Page 1 of 9 General Cosine and Sine Integral of Powers ╬ Francis J. O’Brien, Jr., Ph.D. Aquidneck Indian Council Newport, RI Francis J. O’Brien, Jr. April 22, 2013 Introduction In this paper we show the derivation for two general algebraic-trigonometric integrals using complex variables and the gamma function1. The new integral expressed in combined form is: m cos x n cos 2 m 1 x sin x n dx n , n , , 0, n 0, 0 sin 2 where is the gamma function. See L. Bers, Calculus (pp. 402-403) for elementary properties. Complex variables often simplify trigonometric integral evaluation2. Proof Outline Familiar elementary complex variable relations and operations: Euler’s Formula eix cos x i sin x e ix cos x i sin x 1 Submitted to 8th edition of Gradshyetn and Rhyzik (GR), Table of Integrals, Series and Products for Section 3.761, “Trigonometric Functions of More Complicated Arguments Combined with Powers”. NOTE: One website, http://quickmath.com, was unable to do the definite form or the indefinite form (a complicated indefinite form solution was given at the Mathematica website, http://integrals.wolfram.com/index.jsp). 2 G. S. Carr, Formulas and Theorems in Pure Mathematics, solves similar integrals using complex variables and gamma function; see Theorems 2577–2579. Theorem 2577 is derived below to demonstrate complexification as a means to solve an integral computers cannot evaluate. Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 2 of 9 By definition from Euler’s Formula, eix e ix cos x . 2 o If the power is 1 and one selects e ix or e ix in Euler’s Formulas for analysis, one may use real relations cos x Re e ix or Re eix Use cos x Re e ix for convergence of integrals of negative exponentials By definition from Euler’s Formula, eix e ix sin x . 2i o If power is 1 and one selects e ix or e ix in Euler’s Formulas for analysis, one may use real relations sin x Re i e ix Re ie ix or Re e ix i Re ie ix e ix Use sin x Re Re ie ix i for convergence of integrals of negative exponentials NOTE: the simplified real only formulas are valid for sine & cosine powers of 1. For integer 1 cos(2 x) powers higher than 1 use identities such as sin 2 x , then power of 1 real 2 relations. A future paper will demonstrate the technique. Euler’s identity, ei 1 i o e2 11 2 i i o e 2 i cos i sin 2 2 i o e 2 i cos i sin 2 2 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 3 of 9 Complex conjugation o a bi a bi a bi a bi a 2 b 2 o 1. i i i 2 Complexifying3 the integral by real parts only analysis The following algebraic-exponential formula from Gradshteyn & Rhyzik is used to simplify the complexified integrals: m 1 x m x n e dx , , , 0, n 0 Formula 3.326.2, GR (7th ed.)4 0 n n Derivation Inserting the above relations and operations, and simplifying, the complexified cosine integral is: dx Re x meix dx Re n i ii n x cos x m n 0 0 i 2 Re e cos , n n 2 m 1 , Re , 0, n 0 n or, calculate sum of integrals by classical definition of cosine, 3 A good YouTube lecture from MIT is at “18.03 Differential Equations” (Lecture 6). m x n 4 Derived in “ x e dx and related integrals, 2nd ed.”, Jan. 3, 2013. http://www.docstoc.com/profile/waabu Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 4 of 9 x cos x n dx m 0 i x n m e n e i x Re x dx 2 0 1 1 i 1 i 2 n Re i i Re i i cos . n 2 ____________ The complexified sine integral is similar. Not all the steps are specified: x sin x n dx m 0 Re i x m e ix dx n 0 sin 2 n or, calculate difference of integrals by classical definition of sine, i x n e i x n Re x m e dx sin 0 2i n 2 cosx dx 2 Example: 0 In the general solution, x cos x n dx m cos , 0 n 2 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 5 of 9 2 1 set m 0, 1, n 2, 1 / 2, cos to give the solution by the “short cut” or 4 2 2 standard method: cosx dx 2 0 ix 2 e e ix 2 dx Re e ix dx 2 Re 0 2 0 1 cos 2 2 4 2 4 1 0.62665706865, 2 2 1 where . 2 Example: The cosine Fresnel integral is defined to be: z 2 cos 2 t dt. 0 It cannot be solved in closed form unless z in which case the general integral x cos x n dx m cos holds (for . 2 0 n 2 1 Set m 0, , n 2, 1 / 2, cos : 2 4 2 2 2 1 cos 2 t dt . 2 0 The sine Fresnel integral Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 6 of 9 z 2 lim z sin 2 t dt 0 gives the same solution. NOTE: Theorem 2577 (p. 384) in G.S Carr, x n 1 ax e cosbx dx and x n 1 ax e sin bx dx . 0 0 Online integrators cannot solve these definite integrals; e.g., http://www.quickmath.com/ These integrals are useful because they use clever manipulations of complex variables and serve as a general form from which specific cases can be derived. Figure 1 provides the essential information needed for complex variables to evaluate the integrals. i a bi re in r a bi n r e n b = r sin 2 a b r 2 2 b tan a a = r cos 1 b tan a Figure 1. Polar form of complex number a bi on a right triangle. The form a bi would lie in Quadrant IV. NOTE: it is more traditional to use x & y vs. a & b for polar coordinates. Re-expressing in complex variable form and applying Form. 3.326.2, x m e x dx, the cosine n 0 integral is: Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 7 of 9 n 1 a bi x n x n 1 ax e cosbx dx x n 1 ax ibx e e dx x e dx 0 0 0 a bi n NOTE: With complex variables we have reduced a complicated algebraic-exponential- trigonometric function to an algebraic-exponential function which is more manageable. The solution must be given in terms of the real number parameters a, b & n. Thus, conjugating, inserting the auxiliary polar equations, and simplifying provides a string of reductions: n 1 a bi x x e cosbxdx Re x e dx n Re n 1 ax 1 0 0 a bi n n Re 1 a bi n n Re a bi n a bi n a bi n a bi a bi n n Re 1 rei n a 2 b2 n n Re rn 2n cos n i sin n n cosnn r r tan b a n b cos n tan 1 . a n 2 2 2 a b n b Thus, x n 1 ax e cosbx dx cos n tan 1 . a 2 b2 a n 0 2 NOTE: Solution for x n 1 ax e sin bx dx is similar, defining sin(bx ) Re ie ibx . 0 NOTE: If b 0, Theor. 2577 gives n n x n 1 ax e dx cos n0 n 0 a an since arctan(0) 0, cos(0) 1. Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 8 of 9 NOTE: If a 0, Theor. 2577 gives x n 1 cosbx dx 0 Re x n 1e bi x dx 0 n b cos n lim tan 1 bn a 0 a n b n cos n tan 1 , and recalling the limit of arctangent at infinity, lim tan 1 x 1.57 , then x 2 n x n 1 cosbx dx cos n 0 bn 2 Figure 2. Plot of y tan 1 x . NOTE: If n 1, the first step in the derivation of Theor. 2577 gives an immediate answer: 1 ( a bi ) cosbx dx Re ax a e a bi Re a bi a bi a b 2 2 0 Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved. Page 9 of 9 where 1 =1. Clearly, Theorem 2577 is a useful integral. Why can’t computers solve it? A future paper will demonstrate the solution to integrals of the form cos n x x m 1e x dx 0 sin n x The approach is modeled on the solution of Theorem 2577. Francis J. O’Brien, Jr., 2013 <> Aquidneck Indian Council<> All rights reserved.
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