If we have a function f(x) which is well-defined for some a x b, its integral over those two values is defined as

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1 Y. D. Chong (26) MH28: Complex Methos for the Sciences 2. Integrls If we hve function f(x) which is well-efine for some x, its integrl over those two vlues is efine s N ( ) f(x) = lim x f(x n ) where x n = + n x, x. () N N n= This is clle efinite integrl, n it represents the re uner the grph of f(x) in the region etween x = n x =, s shown in the figure elow: The intervl etween two fixe points, n, is ivie into N segments, of length ( )/N ech. Ech term in the sum represents the re of rectngle. As N, the sum converges to the re uner the curve. For the purposes of imensionl nlysis, n integrl hs the units of the integrn times the units of x. This is esy to rememer if you think of s multiplictive fctor with units of x. From the efintion of the erivtive, we cn show tht f(x) = f(), f(x) = f(). (2) Hence, n integrl is the inverse of erivtive opertion. Notice tht the right-hn-sie of the first eqution oes not involve, the opposite integrl limit. Bse on this, we cn efine n inefinite integrl, or ntierivtive: x f(x ) F (x) such tht F (x) = f(x). (3) Unlike efinite integrl, n ntierivtive is not unique, ut is only efine up to n itive constnt, clle n integrtion constnt. As you my recll from previous mth clsses, integrtion is much hrer thn ifferentition. Once you know how to ifferentite few specil functions, ifferentiting some comintion of those functions just involves strightforwr (though possily teious) ppliction of composition rules. By contrst, there is no generl systemtic proceure for oing n integrl symoliclly. This is clle the ntierivtive prolem. Integrtion often involves mking series of inspire choices, like guessing solution n checking if its erivtive gives the esire integrl expression. Some of the more commonly-use tricks re summrize elow. 2. Integrtion y prts If the integrn consists of two fctors, n you know the ntierivtive of one of the fctors, you cn integrte y prts to shift the erivtive onto the other fctor. Specificlly, f(x) g [ = f(x) g(x) f g(x). (4) 3

2 Y. D. Chong (26) MH28: Complex Methos for the Sciences The first term on the right hn sie is constnt enoting [f()g() f()g(). The secon term is n integrl, which might e esier to o thn the originl integrl. Juicious use of integrtion y prts is key step for solving mny integrls. For exmple, consier x e x. (5) The integrn consists of two fctors, x n e x ; we hppen to know the ntierivtive of oth fctors. Integrting y prts lets us replce one of these fctors with its ntierivtive, while pplying n itionl erivtive on the other fctor. The smrt thing to o is to pply the erivtive on the x fctor, n the ntierivtive on the e x. Then the first fctor turns into unity: x e x = = [ x ex [ x ex ex 2 ex (6). (7) Whenever we finish oing n integrl, it is goo prctice to oule-check the result y mking sure the imensions mtch up. Note tht hs units of inverse x, so the integrl on the left-hn sie hs units of x 2. The solution on the right hn sie hs two terms, with units x/ n / 2 ; oth of these re equivlent to units of x 2, which is wht we nee! 2.2 Chnge of vriles Another useful technique for solving integrls is to chnge vriles. Consier the integrl x 2 +. (8) We cn solve this y mking chnge of vriles x = tn(u). This involves (i) replcing ll occurences of x in the integrn with tn(u), (ii) replcing the integrl limits, n (iii) replcing with (/u) u = /[cos(u) 2 u: π/2 x 2 + = = π/2 [tn(u) 2 + Due to the Pythgoren theorem, the integrn reuces to, so x 2 + = u (9) [cos(u) 2 [sin(u) 2 u. () + [cos(u) 2 π/2 u = π 2. () Clerly, this technique often requires some cleverness n/or tril-n-error in choosing the right chnge of vriles. 2.3 The Gussin integrl Here s fmous integrl: e x2. (2) The integrn is clle Gussin n is plotte elow for the cse of = : 4

3 Y. D. Chong (26) MH28: Complex Methos for the Sciences The lrger the vlue of, the more nrrowly-peke the curve. Hence, the vlue of the efinite integrl epens on. The integrl ws solve y Crl Frierich Guss in prticulrly rillint wy. If I() enote the vlue of the integrl, [I() 2 is just two inepenent copies of the integrl, multiplie together: [ I 2 () = e x2 [ e y2 y. (3) Note tht in the secon integrl, we hve chnge the ummy lel x (the integrtion vrile) into y, to voi miguity. Now, this ecomes two-imensionl integrl, tken over the entire 2D plne: I 2 () = Next, we chnge from Crtesin to polr coorintes: 2π [ I 2 () = r r φ e r2 = r r e r2 Now, y tking the squre root we rrive t the result e x2 = y e (x2 +y 2). (4) [ 2π φ = 2π. 2 (5) π. (6) One very interesting thing to notice out this result is tht it reltes the two trnscenentl constnts e = n π = , y mens of n integrl. The ppernce of π cn e trce ck to the use of polr coorintes to solve the integrl. (As n sie, when stuying the gmm function, we will come cross the fct tht Γ(/2) = π. This is very closely relte result which likewise reltes e which is incorporte into the efinition of the gmm function n π.) 2.4 Differentiting uner the integrl sign In the previous section, we note tht if n integrn contins prmeter (enote ) which is inepenent of the integrtion vrile (enote x), then the efinite integrl cn itself e regre s function of. It cn then e shown tht tking the erivtive of the efinite integrl with respect to is equivlent to tking the prtil erivtive of the integrn: f(x, ) = f (x, ). (7) 5

4 Y. D. Chong (26) MH28: Complex Methos for the Sciences This is clle ifferentiting uner the integrl sign, n ws originlly invente y Gottfrie Wilhelm Leiniz, one of the inventors of clculus. It cn e pplie s technique for solving integrls, which ws populrize y Richr Feynmn in his ook Surely You re Joking, Mr. Feynmn!. Given efinite integrl I, the technique procees s follows: (i) come up with wy to generlize the integrn, y introucing prmeter, such tht the generlize integrl ecomes function I() which reuces to the originl integrl I for prticulr prmeter vlue, sy =. Then, (ii) ifferentite uner the integrl sign. If you hve chosen the generliztion right, the resulting integrl will e esier to solve, so (iii) solve the integrl to otin I (). Finlly, (iv) integrte this over to otin the esire integrl I(), n evlute it t to otin the esire integrl I. An exmple will mke the ove proceure clerer. Consier the integrl First, (i) we generlize the integrl s follows (we ll soon see why): sin(x) x. (8) sin(x) x e x. (9) The esire integrl is I(). Next, (ii) ifferentiting uner the integrl gives I () = sin(x) e x. (2) Tking the prtil erivtive of the integrn with respect to rought own fctor of x, cncelling out the troulesome enomintor. Now, (iii) we solve the new integrl, which cn e one y integrting y prts twice: Hence, I () = [ cos(x) e x + cos(x) e x (2) = + [ sin(x) e x + 2 sin(x) e x (22) = 2 I (). (23) I () = + 2. (24) Finlly, (iv) we nee to integrte this over. But we lrey know how to o this prticulr integrl in Section 2.2, n the result is A tn (), (25) where A is constnt of integrtion. When, the integrl must vnish, which implies tht A = tn (+ ) = π/2. Finlly, we rrive t the result sin(x) x = I() = π 2. (26) It my seem to you tht figuring out this sequence of steps woul tke n lrge mount of ingenuity. However, when we iscuss contour integrtion, we will see more strightforwr wy to o this prticulr integrl. 6

5 Y. D. Chong (26) MH28: Complex Methos for the Sciences 2.5 Exercises. Consier the step function Θ(x) = {, for x, otherwise. (27) Write own n expression for the ntierivtive of Θ(x), n sketch its grph. 2. Show tht 2π [sin(x) 2 = 3. Clculte the following efinite integrls: π x 2 sin(2x) α x ln(x) e x cos(x) e x x cos(x) e x 2π 4. By ifferentiting uner the integrl sign, solve [cos(x) 2 = π. (28) x2 ln(x). (29) Hint: to generlize the integrl, replce x 2 in the numertor with x. 7