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The Integrals in Gradshteyn and Ryzhik. Part 7: Elementary Examples

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The Integrals in Gradshteyn and Ryzhik. Part 7: Elementary Examples THE INTEGRALS IN GRADSHTEYN AND RYZHIK. PART 7: ELEMENTARY EXAMPLES TEWODROS AMDEBERHAN AND VICTOR H. MOLL Abstract. The table of Gradshteyn and Ryzhik contains some elementary integrals. Some of them are derived here. 1. Introduction Elementary mathematics leave the impression that there is marked difference between the two branches of calculus. Differentiation is a subject that is systematic: every evaluation is a consequence of a number of rules and some basic examples. However, integration is a mixture of art and science. The successful evaluation of an integral depends on the right approach, the right change of variables or a patient search in a table of integrals. In fact, the theory of indefinite integrals of elementary functions is complete [3, 4]. Risch’s algorithm determines whether a given function has an antiderivative within a given class of functions. However, the theory of definite integrals is far from complete and there is no general theory available. The level of complexity in the evaluation of a definite integral is hard to predict as can be seen in 2 3 ∞ x ∞ x √π ∞ x 4 (1.1) e− dx =1, e− dx = , and e− dx =Γ . 2 3 Z0 Z0 Z0 The first integrand has an elementary primitive, the second one is the classical Gaussian integral and the evaluation of the third requires Euler’s gamma function defined by ∞ a 1 x (1.2) Γ(a)= x − e− dx. Z0 arXiv:0707.2122v1 [math.CA] 14 Jul 2007 The table of integrals [5] contains a large variety of integrals. This paper contin- ues the work initiated in [1, 7, 8, 9, 10, 11] with the objective of providing proofs and context of all the formulas in the table [5]. Some of them are truly elementary. In this paper we present a derivation of a small number of them. 2. A simple example The first evaluation considered here is that of 3.249.6: 1 p 1 2 (2.1) (1 √x) − dx = . − p(p + 1) Z0 2000 Mathematics Subject Classification. Primary 33. Key words and phrases. Elementary integrals. The second author wishes to the partial support of nsf-dms 0409968. 1 2 TEWODROSAMDEBERHANANDVICTORH.MOLL The evaluation is completely elementary. The change of variables y = 1 √x produces − 1 1 p p 1 (2.2) I = 2 y dy +2 y − dy, − Z0 Z0 and each of these integrals can be evaluated directly to produce the result. This example can be generalized to consider 1 a p 1 (2.3) I(a)= (1 x ) − dx. − Z0 The change of variables t = xa produces 1 1 1/a 1 p 1 (2.4) I(a)= a− t − (1 t) − dt. − Z0 This integral appears as 3.251.1 and it can be evaluated in terms of the beta function 1 a 1 b 1 (2.5) B(a,b)= x − (1 x) − dx, − Z0 as 1 1 (2.6) I(a)= a− B p,a− . The reader will find in [11] details about this evaluation. A further generalization is provided in the next lemma. Lemma 2.1. Let n N, a, b, c R with bc > 0. Define u = ac b2. Then ∈ ∈ − 1 n+1 n j j a + b√x n/2 2u( b) 2u ( 1) b 2b x dx = −n+3 ln(1 + c/b)+ 2 − + . 0 b + c√x c c n j +1 c (n + 2)c j=0 Z X − Proof. Substitute y = b + c√x and expand the new term (y b)n. − 3. A generalization of an algebraic example The evaluation ∞ dx π (3.1) = (1 + x2) √4+3x2 3 Z−∞ appears as 3.248.4 in [5]. We consider here the generalization ∞ dx (3.2) q(a,b)= . (1 + x2) √b + ax2 Z−∞ We assume that a, b> 0. The change of variables x = √bt/√a yields ∞ dt (3.3) q(a,b)=2√a 2 2 0 (a + bt ) √1+ t Z where we have used the symmetry of the integrand to write it over [0, ). The standard trigonometric change of variables t = tan ϕ produces ∞ π/2 cos ϕ dϕ √ (3.4) q(a,b)=2 a 2 2 . 0 a cos ϕ + b sin ϕ Z ELEMENTARY INTEGRALS 3 Finally, u = sin ϕ, produces 1 du (3.5) q(a,b)=2√a . a + (b a)u2 Z0 − The evaluation of this integral is divided into three cases: Case 1. a = b. Then we simply get q(a,b)=2/√a. Case 2. a<b. The change of variables s = u√b a/√a produces (b a)u2 = s2a, so that − − c 2 ds 2 1 (3.6) q(a,b)= 2 = tan− c, √b a 0 1+ s √b a − Z − with c = √b a/√a. − Case 3. a>b. Then we write 1 du (3.7) q(a,b)=2√a . a (a b)u2 Z0 − − The change of variables u = √as/√a b yields − 2 c ds (3.8) q(a,b)= 2 , √a b 0 1 s − Z − where c = √a b/√a. The partial fraction decomposition − 1 1 1 1 (3.9) = + 1 s2 2 1+ s 1 s − − now produces 1 √a + √a b (3.10) q(a,b)= ln − . √a b √a √a b − − − The special case in 3.248.4 corresponds to a = 3 and b = 4. The value of the 1 π integral is 2 tan− (1/√3) = 3 , as claimed. This generalization has been included as 3.248.6 in the latest edition of [5]. We now consider a generalization of this integral. The proof requires several elementary steps, given first for the convenience of the reader. Let a, b R with a<b and n N. Introduce the notation ∈ ∈ ∞ dt (3.11) I = I (a,b) := . n 2 2 0 (a + bt )n √1+ t Z Then we have: Lemma 3.1. The integral In(a,b) is given by 1 2 n 1 (1 v ) − dv (3.12) I (a,b)= − , n (a + αv2)n Z0 with α = b a. − Proof. The change of variables v = t/√1+ t2 gives the result. 4 TEWODROSAMDEBERHANANDVICTORH.MOLL The identity 2 n 1 2 n 2 n 1 1 2 a (3.13) (1 v ) − = (1 v ) + (1 v ) − (a + αv ) − − − α − α produces 1 2 n 1 2 n 1 α (1 v ) 1 (1 v ) − (3.14) I (a,b)= − dv + − dv. n b (a + αv2)n b (a + αv2)n 1 Z0 Z0 − The evaluation of these integrals requieres an intermediate result, that is also of independent interest. Lemma 3.2. Assume z R and n N 0 . Then ∈ ∈ ∪{ } n 1 dx 1 2n tan 1 z 22k 1 (3.15) = − + . (1 + z2x2)n+1 22n n z 2k (1 + z2)k 0 =1 2k k ! Z kX Proof. Define 1 dx 1 z dy (3.16) F (z) := = . n (1 + z2x2)n+1 z (1 + y2)n+1 Z0 Z0 Take derivatives with respect to z on both sides of (3.16). The outcome is a system of differential-difference equations dFn(z) 2(n + 1) 2(n + 1) (3.17) = F +1(z) F (z) dz z n − z n dF 1 1 n = F (z)+ . dz −z n z(1 + z2)n+1 Solving for a purely recursive relation we obtain (after re-indexing n n 1): 7→ − 1 (3.18) 2nFn(z)=(2n 1)Fn 1(z)+ , − − (1 + z2)n 1 1 with the initial condition F0(z) = z tan− z. This recursion is solved using the procedure described in Lemma 2.7 of [1]. This produces the stated expression for Fn(z). The next required evaluation is that of the powers of a simple rational function. Lemma 3.3. Let a, b, c, d be real numbers such that cd > 0. Then 1 2 n n n n k ax + b a 4a d 1 bc ad n 2k 2 2 dx = n + n tan− c/d − − 0 cx + d c c c 4ad k k 1 r k=1 Z p X − n k k 1 2 4an bc ad n 2k 2 − 2 j dj + − − . cn 4ad k k 1 2j (c + d)j =1 j=1 2j j Xk − X Proof. Start with the partial fraction expansion ax2 + b a bc ad 1 (3.19) G(x) := = + − , cx2 + d c cd cx2/d +1 and expand G(x)n by the binomial theorem. The result follows by using Lemma 3.2. The next result follows by combining the statements of the previous three lem- mas. ELEMENTARY INTEGRALS 5 Theorem 3.4. Let a, b R+ with a<b. Then ∈ ∞ dt I +1(a,b) := n 2 +1 2 0 (a + bt )n √1+ t Z n j 1 j 2k 1 n b 2j tan− b/a 1 2 a k = − − + . a(a b)n j 4a j × 2k b j=0 b/ap 1 k=1 2k k ! − X − X p 4. Some integrals involving the exponential function In [5] we find 3.310: ∞ px 1 (4.1) e− dx = , for p> 0, p Z0 that is probably the most elementary evaluation in the table. The example 3.311.1 ∞ dx ln 2 (4.2) = , 1+ epx p Z0 can also be evaluated in elementary terms. Observe first that the change of variables t = px, shows that (4.2) is equivalent to the case p = 1: ∞ dx (4.3) = ln 2. 1+ ex Z0 This can be evaluated by the change of variables u = ex that yields ∞ du (4.4) I = , u(1 + u) Z1 and this can be integrated by partial fractions to produce the result.
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