DIFFERENTIATING UNDER THE INTEGRAL SIGN

DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] showed how to differentiate parameters under the integral sign it s a certain operation. It turns out that s not taught very much in the universities; they don t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. [If] guys at MIT or Princeton had trouble doing a certain integral, [then] I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my bo of tools was different from everybody else s, and they had tried all their tools on it before giving the problem to me. ichard Feynman [, pp. 71 7] 1. Introduction The method of differentiation under the integral sign, due originally to Leibniz, concerns integrals depending on a parameter, such as 1 e t d. Here t is the etra parameter. (Since is the variable of integration, is not a parameter.) In general, we might write such an integral as (1.1) b a f(, t) d, where f(, t) is a function of two variables like f(, t) e t. Eample 1.1. Let f(, t) ( + t 3 ). Then 1 f(, t) d 1 ( + t 3 ) d. An anti-derivative 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 answer is a function of t, which makes sense since the integrand depends on t. We integrate over and are left with something that depends only on t, not. An integral like b a f(, t) d is a function of t, so we can ask about its t-derivative, assuming that f(, t) is nicely behaved. The rule is: the t-derivative of the integral of f(, t) is the integral of the t-derivative of f(, t): (1.) d dt b a f(, t) d 1 b a f(, t) d. t

KEITH CONAD This is called differentiation under the integral sign. If you are used to thinking mostly about functions with one variable, not two, keep in mind that (1.) involves integrals and derivatives with respect to separate variables: integration with respect to and differentiation with respect to t. Eample 1.. We saw in Eample 1.1 that 1 ( + t3 ) d 4/3 + t 3 + t 6, whose t-derivative is 6t + 6t 5. According to (1.), we can also compute the t-derivative of the integral like this: The answers agree. (.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 factorial integral in a new light For integers n, Euler s integral formula for n! is n e d n!, which can be obtained by repeated integration by parts starting from the formula (.) e d 1 when n. Now we are going to derive Euler s formula in another way, by repeated differentiation after introducing a parameter t into (.). For any t >, let tu. Then d t du and (.) becomes te tu du 1. Dividing by t and writing u as (why is this not a problem?), we get (.3) e t d 1 t. This is a parametric form of (.), where both sides are now functions of t. We need t > in order that e t is integrable over the region. Now we bring in differentiation under the integral sign. Differentiate both sides of (.3) with respect to t, using (1.) to treat the left side. We obtain so (.4) e t d 1 t, e t d 1 t.

DIFFEENTIATING UNDE THE INTEGAL SIGN 3 Differentiate both sides of (.4) with respect to t, again using (1.) to handle the left side. We get Taking out the sign on both sides, (.5) e t d t 3. e t d t 3. If we continue to differentiate each new equation with respect to t a few more times, we obtain and Do you see the pattern? 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 have used the presence of the etra variable t to get these equations by repeatedly applying d/dt. Now specialize t to 1 in (.6). We obtain n e d n!, which is our old friend (.1). Voilá! The idea that made this work is introducing a parameter t, using calculus on t, and then setting t to a particular value so it disappears from the final formula. In other words, sometimes to solve a problem it is useful to solve a more general problem. Compare (.1) to (.6). 3. A damped sine integral We are going to use differentiation under the integral sign to prove t sin e d π arctan t for t >. Call this integral F (t) and set f(, t) e t (sin )/, so ( / t)f(, t) e t sin. Then F (t) e t (sin ) d. The integrand e t sin, as a function of, can be integrated by parts: e a (a sin cos ) sin d 1 + a e a. Applying this with a t and turning the indefinite integral into a definite integral, F (t) e t (t sin + cos ) (sin ) d 1 + t e t.

4 KEITH CONAD As, t sin + cos oscillates a lot, but in a bounded way (since sin and cos are bounded functions), while the term e t decays eponentially to since t >. So the value at is. Therefore F (t) e t (sin ) d 1 1 + t. We know an eplicit antiderivative of 1/(1 + t ), namely arctan t. Since F (t) has the same t-derivative as arctan t, they differ by a constant: for some number C, t sin (3.1) e d arctan t + C for t >. We ve computed the integral, up to an additive constant, without finding an antiderivative of e t (sin )/. To compute C in (3.1), let t on both sides. Since (sin )/ 1, the absolute value of the integral on the left is bounded from above by e t d 1/t, so the integral on the left in (3.1) tends to as t. Since arctan t π/ as t, equation (3.1) as t becomes π + C, so C π/. Feeding this back into (3.1), t sin (3.) e d π arctan t for t >. If we let t + in (3.), this equation suggests that sin (3.3) d π, which is true and it is important in signal processing and Fourier analysis. It is a delicate matter to derive (3.3) from (3.) since the integral in (3.3) is not absolutely convergent. Details are provided in an appendi. (4.1) The improper integral formula 4. The Gaussian integral e / d π 1 is fundamental to probability theory and Fourier analysis. The function π e / is called a Gaussian, and (4.1) says the integral of the Gaussian over the whole real line is 1. The physicist Lord Kelvin (after whom the Kelvin temperature scale is named) once wrote (4.1) on the board in a class and said A mathematician is one to whom that [pointing at the formula] is as obvious as twice two makes four is to you. We will prove (4.1) using differentiation under the integral sign. The method will not make (4.1) as obvious as 4. If you take further courses you may learn more natural derivations of (4.1) so that the result really does become obvious. For now, just try to follow the argument here step-by-step. We are going to aim not at (4.1), but at an equivalent formula over the range : π π (4.) e / d. For t >, set ( t A(t) e d) /. We want to calculate A( ) and then take a square root.

Differentiating with respect to t, Let ty, so DIFFEENTIATING UNDE THE INTEGAL SIGN 5 A (t) t A (t) e t / e / d e t / e t / 1 te t y / dy 1 t e / d. te (1+y )t / dy. The function under the integral sign is easily antidifferentiated with respect to t: A (t) 1 e (1+y )t / t 1 + y dy d dt 1 e (1+y )t / 1 + y dy. Letting 1 e (1+ )t / B(t) 1 + d, we have A (t) B (t) for all t >, so there is a constant C such that (4.3) A(t) B(t) + C for all t >. To find C, we let t + in (4.3). The left side tends to ( e d) while the right side tends to 1 d/(1 + ) + C π/ + C. Thus C π/, so (4.3) becomes ( t ) e / d π 1 e (1+ )t / 1 + d. Letting t here, we get ( e / d) π/, so e / d π/. That is (4.). (5.1) 5. Higher moments of the Gaussian For every integer n we want to compute a formula for n e / d. (Integrals of the type n f() d for n, 1,,... are called the moments of f(), so (5.1) is the n-th moment of the Gaussian.) When n is odd, (5.1) vanishes since n e / is an odd function. What if n,, 4,... is even? The first case, n, is the Gaussian integral (4.1): (5.) e / d π. To get formulas for (5.1) when n, we follow the same strategy as our treatment of the factorial integral in Section : stick a t into the eponent of e / and then differentiate repeatedly with respect to t. For t >, replacing with t in (5.) gives π (5.3) e t / d. t Differentiate both sides of (5.3) with respect to t, using differentiation under the integral sign on the left: π / e t d t 3/,

6 KEITH CONAD so (5.4) e t / d π t 3/. Differentiate both sides of (5.4) with respect to t. After removing a common factor of 1/ on both sides, we get (5.5) 4 e t / d 3 π t 5/. Differentiating both sides of (5.5) with respect to t a few more times, we get and Quite generally, 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 numerator is the product of the positive odd integers from 1 to n 1 (understood to be the empty product 1 when n ). In particular, taking t 1 we have computed (5.1): n e / d 1 3 5 (n 1) π. As an application of (5.4), we now compute ( 1 )! : 1/ e d, where the notation ( 1 )! and its definition are inspired by Euler s integral formula (.1) for n! when n is a nonnegative integer. Using the substitution u 1/ in 1/ e d, we have ( ) 1! 1/ e d ue u (u) du u e u du u e u du π 3/ by (5.4) at t π.

We are going to compute by looking at its t-derivative: DIFFEENTIATING UNDE THE INTEGAL SIGN 7 6. A cosine transform of the Gaussian F (t) (6.1) F (t) cos(t)e / d sin(t)e / d. This is good from the viewpoint of integration by parts since e / is the derivative of e /. So we apply integration by parts to (6.1): u sin(t), dv e d and Then 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) stays bounded, so sin(t)/e / goes to. Therefore F (t) tf (t). We know the solutions to this differential equation: constant multiples of e t /. So cos(t)e / d Ce t / for some constant 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 are done: for all real t, π cos(t)e / d / e t. emark 6.1. If we want to compute G(t) sin(t)e / d, with sin(t) in place of cos(t), then in place of F (t) tf (t) we have G (t) 1 tg(t), and G(). From the differential equation, (e t / G(t)) e t /, so G(t) e t / t e / d. So while cos(t)e / d π e t /, the integral sin(t)e / d is impossible to epress in terms of elementary functions.

8 KEITH CONAD 7. Logs in the denominator, part I Consider the following integral over [, 1], where t > : 1 t 1 log d. Since 1/ log as +, the integrand vanishes at. As 1, ( t 1)/ log. Therefore when t is fied the integrand is a continuous function of on [, 1], so the integral is not an improper integral. The t-derivative of this integral is 1 t log log d 1 t d 1 1 + t, which we recognize as the t-derivative of log(1 + t). Therefore 1 t 1 log d log(1 + t) + C. To find C, let t +. On the right side the term log(1 + t) tends to. On the left side, the integrand tends to : ( t 1)/ log (e t log 1)/ log t because e a 1 a when a. Therefore the integral on the left tends to. So C, which implies for all t >. For eample, 1 t 1 log 1 d log(1 + t) 1 d log. log We computed this definite integral without computing an anti-derivative of ( 1)/ log. We now consider the integral 8. Logs in the denominator, part II F (t) d t log for t > 1. The integral converges by comparison with d/ t. We know that at t 1 the integral diverges to : d b log lim d b log b lim log log b lim b log log b log log. So we epect that as t 1 +, F (t) should blow up. But how does it blow up? By analyzing F (t) and then integrating back, we are going to show F (t) behaves essentially like log(t 1) as t 1 +.

DIFFEENTIATING UNDE THE INTEGAL SIGN 9 Using differentiation under the integral sign, for t > 1 F (t) t ( 1 t log ) d t ( log ) d log d t t+1 t + 1 1 t 1 t. We want to bound this derivative from above and below when t > 1. Then we will integrate to get bounds on the size of F (t). For t > 1, the difference 1 t is negative, so 1 t < 1. Dividing both sides of this by 1 t, which is negative, reverses the sense of the inequality and gives 1 t 1 t > 1 1 t. This is a lower bound on F (t). To get an upper bound on F (t), we want to use a lower bound on 1 t. Since e a a + 1 for all a (the graph of y e lies on or above its tangent line at, which is y + 1), e log (log ) + 1 for all. Taking 1 t, (8.1) 1 t (log )(1 t) + 1. When t > 1, 1 t is negative, so dividing (8.1) by 1 t reverses the sense of the inequality: 1 t t 1 log + 1 1 t. This is an upper bound on F (t). Putting the upper and lower bounds on F (t) together, 1 (8.) 1 t < F (t) log + 1 1 t for all t > 1. We are concerned with the behavior of F (t) as t 1 +. Let s integrate (8.) from a to, where 1 < a < : dt ( a 1 t < F (t) dt log + 1 ) dt. a a 1 t Using the Fundamental Theorem of Calculus, log(t 1) < F (t) ((log )t log(t 1)), so a a log(a 1) < F () F (a) (log )( a) + log(a 1). Manipulating to get inequalities on F (a), we have (log )(a ) log(a 1) + F () F (a) < log(a 1) + F () a

1 KEITH CONAD Since a > 1 for 1 < a <, (log )(a ) is greater than log. This gives the bounds Writing a as t, we get log(a 1) + F () log F (a) < log(a 1) + F () log(t 1) + F () log F (t) < log(t 1) + F (), so F (t) is a bounded distance from log(t 1) when 1 < t <. In particular, F (t) as t 1 +. 9. Smoothly dividing by t Let h(t) be an infinitely differentiable function for all real t such that h(). The ratio h(t)/t makes sense for t, and it also can be given a reasonable meaning at t : from the very definition of the derivative, when t we have Therefore the function h(t) t r(t) h(t) h() t h (). { h(t)/t, if t, h (), if t is continuous for all t. We can see immediately from the definition of r(t) that it is better than continuous when t : it is infinitely differentiable when t. The question we want to address is this: is r(t) infinitely differentiable at t too? If h(t) has a power series representation around t, then it is easy to show that r(t) is infinitely differentiable at t by working with the series for h(t). Indeed, write h(t) c 1 t + c t + c 3 t 3 + for all small t. Here c 1 h (), c h ()/! and so on. For small t, we divide by t and get (9.1) r(t) c 1 + c t + c 3 t 3 +, which is a power series representation for r(t) for all small t. The value of the right side of (9.1) at t is c 1 h (), which is also the defined value of r(), so (9.1) is valid for all small (including t ). Therefore r(t) has a power series representation around (it s just the power series for h(t) at divided by t). Since functions with power series representations around a point are infinitely differentiable at the point, r(t) is infinitely differentiable at t. However, this is an incomplete answer to our question about the infinite differentiability of r(t) at t because we know by the key eample of e 1/t (at t ) that a function can be infinitely differentiable at a point without having a power series representation at the point. How are we going to show r(t) h(t)/t is infinitely differentiable at t if we don t have a power series to help us out? Might there actually be a countereample? The solution is to write h(t) in a very clever way using differentiation under the integral sign. Start with h(t) t h (u) du. (This is correct since h().) For t, introduce the change of variables u t, so du t d. At the boundary, if u then. If u t then 1 (we can divide the equation t t by t because t ). Therefore h(t) 1 h (t)t d t 1 h (t) d.

Dividing by t when t, we get DIFFEENTIATING UNDE THE INTEGAL SIGN 11 r(t) h(t) t 1 h (t) d. The left and right sides don t have any t in the denominator. Are they equal at t too? The left side at t is r() h (). The right side is 1 h () d h () too, so (9.) r(t) 1 h (t) d for all t, including t. This is a formula for h(t)/t where there is no longer a t being divided! Now we re set to use differentiation under the integral sign. The way we have set things up here, we want to differentiate with respect to t; the integration variable on the right is. We can use differentiation under the integral sign on (9.) when the integrand is differentiable. Since the integrand is infinitely differentiable, r(t) is infinitely differentiable! Eplicitly, and and more generally r (t) r (t) r (k) (t) 1 1 1 In particular, 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. Countereamples We have seen many eamples where differentiation under the integral sign can be carried out with interesting results, but we have not actually stated conditions under which (1.) is valid. Something does need to be checked. In [6], an incorrect use of differentiation under the integral sign due to Cauchy is discussed, where a divergent integral is evaluated as a finite epression. Here are two other eamples where differentiation under the integral sign does not work. Eample 1.1. It is pointed out in [3] that the formula sin d π, which we discussed at the end of Section 3, can be rewritten as (1.1) sin(ty) y dy π, for any t >, by the change of variables ty. Then differentiation under the integral sign implies which doesn t make sense. cos(ty) dy, The net eample shows that even if both sides of (1.) make sense, they need not be equal.

1 KEITH CONAD Eample 1.. For any real numbers and t, let t 3 f(, t) ( + t, if or t, ), if and t. Let F (t) 1 f(, t) d. For instance, F () 1 f(, ) d 1 d. When t, F (t) 1 t 3 ( + 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 formula also works at t, so F (t) t/((1 + t )) for all t. Therefore F (t) is differentiable and F (t) 1 t (1 + t ) f(, t) d. Since f(, t) for all t, f(, t) is differen- f(, t). For, f(, t) is differentiable in t and for all t. In particular, F () 1. Now we compute t f(, t) and then 1 tiable in t and t t Combining both cases ( and ), (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 particular t t f(, t). Therefore at t the left side of the formula d dt 1 is F () 1/ and the right side is 1 f(, t) d t 1 f(, t) d. t t f(, t) d. The two sides are unequal!

DIFFEENTIATING UNDE THE INTEGAL SIGN 13 The problem in this eample is that tf(, t) is not a continuous function of (, t). Indeed, the denominator in the formula in (1.) is ( + t ) 3, which has a problem near (, ). Specifically, while this derivative vanishes at (, ), it we let (, t) (, ) along the line t, then on this line tf(, t) has the value 1/(4), which does not tend to as (, t) (, ). Theorem 1.3. The two sides of (1.) both eist and are equal at a point t t provided the following two conditions hold: f(, t) and tf(, t) are continuous functions of two variables when is in the range of integration and t is in some interval around t, there are upper bounds f(, t) A() and tf(, t) B(), both being independent of t, such that b a A() d and b a B() d converge. Proof. See [4, pp. 337 339]. In Table 1 we include choices for A() and B() for each of the functions we have treated. Since the calculation of a derivative at a point only depends on an interval around the point, we have replaced a t-range such as t > with t c > in some cases to obtain choices for A() and B(). Section f(, t) range t range t we want 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 / [, ) all 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 ma Table 1. Summary y c h(k+1) (y) 1 c ma y c h(k+) (y) A version of differentiation under the integral sign for t a comple variable is in [5, pp. 39 393]. 11. An eample needing a change of variables Our net eample is taken from [1, pp. 78,84]. For all t, we will show by differentiation under the integral sign that cos(t) (11.1) 1 + d πe t. Here f(, t) cos(t)/(1 + ). Since f(, t) is continuous and f(, t) 1/(1 + ), the integral eists for all t. The function πe t is not differentiable at t, so we shouldn t epect to be able to prove (11.1) at t using differentiation under the integral sign; this special case can be treated with elementary calculus: d 1 + arctan π.

14 KEITH CONAD The integral in (11.1) is an even function of t, so to compute it for t it suffices to treat the case t >. 1 Let cos(t) F (t) 1 + d. If we try to compute F (t) for t > using differentiation under the integral sign, we get ( ) (11.) F (t)? cos(t) sin(t) t 1 + d 1 + d. Unfortunately, there is no upper bound tf(, t) B() that justifies differentiating F (t) under the integral sign (or even justifies that F (t) is differentiable). Indeed, when is near a large odd multiple of (π/)/t, the integrand in (11.) has values that are approimately /(1 + ) 1/, which is not integrable for large. That does not mean (11.) is actually false, although if we weren t already told the answer on the right side of (11.1) then we might be suspicious about whether the integral is differentiable for all t > ; after all, you can t easily tell from the integral that it is not differentiable at t. Having already raised suspicions about (11.), we can get something really crazy if we differentiate under the integral sign a second time: If this made sense then (11.3) F (t) F (t) F (t)? cos(t) 1 + d. ( + 1) cos(t) 1 + d cos(t) d???. All is not lost! Let s make a change of variables. Fiing t >, set y t, so dy t d and cos y dy t cos y F (t) 1 + y /t t t + y dy. This new integral will be accessible to differentiation under the integral sign. (Although the new integral is an odd function of t while F (t) is an even function of t, there is no contradiction because this new integral was derived only for t >.) Fi c > c >. For t (c, c ), the integrand in t cos y t + y dy is bounded above in absolute value by t/(t + y ) c /(c + y ), which is independent of t and integrable over. The t-partial derivative of the integrand is (y t )(cos y)/(t + y ), which is bounded above in absolute value by (y + t )/(t + y ) 1/(t + y ) 1/(c + y ), which is independent of t and integrable over. This justifies the use differentiation under the integral sign according to Theorem 1.3: for c < t < c, and hence for all t > since we never specified c or c, F (t) t ( t cos y t + y ) dy y t (t + y cos y dy. ) 1 A reader who knows comple analysis can derive (11.1) for t > by the residue theorem, viewing cos(t) as the real part of e it.

DIFFEENTIATING UNDE THE INTEGAL SIGN 15 We want to compute F (t) using differentiation under the integral sign. For < c < t < c, the t-partial derivative of the integrand for F (t) is bounded above in absolute value by a function of y that is independent of t and integrable over (eercise), so for all t > we have F ( ) t cos y ( ) t (t) t t + y dy t t + y cos y dy. It turns out that ( / t )(t/(t + y )) ( / y )(t/(t + y )), so F ( ) t (t) y t + y cos y dy. Using integration by parts on this formula for F (t) twice (starting with u cos y and dv ( / y )(t/(t + y )), we obtain ( ) ( ) F t t (t) y t + y sin y dy t + y cos y dy F (t). The equation F (t) F (t) is a second order linear ODE whose general solution is ae t + be t, so cos(t) (11.4) 1 + d aet + be t for all t > and some real constants a and b. To determine a and b we look at the behavior of the integral in (11.4) as t + and as t. As t +, the integrand in (11.4) tends pointwise to 1/(1 + ), so we epect the integral tends to d/(1 + ) π as t +. To justify this, we will bound the absolute value of the difference cos(t) 1 + d d 1 + N cos(t) 1 1 + d by an epression that is arbitrarily small as t +. For any N >, break up the integral over into the regions N and N. We have 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 arctan N. Taking N sufficiently large, we can make π/ arctan N as small as we wish, and after doing that we can make the first term as small as we wish by taking t sufficiently small. eturning to (11.4), letting t + we obtain π a + b, so (11.5) cos(t) 1 + d aet + (π a)e t for all t >. Now let t in (11.5). The integral tends to by the iemann Lebesgue lemma from Fourier analysis, although we can eplain this concretely in our special case: using integration by parts with u 1/(1 + ) and dv cos(t) d, we get cos(t) 1 + d 1 t sin(t) (1 + ) d.

16 KEITH CONAD The absolute value of the term on the right is bounded above by a constant divided by t, which tends to as t. Therefore ae t + (π a)e t as t. This forces a, which completes the proof that F (t) πe t for t >. 1. Eercises t sin 1. From the formula e d π arctan t for t >, in Section 3, use a change a sin(b) of variables to obtain a formula for e d when a and b are positive. Then use differentiation under the integral sign with respect to b to find a formula for e a cos(b) d when a and b are positive. (Differentiation under the integral sign with respect to a will produce a formula for e a sin(b) d, but that would be circular in our approach since we used that integral in our derivation of the formula for. From the formula with a > implies t sin e d.) t sin e d π arctan t for t >, the change of variables ay tay sin(ay) e dy π arctan t, y so the integral on the left is independent of a and thus has a-derivative. Differentiation under the integral sign, with respect to a, implies e tay (cos(ay) t sin(ay)) dy. Verify that this application of differentiation under the integral sign is valid when a > and t >. What happens if t? ( ) t cos 1 t 3. Prove for t > that e d log. What happens to the integral 1 + t as t +? 4. In calculus tetbooks, formulas for the indefinite integrals n sin d and n cos d are derived recursively using integration by parts. Find formulas for these integrals when n 1,, 3, 4 using differentiation under the integral sign starting with the formulas cos(t) d sin(t), sin(t) d cos(t) t t for t >. 5. If you are familiar with integration of comple-valued functions, show e (+iy) d π

DIFFEENTIATING UNDE THE INTEGAL SIGN 17 for all y. In other words, show the integral on the left side is independent of y. (Hint: Use differentiation under the integral sign to compute the y-derivative of the left side.) Appendi A. Justifying passage to the limit in a sine integral In Section 3 we derived the equation (A.1) t sin e d π arctan t for t >, which by naive passage to the limit as t + suggests that (A.) To prove (A.) is correct, we will show (A.3) sin d sin d π. sin d eists and then show the difference t sin e d (1 e t ) sin tends to as t +. The key in both cases is alternating series. On the interval [kπ, (k + 1)π], where k is an integer, we can write sin ( 1) k sin, so convergence of sin d lim b sin b d is equivalent to convergence of the series (k+1)π sin kπ d (k+1)π ( 1) k sin d. k kπ k This is an alternating series in which the terms a k (k+1)π a k+1 (k+)π (k+1)π sin d (k+1)π kπ d kπ sin( + π) + π sin (k+1)π d d are monotonically decreasing: kπ sin + π d < a k. By a simple estimate a k 1 kπ for k 1, so a k. Thus sin d k ( 1)k a k converges. To show the right side of (A.3) tends to as t +, we write it as an alternating series. Breaking up the interval of integration [, ) into a union of intervals [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 > and >, the series k ( 1)k I k (t) is alternating. The upper bound 1 e t < 1 tells us I k (t) 1 kπ for k 1, so I k(t) as k. To show the terms I k (t) are monotonically decreasing with k, set this up as the inequality (A.5) I k (t) I k+1 (t) > for t >. Each I k (t) is a function of t for all t, not just t > (note I k (t) only involves integration on a bounded interval). The difference I k (t) I k+1 (t) vanishes when t (in fact both terms are then ), and I k (t) (k+1)π kπ e t sin d for all t by differentiation under the integral sign, so (A.5) would follow from the derivative inequality I k (t) I k+1 (t) > for t >. By a change of variables y π in the integral 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 that the series in (A.4) for t > satisfies the alternating series test. kπ

18 KEITH CONAD If we truncate the series k ( 1)k I k (t) after the Nth term, the magnitude of the error is no greater than the absolute value 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 any ε > we can make the second term at most ε/ by a suitable choice of N. Then the first term is at most ε/ for all small enough t (depending on N), and that shows (A.3) tends to as t +. eferences [1] W. Appel, Mathematics for Physics and Physicists, Princeton Univ. Press, Princeton, 7. []. P. Feynman, Surely You re Joking, Mr. Feynman!, Bantam, New York, 1985. [3] S. K. Goel and A. J. Zajta, Parametric Integration Techniques, Math. Mag. 6 (1989), 318 3. [4] S. Lang, Undergraduate Analysis, nd ed., Springer-Verlag, New York, 1997. [5] S. Lang, Comple Analysis, 3rd ed., Springer-Verlag, New York, 1993. [6] E. Talvila, Some Divergent Trigonometric Integrals, Amer. Math. Monthly 18 (1), 43 436.