DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I hd lerned to do integrls by vrious methods shown in book tht my high school physics techer Mr. Bder hd given me. [It] showed how to differentite prmeters under the integrl sign it s certin opertion. It turns out tht s not tught very much in the universities; they don t emphsize it. But I cught on how to use tht method, nd I used tht one dmn tool gin nd gin. [If] guys t MIT or Princeton hd trouble doing certin integrl, [then] I come long nd try differentiting under the integrl sign, nd often it worked. So I got gret reputtion for doing integrls, only becuse my bo of tools ws different from everybody else s, nd they hd tried ll their tools on it before giving the problem to me. ichrd Feynmn [, pp. 71 7] 1. Introduction The method of differentition under the integrl sign, due originlly to Leibniz, concerns integrls depending on prmeter, such s 1 e t d. Here t is the etr prmeter. (Since is the vrible of integrtion, is not prmeter.) In generl, we might write such n integrl s (1.1) b f(, t) d, where f(, t) is function of two vribles like f(, t) e t. Emple 1.1. Let f(, t) ( + t 3 ). Then 1 f(, t) d 1 ( + t 3 ) d. An nti-derivtive of ( + t 3 ) with respect to is 1 6 ( + t3 ) 3, so 1 ( + t 3 ) d ( + t3 ) 3 1 6 ( + t3 ) 3 t 9 6 4 3 + t3 + t 6. This nswer is function of t, which mkes sense since the integrnd depends on t. We integrte over nd re left with something tht depends only on t, not. An integrl like b f(, t) d is function of t, so we cn sk bout its t-derivtive, ssuming tht f(, t) is nicely behved. The rule is: the t-derivtive of the integrl of f(, t) is the integrl of the t-derivtive of f(, t): (1.) d dt b f(, t) d 1 b f(, t) d. t
KEITH CONAD This is clled differentition under the integrl sign. If you re used to thinking mostly bout functions with one vrible, not two, keep in mind tht (1.) involves integrls nd derivtives with respect to seprte vribles: integrtion with respect to nd differentition with respect to t. Emple 1.. We sw in Emple 1.1 tht 1 ( + t3 ) d 4/3 + t 3 + t 6, whose t-derivtive is 6t + 6t 5. According to (1.), we cn lso compute the t-derivtive of the integrl like this: The nswers gree. (.1) d dt 1 ( + t 3 ) d 1 1 1 t ( + t3 ) d ( + t 3 )(3t ) d (1t + 6t 5 ) d 6t + 6t 5 1 6t + 6t 5.. Euler s fctoril integrl in new light For integers n, Euler s integrl formul for n! is n e d n!, which cn be obtined by repeted integrtion by prts strting from the formul (.) e d 1 when n. Now we re going to derive Euler s formul in nother wy, by repeted differentition fter introducing prmeter t into (.). For ny t >, let tu. Then d t du nd (.) becomes te tu du 1. Dividing by t nd writing u s (why is this not problem?), we get (.3) e t d 1 t. This is prmetric form of (.), where both sides re now functions of t. We need t > in order tht e t is integrble over the region. Now we bring in differentition under the integrl sign. Differentite both sides of (.3) with respect to t, using (1.) to tret the left side. We obtin so (.4) e t d 1 t, e t d 1 t.
DIFFEENTIATING UNDE THE INTEGAL SIGN 3 Differentite both sides of (.4) with respect to t, gin using (1.) to hndle the left side. We get Tking out the sign on both sides, (.5) e t d t 3. e t d t 3. If we continue to differentite ech new eqution with respect to t few more times, we obtin nd Do you see the pttern? It is (.6) 3 e t d 6 t 4, 4 e t d 4 t 5, 5 e t d 1 t 6. n e t d n! t n+1. We hve used the presence of the etr vrible t to get these equtions by repetedly pplying d/dt. Now specilize t to 1 in (.6). We obtin n e d n!, which is our old friend (.1). Voilá! The ide tht mde this work is introducing prmeter t, using clculus on t, nd then setting t to prticulr vlue so it disppers from the finl formul. In other words, sometimes to solve problem it is useful to solve more generl problem. Compre (.1) to (.6). 3. A dmped sine integrl We re going to use differentition under the integrl sign to prove t sin e d π rctn t for t >. Cll this integrl F (t) nd set f(, t) e t (sin )/, so (/t)f(, t) e t sin. Then F (t) e t (sin ) d. The integrnd e t sin, s function of, cn be integrted by prts: e ( sin cos ) sin d 1 + e. Applying this with t nd turning the indefinite integrl into definite integrl, F (t) e t (t sin + cos ) (sin ) d 1 + t e t.
4 KEITH CONAD As, t sin + cos oscilltes lot, but in bounded wy (since sin nd cos re bounded functions), while the term e t decys eponentilly to since t >. So the vlue t is. Therefore F (t) e t (sin ) d 1 1 + t. We know n eplicit ntiderivtive of 1/(1 + t ), nmely rctn t. Since F (t) hs the sme t-derivtive s rctn t, they differ by constnt: for some number C, (3.1) t sin e d rctn t + C for t >. We ve computed the integrl, up to n dditive constnt, without finding n ntiderivtive of e t (sin )/. To compute C in (3.1), let t on both sides. Since (sin )/ 1, the bsolute vlue of the integrl on the left is bounded from bove by e t d 1/t, so the integrl on the left in (3.1) tends to s t. Since rctn t π/ s t, eqution (3.1) s t becomes π + C, so C π/. Feeding this bck into (3.1), t sin (3.) e d π rctn t for t >. If we let t + in (3.), this eqution suggests tht (3.3) sin d π, which is true nd it is importnt in signl processing nd Fourier nlysis. It is delicte mtter to derive (3.3) from (3.) since the integrl in (3.3) is not bsolutely convergent. Detils re provided in n ppendi. (4.1) The improper integrl formul 4. The Gussin integrl e / d π 1 is fundmentl to probbility theory nd Fourier nlysis. The function π e / is clled Gussin, nd (4.1) sys the integrl of the Gussin over the whole rel line is 1. The physicist Lord Kelvin (fter whom the Kelvin temperture scle is nmed) once wrote (4.1) on the bord in clss nd sid A mthemticin is one to whom tht [pointing t the formul] is s obvious s twice two mkes four is to you. We will prove (4.1) using differentition under the integrl sign. The method will not mke (4.1) s obvious s 4. If you tke further courses you my lern more nturl derivtions of (4.1) so tht the result relly does become obvious. For now, just try to follow the rgument here step-by-step. We re going to im not t (4.1), but t n equivlent formul over the rnge : π π (4.) e / d. Cll the integrl on the left I. For t, set F (t) e t (1+ )/ 1 + d.
DIFFEENTIATING UNDE THE INTEGAL SIGN 5 Then F () d/(1 + ) π/ nd F ( ). Differentiting under the integrl sign, F (t) te t (1+ )/ d te t / Mke the substitution y t, with dy t d, so F (t) e t / e y / dy Ie t /. e (t) / d. For b >, integrte both sides from to b nd use the Fundmentl Theorem of Clculus: Letting b, (5.1) b F (t) dt I b e t / dt F (b) F () I π I I π I π. b e t / dt. I lerned this from Michel ozmn [6], who modified n ide on mth.stckechnge [9]. 5. Higher moments of the Gussin For every integer n we wnt to compute formul for n e / d. (Integrls of the type n f() d for n, 1,,... re clled the moments of f(), so (5.1) is the n-th moment of the Gussin.) When n is odd, (5.1) vnishes since n e / is n odd function. Wht if n,, 4,... is even? The first cse, n, is the Gussin integrl (4.1): (5.) e / d π. To get formuls for (5.1) when n, we follow the sme strtegy s our tretment of the fctoril integrl in Section : stick t into the eponent of e / nd then differentite repetedly with respect to t. For t >, replcing with t in (5.) gives π (5.3) e t / d. t Differentite both sides of (5.3) with respect to t, using differentition under the integrl sign on the left: π / e t d t 3/, so π (5.4) e t / d t 3/. Differentite both sides of (5.4) with respect to t. After removing common fctor of 1/ on both sides, we get (5.5) 4 e t / d 3 π t 5/.
6 KEITH CONAD Differentiting both sides of (5.5) with respect to t few more times, we get nd Quite generlly, when n is even 6 e t / d 3 5 π t 7/, 8 e t / d 3 5 7 π t 9/, 1 e t / d 3 5 7 9 π t 11/. n e t / d 1 3 5 (n 1) π t n/ t, where the numertor is the product of the positive odd integers from 1 to n 1 (understood to be the empty product 1 when n ). In prticulr, tking t 1 we hve computed (5.1): n e / d 1 3 5 (n 1) π. As n ppliction of (5.4), we now compute ( 1 )! : 1/ e d, where the nottion ( 1 )! nd its definition re inspired by Euler s integrl formul (.1) for n! when n is nonnegtive integer. Using the substitution u 1/ in 1/ e d, we hve ( ) 1! 1/ e d We re going to compute by looking t its t-derivtive: ue u (u) du u e u du u e u du π 3/ by (5.4) t t π. 6. A cosine trnsform of the Gussin F (t) (6.1) F (t) cos(t)e / d sin(t)e / d.
DIFFEENTIATING UNDE THE INTEGAL SIGN 7 This is good from the viewpoint of integrtion by prts since e / is the derivtive of e /. So we pply integrtion by prts to (6.1): nd Then u sin(t), dv e d du t cos(t) d, v e /. F (t) uv u dv sin(t) e / sin(t) e / v du t tf (t). cos(t)e / d As, e / blows up while sin(t) stys bounded, so sin(t)/e / goes to. Therefore F (t) tf (t). We know the solutions to this differentil eqution: constnt multiples of e t /. So cos(t)e / d Ce t / for some constnt C. To find C, set t. The left side is e / d, which is π/ by (4.). The right side is C. Thus C π/, so we re done: for ll rel t, π cos(t)e / d / e t. emrk 6.1. If we wnt to compute G(t) sin(t)e / d, with sin(t) in plce of cos(t), then in plce of F (t) tf (t) we hve G (t) 1 tg(t), nd G(). From the differentil eqution, (e t / G(t)) e t /, so G(t) e t / t e / d. So while cos(t)e / d π e t /, the integrl sin(t)e / d is impossible to epress in terms of elementry functions. 7. Logs in the denomintor, prt I Consider the following integrl over [, 1], where t > : 1 t 1 log d. Since 1/ log s +, the integrnd vnishes t. As 1, ( t 1)/ log t. Therefore when t is fied the integrnd is continuous function of on [, 1], so the integrl is not n improper integrl. The t-derivtive of this integrl is 1 t log log d 1 t d 1 t + 1,
8 KEITH CONAD which we recognize s the t-derivtive of log(t + 1). Therefore 1 t 1 log d log(t + 1) + C for some C. To find C, let t +. On the right side, log(1 + t) tends to. On the left side, the integrnd tends to : ( t 1)/ log (e t log 1)/ log t becuse e 1 when. Therefore the integrl on the left tends to s t +. So C, which implies (7.1) 1 t 1 log d log(t + 1) for ll t >, nd it s obviously lso true for t. Another wy to compute this integrl is to write t e t log s power series nd integrte term by term, which is vlid for 1 < t < 1. Under the chnge of vribles e y, (7.1) becomes ( (7.) e y e (t+1)y) dy log(t + 1). y 8. Logs in the denomintor, prt II We now consider the integrl d F (t) t log for t > 1. The integrl converges by comprison with d/ t. We know tht t t 1 the integrl diverges to : d b log lim d b log b lim log log b lim b log log b log log. So we epect tht s t 1 +, F (t) should blow up. But how does it blow up? By nlyzing F (t) nd then integrting bck, we re going to show F (t) behves essentilly like log(t 1) s t 1 +. Using differentition under the integrl sign, for t > 1 ( ) F 1 (t) t t d log t ( log ) d log d t t+1 t + 1 1 t 1 t. We wnt to bound this derivtive from bove nd below when t > 1. Then we will integrte to get bounds on the size of F (t).
DIFFEENTIATING UNDE THE INTEGAL SIGN 9 For t > 1, the difference 1 t is negtive, so 1 t < 1. Dividing both sides of this by 1 t, which is negtive, reverses the sense of the inequlity nd gives 1 t 1 t > 1 1 t. This is lower bound on F (t). To get n upper bound on F (t), we wnt to use lower bound on 1 t. Since e + 1 for ll (the grph of y e lies on or bove its tngent line t, which is y + 1), for ll. Tking 1 t, e log (log ) + 1 (8.1) 1 t (log )(1 t) + 1. When t > 1, 1 t is negtive, so dividing (8.1) by 1 t reverses the sense of the inequlity: 1 t t 1 log + 1 1 t. This is n upper bound on F (t). Putting the upper nd lower bounds on F (t) together, (8.) 1 1 t < F (t) log + 1 1 t for ll t > 1. We re concerned with the behvior of F (t) s t 1 +. Let s integrte (8.) from to, where 1 < < : dt 1 t < F (t) dt Using the Fundmentl Theorem of Clculus, log(t 1) < F (t) so ( log + 1 ) dt. 1 t ((log )t log(t 1)) log( 1) < F () F () (log )( ) + log( 1). Mnipulting to get inequlities on F (), we hve (log )( ) log( 1) + F () F () < log( 1) + F () Since > 1 for 1 < <, (log )( ) is greter thn log. This gives the bounds Writing s t, we get log( 1) + F () log F () < log( 1) + F () log(t 1) + F () log F (t) < log(t 1) + F (), so F (t) is bounded distnce from log(t 1) when 1 < t <. In prticulr, F (t) s t 1 +.,
1 KEITH CONAD 9. Smoothly dividing by t Let h(t) be n infinitely differentible function for ll rel t such tht h(). The rtio h(t)/t mkes sense for t, nd it lso cn be given resonble mening t t : from the very definition of the derivtive, when t we hve Therefore the function h(t) t r(t) h(t) h() h (). t { h(t)/t, if t, h (), if t is continuous for ll t. We cn see immeditely from the definition of r(t) tht it is better thn continuous when t : it is infinitely differentible when t. The question we wnt to ddress is this: is r(t) infinitely differentible t t too? If h(t) hs power series representtion round t, then it is esy to show tht r(t) is infinitely differentible t t by working with the series for h(t). Indeed, write h(t) c 1 t + c t + c 3 t 3 + for ll smll t. Here c 1 h (), c h ()/! nd so on. For smll t, we divide by t nd get (9.1) r(t) c 1 + c t + c 3 t 3 +, which is power series representtion for r(t) for ll smll t. The vlue of the right side of (9.1) t t is c 1 h (), which is lso the defined vlue of r(), so (9.1) is vlid for ll smll (including t ). Therefore r(t) hs power series representtion round (it s just the power series for h(t) t divided by t). Since functions with power series representtions round point re infinitely differentible t the point, r(t) is infinitely differentible t t. However, this is n incomplete nswer to our question bout the infinite differentibility of r(t) t t becuse we know by the key emple of e 1/t (t t ) tht function cn be infinitely differentible t point without hving power series representtion t the point. How re we going to show r(t) h(t)/t is infinitely differentible t t if we don t hve power series to help us out? Might there ctully be counteremple? The solution is to write h(t) in very clever wy using differentition under the integrl sign. Strt with h(t) t h (u) du. (This is correct since h().) For t, introduce the chnge of vribles u t, so du t d. At the boundry, if u then. If u t then 1 (we cn divide the eqution t t by t becuse t ). Therefore Dividing by t when t, we get h(t) 1 h (t)t d t 1 h (t) d. r(t) h(t) h (t) d. t The left nd right sides don t hve ny t in the denomintor. Are they equl t t too? The left side t t is r() h (). The right side is 1 h () d h () too, so (9.) r(t) 1 1 h (t) d
DIFFEENTIATING UNDE THE INTEGAL SIGN 11 for ll t, including t. This is formul for h(t)/t where there is no longer t being divided! Now we re set to use differentition under the integrl sign. The wy we hve set things up here, we wnt to differentite with respect to t; the integrtion vrible on the right is. We cn use differentition under the integrl sign on (9.) when the integrnd is differentible. Since the integrnd is infinitely differentible, r(t) is infinitely differentible! Eplicitly, nd nd more generlly r (t) r (t) r (k) (t) 1 1 1 In prticulr, r (k) () 1 k h (k+1) () d h(k+1) () k+1. vh (t) d vh (t) d k h (k+1) (t) d. 1. Counteremples We hve seen mny emples where differentition under the integrl sign cn be crried out with interesting results, but we hve not ctully stted conditions under which (1.) is vlid. Something does need to be checked. In [8], n incorrect use of differentition under the integrl sign due to Cuchy is discussed, where divergent integrl is evluted s finite epression. Here re two other emples where differentition under the integrl sign does not work. Emple 1.1. It is pointed out in [3, Emple 6] tht the formul sin d π, which we discussed t the end of Section 3, leds to n erroneous instnce of differentition under the integrl sign. ewrite the formul s (1.1) sin(ty) y dy π for ny t >, by the chnge of vribles ty. Then differentition under the integrl sign implies which doesn t mke sense. cos(ty) dy, The net emple shows tht even if both sides of (1.) mke sense, they need not be equl. Emple 1.. For ny rel numbers nd t, let t 3 f(, t) ( + t, if or t, ), if nd t. Let F (t) 1 f(, t) d.
1 KEITH CONAD For instnce, F () 1 f(, ) d 1 d. When t, 1 t 3 F (t) ( + t ) d 1+t t 3 t t3 u u du (where u + t ) u1+t ut t 3 (1 + t ) + t3 t t (1 + t ). This formul lso works t t, so F (t) t/((1 + t )) for ll t. Therefore F (t) is differentible nd F (t) 1 t (1 + t ) f(, t) d. Since f(, t) for ll t, f(, t) is differen- f(, t). For, f(, t) is differentible in t nd for ll t. In prticulr, F () 1. Now we compute t f(, t) nd then 1 tible in t nd t t Combining both cses ( nd ), (1.) t f(, t) ( + t ) (3t ) t 3 ( + t )t ( + t ) 4 t ( + t )(3( + t ) 4t ) ( + t ) 4 t (3 t ) ( + t ) 3. f(, t) t { t (3 t ), ( +t ) 3 if,, if. In prticulr t t f(, t). Therefore t t the left side of the formul d dt 1 is F () 1/ nd the right side is 1 f(, t) d t 1 f(, t) d. t t f(, t) d. The two sides re unequl! The problem in this emple is tht tf(, t) is not continuous function of (, t). Indeed, the denomintor in the formul in (1.) is ( + t ) 3, which hs problem ner (, ). Specificlly, while this derivtive vnishes t (, ), it we let (, t) (, ) long the line t, then on this line tf(, t) hs the vlue 1/(4), which does not tend to s (, t) (, ). Theorem 1.3. The eqution d dt b f(, t) d b f(, t) d t
DIFFEENTIATING UNDE THE INTEGAL SIGN 13 is vlid t t t, in the sense tht both sides eist nd re equl, provided the following two conditions hold: f(, t) nd tf(, t) re continuous functions of two vribles when is in the rnge of integrtion nd t is in some intervl round t, there re upper bounds f(, t) A() nd tf(, t) B(), both being independent of t, such tht b A() d nd b B() d eist. Proof. See [4, pp. 337 339]. If the intervl of integrtion is infinite, b A() d nd b B() d re improper. In Tble 1 we include choices for A() nd B() for ech of the functions we hve treted. Since the clcultion of derivtive t point only depends on n intervl round the point, we hve replced t-rnge such s t > with t c > in some cses to obtin choices for A() nd B(). Section f(, t) rnge t rnge t we wnt A() B() n e t [, ) t c > 1 n e c n+1 e c 3 e t sin (, ) t c > e c e c e 4 t (1+ ) 1 [, 1] t c t 1+ 1+ c 5 n e t t c > 1 n e c n+ e c 6 cos(t)e / [, ) ll t e / e / 7 t 1 log (, 1] < t < c 1 1 c log 1 1 1 8 t log [, ) t c > 1 t > 1 log 9 k h (k+1) (t) [, 1] t < c m Tble 1. Summry y c h(k+1) (y) 1 c m y c h(k+) (y) Corollry 1.4. If (t) nd b(t) re both differentible, then d dt b(t) (t) f(, t) d b(t) (t) if the following conditions re stisfied: t f(, t) d + f(b(t), t)b (t) f((t), t) (t) there re α < β nd c 1 < c such tht f(, t) nd t f(, t) re continuous on [α, β] (c 1, c ), we hve (t) [α, β] nd b(t) [α, β] for ll t (c 1, c ), there re upper bounds f(, t) A() nd t f(, t) B() for (, t) [α, β] (c 1, c ) such tht β α A() d nd β α B() d eist. Proof. This is consequence of Theorem 1.3 nd the chin rule for multivrible functions. Set function of three vribles I(t,, b) b f(, t) d for (t,, b) (c 1, c ) [α, β] [α, β]. (Here nd b re not functions of t.) Then (1.3) I (t,, b) t b I f(, t) d, t I (t,, b) f(, t), (t,, b) f(b, t), b
14 KEITH CONAD where the first formul follows from Theorem 1.3 (its hypotheses re stisfied for ech nd b in [α, β]) nd the second nd third formuls re the Fundmentl Theorem of Clculus. For differentible functions (t) nd b(t) with vlues in [α, β] for c 1 < t < c, by the chin rule d dt b(t) (t) f(, t) d d I(t, (t), b(t)) dt I dt (t, (t), b(t)) t dt + I d (t, (t), b(t)) dt + I db (t, (t), b(t)) b dt b(t) f t (, t) d f((t), t) (t) + f(b(t), t)b (t) by (1.3). (t) A version of differentition under the integrl sign for t comple vrible is in [5, pp. 39 393]. Emple 1.5. For prmetric integrl t f(, t) d, where is fied, Corollry 1.4 tells us tht (1.4) d dt t f(, t) d t f(, t) d + f(t, t) t provided the hypotheses of the corollry re stisfied: (i) there re α < β nd c 1 < c such tht f nd f/t re continuous for (, t) [α, β] (c 1, c ), (ii) α β nd (c 1, c ) [α, β], nd (iii) there re bounds f(, t) A() nd t f(, t) B() for (, t) [α, β] (c 1, c ) such tht the integrls β α A() d nd β α B() d both eist. Let s pply this to the integrl F (t) t log(1 + t) 1 + d where t >. Here f(, t) log(1 + t)/(1 + ) nd t f(, t) 1 (1+t)(1+ ). Fi c > nd use α c 1 nd β c c. Then we cn justify (1.4) for (, t) [, c] (, c) using A() log(1 + c )/(1 + ) nd B() c/(1 + ), so F (t) t t (1 + t)(1 + ) d + log(1 + t ) 1 + t ( 1 t 1 + t 1 + t + t + ) 1 + d + log(1 + t ) 1 + t. Splitting up the integrl into three terms, ( 1 F t (t) log(1 + t) + 1 + t 1 + t rctn() + log(1 + ) ) t (1 + t ) log(1 + t ) 1 + t + t rctn(t) 1 + t + log(1 + t ) (1 + t ) + log(1 + t ) 1 + t t rctn(t) 1 + t + log(1 + t ) (1 + t ). Clerly F (), so by the Fundmentl Theorem of Clculus t t ( y rctn(y) F (t) F (y) dy 1 + y + log(1 + ) y ) (1 + y dy. ) + log(1 + t ) 1 + t
DIFFEENTIATING UNDE THE INTEGAL SIGN 15 Using integrtion by prts on the first integrnd with u rctn(y) nd dv t t t F (t) uv log(1 + y ) v du + (1 + y ) dy rctn(y) log(1 + y t ) t log(1 + y ) t (1 + y ) dy + 1 rctn(t) log(1 + t ). y 1+y dy, log(1 + y ) (1 + y ) dy Thus for < t < c we hve t log(1 + t) 1 + d 1 rctn(t) log(1 + t ). Since c ws rbitrry, this eqution holds for ll t > (nd trivilly t t too). Setting t 1, 1 log(1 + ) 1 + d 1 π log rctn(1) log. 8 11. The Fundmentl Theorem of Algebr By differentiting under the integrl sign we will deduce the fundmentl theorem of lgebr: nonconstnt polynomil p(z) with coefficients in C hs root in C. The proof is due to Schep [7]. Arguing by contrdiction, ssume p(z) for ll z C. For r, consider the following integrl round circle of rdius r centered t the origin: I(r) π dθ p(re iθ ). This integrl mkes sense since the denomintor is never, so 1/p(z) is continuous on C. Let f(θ, r) 1/p(re iθ ), so I(r) π f(θ, r) dθ. We will prove three properties of I(r): (1) Theorem 1.3 cn be pplied to I(r) for r >, () I(r) s r, (3) I(r) I() s r + (continuity t r ). Tking these for grnted, let s see how contrdiction occurs. For r >, Since for r > we hve I (r) π I (r) f(θ, r) dθ r π f(θ, r) dθ r π p (re iθ )e iθ p(re iθ ) θ f(θ, r) p (re iθ ) p(re iθ ) ireiθ ir f(θ, r), r π 1 ir θ f(θ, r) dθ 1 f(θ, r) ir θπ θ dθ. 1 ir ( 1 p(r) 1 ). p(r) Thus I(r) is constnt for r >. Since I(r) s r, the constnt is zero: I(r) for r >. Since I(r) I() s r + we get I(), which is flse since I() π/p(). It remins to prove the three properties of I(r). (1) Theorem 1.3 cn be pplied to I(r) for r > :
16 KEITH CONAD Since p(z) nd p (z) re both continuous on C, the functions f(θ, r) nd (/r)f(θ, r) re continuous for θ [, π] nd ll r. This confirms the first condition in Theorem 1.3. For ech r > the set {(θ, r) : θ [, π], r [, r ] is closed nd bounded, so the functions f(θ, r) nd (/r)f(θ, r) re both bounded bove by constnt (independent of r nd θ) on this set. The rnge of integrtion [, π] is finite, so the second condition in Theorem 1.3 is stisfied using constnts for A(θ) nd B(θ). () I(r) s r : Let p(z) hve leding term cz d, with d deg p(z) 1. As r, p(re iθ ) / re iθ d c >, so for ll lrge r we hve p(re iθ ) c r d /. For such lrge r, I(r) π dθ p(re iθ ) π dθ c r d / 4π c r d, nd the upper bound tends to s r since d >, so I(r) s r. (3) I(r) I() s r + : For r >, π ( 1 (11.1) I(r) I() p(re iθ ) 1 ) π dθ I(r) I() 1 p() p(re iθ ) 1 p() dθ. Since 1/p(z) is continuous t, for ny ε > there is δ > such tht z < δ 1/p(z) 1/p() < ε. Therefore if < r < δ, (11.1) implies I(r) I() π ε dθ πε. 1. An emple needing chnge of vribles Our net emple is tken from [1, pp. 78,84]. For ll t, we will show by differentition under the integrl sign tht cos(t) (1.1) 1 + d πe t. Here f(, t) cos(t)/(1 + ). Since f(, t) is continuous nd f(, t) 1/(1 + ), the integrl eists for ll t. The function πe t is not differentible t t, so we shouldn t epect to be ble to prove (1.1) t t using differentition under the integrl sign; this specil cse cn be treted with elementry clculus: d 1 + rctn π. The integrl in (1.1) is n even function of t, so to compute it for t it suffices to tret the cse t >. 1 Let cos(t) F (t) 1 + d. If we try to compute F (t) for t > using differentition under the integrl sign, we get ( ) (1.) F (t)? cos(t) sin(t) t 1 + d 1 + d. Unfortuntely, there is no upper bound tf(, t) B() tht justifies differentiting F (t) under the integrl sign (or even justifies tht F (t) is differentible). Indeed, when is ner lrge odd multiple of (π/)/t, the integrnd in (1.) hs vlues tht re pproimtely /(1 + ) 1/, which is not integrble for lrge. Tht does not men (1.) is ctully flse, lthough if we weren t lredy told the nswer on the right side of (1.1) then we might be suspicious bout 1 A reder who knows comple nlysis cn derive (1.1) for t > by the residue theorem, viewing cos(t) s the rel prt of e it.
DIFFEENTIATING UNDE THE INTEGAL SIGN 17 whether the integrl is differentible for ll t > ; fter ll, you cn t esily tell from the integrl tht it is not differentible t t. Hving lredy rised suspicions bout (1.), we cn get something relly crzy if we differentite under the integrl sign second time: F (t)? cos(t) 1 + d. If this mde sense then (1.3) F ( + 1) cos(t) (t) F (t) 1 + d cos(t) d???. All is not lost! Let s mke chnge of vribles. Fiing t >, set y t, so dy t d nd F (t) cos y dy 1 + y /t t t cos y t + y dy. This new integrl will be ccessible to differentition under the integrl sign. (Although the new integrl is n odd function of t while F (t) is n even function of t, there is no contrdiction becuse this new integrl ws derived only for t >.) Fi c > c >. For t (c, c ), the integrnd in t cos y t + y dy is bounded bove in bsolute vlue by t/(t + y ) c /(c + y ), which is independent of t nd integrble over. The t-prtil derivtive of the integrnd is (y t )(cos y)/(t + y ), which is bounded bove in bsolute vlue by (y + t )/(t + y ) 1/(t + y ) 1/(c + y ), which is independent of t nd integrble over. This justifies the use differentition under the integrl sign ccording to Theorem 1.3: for c < t < c, nd hence for ll t > since we never specified c or c, ( ) F t cos y y t (t) t t + y dy (t + y cos y dy. ) We wnt to compute F (t) using differentition under the integrl sign. For < c < t < c, the t-prtil derivtive of the integrnd for F (t) is bounded bove in bsolute vlue by function of y tht is independent of t nd integrble over (eercise), so for ll t > we hve F (t) t ( t cos y t + y ) dy t ( t t + y It turns out tht ( /t )(t/(t + y )) ( /y )(t/(t + y )), so F ( ) t (t) y t + y cos y dy. ) cos y dy. Using integrtion by prts on this formul for F (t) twice (strting with u cos y nd dv ( /y )(t/(t + y )), we obtin ( ) ( ) F t t (t) y t + y sin y dy t + y cos y dy F (t). The eqution F (t) F (t) is second order liner ODE whose generl solution is e t + be t, so cos(t) (1.4) 1 + d et + be t
18 KEITH CONAD for ll t > nd some rel constnts nd b. To determine nd b we look t the behvior of the integrl in (1.4) s t + nd s t. As t +, the integrnd in (1.4) tends pointwise to 1/(1 + ), so we epect the integrl tends to d/(1 + ) π s t +. To justify this, we will bound the bsolute vlue of the difference cos(t) 1 + d d 1 + N cos(t) 1 1 + d by n epression tht is rbitrrily smll s t +. For ny N >, brek up the integrl over into the regions N nd N. We hve cos(t) 1 cos(t) 1 1 + d N 1 + d + N 1 + d t 1 + d + 1 + d t N N ( π ) 1 + d + 4 rctn N. Tking N sufficiently lrge, we cn mke π/ rctn N s smll s we wish, nd fter doing tht we cn mke the first term s smll s we wish by tking t sufficiently smll. eturning to (1.4), letting t + we obtin π + b, so (1.5) cos(t) 1 + d et + (π )e t for ll t >. Now let t in (1.5). The integrl tends to by the iemnn Lebesgue lemm from Fourier nlysis, lthough we cn eplin this concretely in our specil cse: using integrtion by prts with u 1/(1 + ) nd dv cos(t) d, we get cos(t) 1 + d 1 t sin(t) (1 + ) d. The bsolute vlue of the term on the right is bounded bove by constnt divided by t, which tends to s t. Therefore e t + (π )e t s t. This forces, which completes the proof tht F (t) πe t for t >. 13. Eercises t sin 1. From the formul e d π rctn t for t >, in Section 3, use chnge sin(b) of vribles to obtin formul for e d when nd b re positive. Then use differentition under the integrl sign with respect to b to find formul for e cos(b) d when nd b re positive. (Differentition under the integrl sign with respect to will produce formul for e sin(b) d, but tht would be circulr in our pproch since we used tht integrl in our derivtion of the formul for t sin e d in Section 3.)
. From the formul with > implies DIFFEENTIATING UNDE THE INTEGAL SIGN 19 t sin e d π rctn t for t >, the chnge of vribles y ty sin(y) e dy π rctn t, y so the integrl on the left is independent of nd thus hs -derivtive. Differentition under the integrl sign, with respect to, implies e ty (cos(y) t sin(y)) dy. Verify tht this ppliction of differentition under the integrl sign is vlid when > nd t >. Wht hppens if t? sin(t) 3. Show ( + 1) d π (1 e t ) for t > by justifying differentition under the integrl sign nd using (1.1). ( ) t cos 1 t 4. Prove tht e d log for t >. Wht hppens to the integrl 1 + t s t +? 5. Prove tht for > nd b >. 6. Prove tht log(1 + t ) 1 + d π log(1 + t) for t > (it is obvious for t ). Then deduce (e e t ) d log(1 + ) b + sign. This is (7.) for t > 1. Deduce tht d π log(1 + b) b log t for t > by justifying differentition under the integrl (e e b ) d 7. In clculus tetbooks, formuls for the indefinite integrls n sin d nd n cos d log(b/) for positive nd b. re derived recursively using integrtion by prts. Find formuls for these integrls when n 1,, 3, 4 using differentition under the integrl sign strting with the formuls cos(t) d sin(t), sin(t) d cos(t) t t for t >. 8. If you re fmilir with integrtion of comple-vlued functions, show for y tht e (+iy) d π. In other words, show the integrl on the left side is independent of y. (Hint: Use differentition under the integrl sign to compute the y-derivtive of the left side.)
KEITH CONAD Appendi A. Justifying pssge to the limit in sine integrl In Section 3 we derived the eqution (A.1) t sin e d π rctn t for t >, which by nive pssge to the limit s t + suggests tht (A.) To prove (A.) is correct, we will show (A.3) sin d sin d π. sin d eists nd then show the difference t sin e d (1 e t ) sin tends to s t +. The key in both cses is lternting series. On the intervl [kπ, (k + 1)π], where k is n integer, we cn write sin ( 1) k sin, so convergence of sin d lim b sin b d is equivlent to convergence of the series k (k+1)π kπ sin d d (k+1)π ( 1) k k kπ This is n lternting series in which the terms k (k+1)π kπ k+1 (k+)π (k+1)π sin d (k+1)π kπ sin( + π) + π sin sin (k+1)π d d. d re monotoniclly decresing: kπ sin + π d < k. By simple estimte k 1 kπ for k 1, so k. Thus sin d k ( 1)k k converges. To show the right side of (A.3) tends to s t +, we write it s n lternting series. Breking up the intervl of integrtion [, ) into union of intervls [kπ, (k + 1)π] for k, (A.4) (1 e t ) sin d ( 1) k I k (t), where I k (t) k (k+1)π kπ (1 e t sin ) d. Since 1 e t > for t > nd >, the series k ( 1)k I k (t) is lternting. The upper bound 1 e t < 1 tells us I k (t) 1 kπ for k 1, so I k(t) s k. To show the terms I k (t) re monotoniclly decresing with k, set this up s the inequlity (A.5) I k (t) I k+1 (t) > for t >. Ech I k (t) is function of t for ll t, not just t > (note I k (t) only involves integrtion on bounded intervl). The difference I k (t) I k+1 (t) vnishes when t (in fct both terms re then ), nd I k (t) (k+1)π kπ e t sin d for ll t by differentition under the integrl sign, so (A.5) would follow from the derivtive inequlity I k (t) I k+1 (t) > for t >. By chnge of vribles y π in the integrl for I k+1 (t), (k+1)π (k+1)π I k+1 (t) e t(y+π) sin(y + π) dy e tπ e ty sin y dy < I k (t). kπ This completes the proof tht the series in (A.4) for t > stisfies the lternting series test. kπ
DIFFEENTIATING UNDE THE INTEGAL SIGN 1 If we truncte the series k ( 1)k I k (t) fter the Nth term, the mgnitude of the error is no greter thn the bsolute vlue of the net term: N ( 1) k I k (t) ( 1) k 1 I k (t) + r N, r N I N+1 (t) (N + 1)π. k Since 1 e t t, N ( 1) k I k (t) k k (N+1)π (1 e t sin (N+1)π ) d t d t(n + 1)π. Thus (1 e t ) sin d t(n + 1)π + 1 (N + 1)π. For ny ε > we cn mke the second term t most ε/ by suitble choice of N. Then the first term is t most ε/ for ll smll enough t (depending on N), nd tht shows (A.3) tends to s t +. eferences [1] W. Appel, Mthemtics for Physics nd Physicists, Princeton Univ. Press, Princeton, 7. []. P. Feynmn, Surely You re Joking, Mr. Feynmn!, Bntm, New York, 1985. [3] S. K. Goel nd A. J. Zjt, Prmetric Integrtion Techniques, Mth. Mg. 6 (1989), 318 3. [4] S. Lng, Undergrdute Anlysis, nd ed., Springer-Verlg, New York, 1997. [5] S. Lng, Comple Anlysis, 3rd ed., Springer-Verlg, New York, 1993. [6] M. ozmn, Evlute Gussin integrl using differentition under the integrl sign, Course notes for Physics 4 (UConn), Spring 16. [7] A.. Schep, A Simple Comple Anlysis nd n Advnced Clculus Proof of the Fundmentl Theorem of Algebr, Amer. Mth. Monthly 116 (9), 67 68. [8] E. Tlvil, Some Divergent Trigonometric Integrls, Amer. Mth. Monthly 18 (1), 43 436. [9] http://mth.stckechnge.com/questions/3985/integrting-int-infty--e--d-using-feynmnsprmetriztion-trick