √ − m + m )(2 + √ − m ) ·· − m + m )(1 − m ) dm. Simplifying we get (6.1). (cid:3)
Remark 6.2.
Taking the interval m ∈ (2 , √
3) we obtain instead of (6.1) I = √ (cid:112) √ (cid:90) √ (cid:16) − x + x x (1 − x )(2 − x ) (cid:17) / dxx − √ . Expression as a logarithmic integral.
Proposition 7.1.
Let C = (cid:112)
21 + 12 √ √ √ , B ( t ) = 1 + 10 t − √ t + 64 t t . (7.1) Then we have I = C + 2 √ (cid:112) − √ (cid:90) ∞ (2+ √ log 2 + √ + √ (cid:112) B ( t ) dt √ t . (7.2) Proof.
The zeros of the radical in (6.1) are symmetric with respect two linesRe x = and Im x = 0. So we change variables by x = − yI = 2 √ (cid:112) − √ (cid:90) √ − (cid:16) (3 + 4 y )(1 − y )(9 − y ) (cid:17) / dy + √ y . Let A ( y ) = (3+4 y )(1 − y )(9 − y ) . Our integral is I = 2 √ (cid:112) − √ (cid:90) √ − (cid:16)(cid:90) A ( y )0 dt √ t (cid:17) dy + √ y . Following Knuth we denote by [ t < A ( y )] a function that is 1 when t < A ( y )and 0 in other case, using this we have I = 2 √ (cid:112) − √ (cid:90) √ − (cid:90) + ∞ [ t < A ( y )] dt √ t dy + √ y . We reverse now the order of integration.For any fixed value of t > y with A ( y ) = t , y = 1 + 10 t − √ t + 64 t t , y = 1 + 10 t + √ t + 64 t t . We are integrating for y ∈ ( √ − , ), so we are only interested in 0 < y < .In this range of y the condition t < A ( y ) is equivalent to 16 y t − (40 t − y +9 t − <
0. This implies that y is contained between the two roots above.The second root gives y > /
8, so y > < y < t < A ( y ) is equivalent to y > B ( t ) := 1 + 10 t − √ t + 64 t t . Therefore we have I = 2 √ (cid:112) − √ (cid:90) √ − (cid:90) + ∞ [ t < A ( y )]2 √ t dy + √ y = (cid:90) ∞ dt √ t (cid:90) J dy + √ y where J is the interval ( √ − , / ∩ ( (cid:112) B ( t ) , ∞ ). (cid:112) B ( t ) increases with t , and (cid:112) B ( t ) = √ − just for t = (2 + √ (cid:112) B ( t ) < for t > (2 + √
3) so that I = 2 √ (cid:112) − √ (cid:16)(cid:90) (2+ √ dt √ t (cid:90) √ − dy + √ y + (cid:90) ∞ (2+ √ dt √ t (cid:90) √ B ( t ) dy + √ y (cid:17) . After simplification we have (cid:90) √ − dy + √ y = log 3 + 2 √ , √ (cid:112) − √ (cid:90) (2+ √ dt √ t = (cid:115)
21 + 12 √ . RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 13 I = (cid:112)
21 + 12 √ √ √
36 + 2 √ (cid:112) − √ (cid:90) ∞ (2+ √ log 2 + √ + √ (cid:112) B ( t ) dt √ t . (cid:3) Reduction to two elliptic integrals.
Proposition 8.1.
We have I = √ √ −√ J , where J = 12 (cid:90) √ − (cid:16) − xx − (cid:17) / √ √
3) + x dx √ − x − (cid:90) √ − (cid:16) − xx − (cid:17) / dx √
3) + x . (8.1)
Proof.
By Proposition 7.1 we have I = C + √ √ −√ U where U = (cid:90) ∞ (2+ √ log 2 + √ + √ (cid:112) B ( t ) dt √ t . (cid:112) B ( t ) contains the inner root √ t + 64 t . We rationalize this squareroot by the change t = m − − m , t + 64 t = (cid:16) − m + m − m (cid:17) . The value of m − − m run through the interval ( (2 + √ , ∞ ) for two intervals of m . First for m ∈ ( − √ , − m ∈ (4 , √ − (cid:112) B ( t ) is different in the two intervals. Weobtain B (cid:16) m − − m (cid:17) = 5 − m , < m < √ − B (cid:16) m − − m (cid:17) = 3(8 + m )4(8 − m ) , − √ < m < − . Therefore U = − (cid:90) √ − log 2 + √ + √ √ − m d (cid:16) − mm − (cid:17) / . This can be written U = (cid:90) √ − log (cid:16) − √
32 ( √ − m − (cid:17) d (cid:16) − mm − (cid:17) / . Integrating by parts U = 1 + √
34 log(4 √ − (cid:90) √ − (cid:16) − mm − (cid:17) /
11 + −√ ( √ − m −
1) 2 − √
32 12 √ − m dm, simplifying and changing the name of the variable U = 1 + √
34 log(4 √ − (cid:90) √ − (cid:16) − xx − (cid:17) /
13 + 2 √ √ − x √ − x dx. This integral is the sum of other two U = 1 + √
34 log(4 √ − (cid:90) √ − (cid:16) − xx − (cid:17) / √ √
3) + x √ − x dx − (cid:90) √ − (cid:16) − xx − (cid:17) / √
3) + x dx, := 1 + √
34 log(4 √ −
6) + J, (notice that 4(4 + 3 √
3) = (3 + 2 √ − I = C + √ √ −√ J where C = C + 2 √ (cid:112) − √ √
34 log(4 √ − . But this constant is equal to 0. To prove it notice first thatlog 3 + 2 √
36 + log(4 √ −
6) = log 1 = 0 . Then we only have to check that (cid:112)
21 + 12 √ √ √ (cid:112) − √ √ . This is equivalent to √ √ √ √ = 1, that is checked by squaring. (cid:3) RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 15 Reduction to normal form (modulus k = 2 − √ ). The integrals in Proposition 8.1 are elliptic integrals. That is integrals (cid:82) R ( x, y ) dx where R is rational and y = (cid:112) p ( x ) where p ( x ) is a polynomial ofdegree 3 or 4 without multiple roots. It is known that these integrals can bereduced to three canonical forms.Our two elliptic integrals are in fact corresponding to the same modulus k = 2 − √ Proposition 9.1.
The integral I = aJ + bJ with J = (cid:90) √ √ x + 1 x + 1 + √ dx (cid:112) ( x − − k x ) ,J = (cid:90) √ x − − √ x + 1 + √ dx (cid:112) ( x − − k x ) , (9.1) where k = 2 − √ , a and b the algebraic numbers a = √ √ , b = 2 √ − √ . (9.2) Proof.
Consider the first integral in (8.1) H = 12 (cid:90) − √ (cid:16) − xx − (cid:17) / √ √
3) + x √ − x dx. We reduce the radical to the usual form. Following the general theory, wechange variables by means of the bilinear transformation x = L ( t ) := 413 (14 + 3 √ t − − √ t − √ L send the points − /k , −
1, 1 and 1 /k into 5, 4, −
4, 8 respectively, where k = 2 − √ x − − x )(5 − x ) → − (2 − √
3) 2 (1 − t )(1 − k t )(3 − √ √ − t ) . The interval of integration 4 < x < − √
3) corresponds to − > t > − − √
3. (3 + 2 √ − x )4(4 + 3 √
3) + x = 3 − √ t − − √ t − (1 + √ .dx = − √ t − (14 + 3 √ dt. After simplification we get H = 11 √ − √ (cid:112)
698 + 391 √ (cid:90) − − −√ (cid:112) ( t − − k t ) (2 + √ − t )( (1 + √ − t ) dt. Therefore the coefficient of √ √ −√ H is equal to2 √ (cid:112) − √ · √ − √ (cid:112)
698 + 391 √ − √ √ . That is easily checked after we notice that (cid:112) − √ (cid:112)
698 + 391 √ √ t = − kx we obtain2 √ (cid:112) − √ H = √ √ (cid:90) √ √ x + 1 x + 1 + √ dx (cid:112) ( x − − k x ) . The second integral in (8.1) is H = 12 (cid:90) √ − (cid:16) − xx − (cid:17) / √
3) + x dx.
The change of variables x = L ( t ) := 4(2 + √ t − √ t − − √ , sends − /k , −
1, 1, 1 /k into 8, 4, − ∞ . We have also for 4 < x < √ − (cid:112) ( x − − x ) = (cid:112) ( t − − k t ) 8(9 + 5 √ t − − √ , − x √
3) + x ( t − − √ √ L (cid:48) ( t ) = 3 √ − t + 2 + √ t − − √ ,L ( −
1) = 4 , L ( − (1 + √ /
2) = 4(3 √ − . Therefore after simplifications we get H = − √ − (cid:90) − − √ t + 2 + √ t − − √ dt (cid:112) − (1 − t )(1 − k t ) . The coefficient here by the factor of J in Proposition 8.1 is b = 2 √ (cid:112) − √ · √ −
58 = 2 √ − √ . RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 17
Changing variables t = − x yields2 √ (cid:112) − √ H = − √ − √ (cid:90) √ x − − √ x + 1 + √ dx (cid:112) ( x − − k x ) , and we get our result I = aJ + bJ . (cid:3) Theorem 9.2.
Let k = 2 − √ , a = (1 + √ / − k and ∆ = ( x − − k x ) then √ I = √ (cid:90) /k dx √ ∆ + ( √ − (cid:90) a dx √ ∆ − (cid:90) /k x + 1 + √ dx √ ∆ . (9.3) Proof.
By Proposition 9.1 we have2 √ I = (cid:90) /ka √ x + 1) x + 1 + √ dx √ ∆ + (cid:90) a (2 √ − x − √ x + 1 + √ dx √ ∆= (cid:90) /k √ x + 1) x + 1 + √ dx √ ∆ + (cid:90) a ( √ − x − √ x + 1 + √ dx √ ∆= (cid:90) /k √ x + 1) x + 1 + √ dx √ ∆ + ( √ − (cid:90) a dx √ ∆= √ (cid:90) /k dx √ ∆ − (cid:90) /k x + 1 + √ dx √ ∆ + ( √ − (cid:90) a dx √ ∆ . (cid:3) Reduction to normal form with modulus / √ ϕ, n , k ) = (cid:90) ϕ da (1 − n sin a ) (cid:112) − k sin a = (cid:90) sin ϕ dx (1 − n x ) (cid:112) (1 − x )(1 − k x ) , ( −∞ < n < ∞ ) .F ( ϕ, k ) = (cid:90) ϕ dα (cid:112) − k sin α = (cid:90) sin ϕ dx (cid:112) (1 − x )(1 − k x ) .E ( ϕ, k ) = (cid:90) ϕ (cid:112) − k sin α dα = (cid:90) sin ϕ √ − k x √ − x dx. Here k is the modulus. The integrals appearing in Theorem 9.2 are real but the integrand contains (cid:112) ∆( x ) with ∆( x ) = ( x − − k x ) by a change of variables we may considerreal integrals with the standard form of the radicand (1 − x )(1 − (cid:96) x ). Inour case (cid:96) will be 1 /
3. All this can be done by standard transformations.In particular I find very useful Byrd and Friedman [4] Handbook of EllipticIntegrals.The first integral in (9.3) is well known in [11, p. 501] we find (cid:90) /k dx √ ∆ = K (cid:48) = K ( k (cid:48) ) , where the complementary modulus k (cid:48) is defined by k + k (cid:48) = 1. For the othertwo integrals we use the tables of Byrd and Friedman. Proposition 10.1.
Let k = 2 − √ , we have (cid:90) a dx √ ∆ = 3 + √ (cid:90) √ k dx (cid:113) (1 − x )(1 − x ) . (10.1) (cid:90) /k x + 1 + √ dx √ ∆= 2(1 + √ (cid:90) dx (1 − k x ) (cid:113) (1 − x )(1 − x ) − √ K (1 / √ . (10.2) Proof.
According to [4, p. 120] we have (cid:90) a dx √ ∆ = 1 k (cid:90) a dx (cid:113) ( k − x )( x − x + 1)( x + k ) = gk F ( ϕ, k ) , where k = ( k − k − k + 1)( k + 1) ∴ k = 1 √ ϕ = arcsin (cid:115) ( k + 1)( a − k − a + 1) ∴ ϕ = arcsin √ k.g = 2 (cid:113) ( k + 1)(1 + k ) ∴ g = 3 − √ , gk = 3 + √ . In the same way [4, p. 124] with m = 1 gives (cid:90) /k x + 1 + √ dx √ ∆ = g (2 + √ k (cid:90) u − α sn u − α sn u du RUE VALUE OF AN INTEGRAL IN GRADSHTEYN AND RYZHIK’S TABLE. 19 where the modulus of the function sn u is k = 3 − / , g is given above, sn u =sin ϕ , α = k − k +1 = √ , ϕ = arcsin (cid:115) ( k + 1)( k − k − k + 1) = arcsin 1 = π , sn u = 1 , ∴ u = K ( k ) . and α = ( − − √ k − − − √ − k + 1) , ∴ α = 2 − √ . The primitive is given in [4, p. 205]. It follows that (cid:90) /k x + 1 + √ dx √ ∆ = 3 − √
33 12 − √ (cid:104) (2 −√ − √ )Π( π , −√ , k )+ √ u (cid:105) . So that (cid:90) /k x + 1 + √ dx √ ∆ = 1 + √ K (1 / √ − √ − π , k, / √ . (cid:3) We state now our main result.
Theorem 10.2.
We have with k = 2 − √ √ I = ( √ − π , k, − / ) − F ( α, − / ) , (10.3) where α = arcsin √ k .Proof. Substituting in (9.3) the values given in Proposition 10.1 yields2 √ I = √ K ( k (cid:48) ) − F ( α, − / ) − (1 + √ K (3 − / ) + 2( √ − π , k, − / ) , But by the descending Landen transformation [10, ] K ( √ − k ) = 21 + k K (cid:16) − k k (cid:17) , with k = 2 − √ √ K ( k (cid:48) ) = (1 + √ K (3 − / ) . Therefore two terms cancel in our equation and (10.3) follows. (cid:3)
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Univ. de Sevilla, Facultad de Matem´aticas, c/Tarfia, sn, 41012-Sevilla,Spain
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