Chapter 9 One dimensional integrals in several variables 9.1 Differentiation under the integral Note: less than 1 lecture Letf(x,y) be a function of two variables and define b g(y):= f(x,y)dx �a Suppose that f is differentiable in y. The question we ask is when can we simply “differentiate under the integral”, that is, when is it true thatg is differentiable and its derivative b ? ∂f g�(y) = (x,y)dx. y �a ∂ Differentiation is a limit and therefore we are really asking when do the two limitting operations of integration and differentiation commute. As we have seen, this is not always possible. In particular, ∂f thefirst question we would face is the integrability of ∂y . Let us prove a simple, but the most useful version of this theorem. Theorem 9.1.1 (Leibniz integral rule). Suppose f:[a,b] [c,d] R is a continuous function, such × → that ∂f exists for all(x,y) [a,b] [c,d] and is continuous. Define ∂y ∈ × b g(y):= f(x,y)dx. �a Then g:[c,d] R is differentiable and → b ∂f g�(y) = (x,y)dx. y �a ∂ 53 54 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES ∂f Note that the continuity requirements for f and ∂y can be weakened but not dropped outright. ∂f The main point is for ∂y to exist and be continuous for a small interval in the y direction. In applications, the [c,d] can be made a very small interval around the point where you need to differentiate. Proof. Fix y [c,d] and let ε>0 be given. As ∂f is continuous on [a,b] [c,d] it is uniformly ∈ ∂y × continuous. In particular, there exists δ>0 such that whenever y [c,d] with y y <δ and all 1 ∈ | 1 − | x [a,b] we have ∈ ∂f ∂f (x,y ) (x,y) <ε. ∂y 1 − ∂y � � � � Now suppose h is such that y+h� [c,d] and h <δ .� Fix x for a moment and apply mean value �∈ | | � theorem tofind ay 1 betweeny andy+h such that f(x,y+h) f(x,y) ∂f − = (x,y ). h ∂y 1 If h <δ then | | f(x,y+h) f(x,y) ∂f ∂f ∂f − (x,y) = (x,y ) (x,y) <ε. h − ∂y ∂y 1 − ∂y � � � � � � � � This argument worked� for everyx [a,b]. Therefore,� as� a function ofx � � ∈ � � � f(x,y+h) f(x,y) ∂f x − converges uniformly tox (x,y) ash 0. �→ h �→ ∂y → We only defined uniform convergence for sequences although the idea is the same. If you wish you can replaceh with 1/n above and letn ∞. → Now consider the difference quotient g(y+h) g(y) b f(x,y+h)dx b f(x,y)dx b f(x,y+h) f(x,y) − = a − a = − dx. h h h � � �a Uniform convergence can be taken underneath the integral and therefore g(y+h) g(y) b f(x,y+h) f(x,y) b ∂f lim − = lim − dx= (x,y)dx. h 0 h a h 0 h a ∂y → � → � Example 9.1.2: Let 1 f(y) = sin(x2 y 2)dx. − �0 Then 1 2 2 f �(y) = 2ycos(x y )dx. − − �0 9.1. DIFFERENTIATION UNDER THE INTEGRAL 55 Example 9.1.3: Suppose we start with 1 x 1 − dx. (x) �0 ln The function under the integral can be checked to be continuous, and in fact extends to be continuous on [0,1], and hence the integral exists. Trouble isfinding it. Introduce a parameter y and define a function. 1 xy 1 f(y):= − dx. (x) �0 ln xy 1 Again it can be checked that ln(−x) is a continuous function (that is extends to be continuous) of x and y for (x,y) [0,1] [0,1] . Therefore f is a continuous function of on [0,1]. In particular f(0) =0 . ∈ × For any ε>0 , the y derivative of the integrand, that is xy is continuous on [0,1] [ε,1] . Therefore, × fory> 0 we can differentiate under the integral sign 1 y 1 ln(x)x y 1 f �(y) = dx= x dx= . (x) y+ �0 ln �0 1 1 We need tofigure out f(1) , knowing f �(y) = y+1 and f(0) =0 . By elementary calculus wefind 1 f(1) = 0 f �(y)dy= ln(2). Therefore � 1 x 1 − dx= ln(2). (x) �0 ln Exercise 9.1.1: Prove the two statements that were asserted in the example. x 1 a) Prove ln−(x) extends to a continuous function of[0,1]. xy 1 b) Prove − extends to be a continuous function on[0,1] [0,1]. ln(x) × 9.1.1 Exercises Exercise 9.1.2: Suppose h:R R is a continuous function. Suppose that g:R R is which is continuously → → differentiable and compactly supported. That is there exists some M>0 such that g(x)=0 whenever x M . | | ≥ Define ∞ f(x):= h(y)g(x y)dy. ∞ − �− Show that f is differentiable. Exercise 9.1.3: Suppose f is an infinitely differentiable function (all derivatives exist) such that f(0) =0 . Then show that there exists another infinitely differentiable function g(x) such that f(x)=xg(x) . Finally show that if f (0)=0 then g(0)=0 . Hint:first write f(x)= x f (s)ds and then rewrite the integral to go � � � 0 � from0 to 1. � 56 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES 1 tx 1 n x Exercise 9.1.4: Compute 0 e dx. Derive the formula for 0 x e dx not using itnegration by parts, but by differentiation underneath� the integral. � n Exercise 9.1.5: Let U R be an open set and suppose f(x,y 1,y 2,...,y n) is a continuous function defined ⊂ ∂f ∂f ∂f on [0,1] U R n+1. Suppose , ,..., exist and are continuous on [0,1] U . Then prove that × ⊂ ∂y 1 ∂y 2 ∂y n × F:U R defined by → 1 F(y1,y 2,...,y n):= f(x,y 1,y 2,...,y n)dx �0 is continuously differentiable. 9.2. PATH INTEGRALS 57 9.2 Path integrals Note: ??? lectures 9.2.1 Piecewise smooth paths Definition 9.2.1. A continuously differentiable function γ:[a,b] R n is called a smooth path or a → continuously differentiable path ifγ is continuously differentiable andγ (t)= 0 for allt [a,b]. ∗ � � ∈ The function γ is called a piecewise smooth path or a piecewise continuously differentiable path if there existfinitely many points t =a<t <t < <t =b such that the restriction of the 0 1 2 ··· k functionγ [tj 1,tj] is smooth path. | − We say γ is a simple path if γ is a one-to-one function. A γ is a closed path if γ(a) =γ(b) , |(a,b) that is if the path starts and ends in the same point. Since γ is a function of one variable, we have seen before that treating γ�(t) as a matrix is equivalent to treating it as a vector since it is an n 1 matrix, that is, a column vector. In fact, by an × exercise, even the operator norm of γ�(t) is equal to the euclidean norm. Therefore, we will write γ�(t) as a vector as is usual. Generally, it is the direct image γ [a,b] that is what we are interested in. We will informally talk about a curve, by which we will generally mean the setγ [a,b] . � � Example 9.2.2: Letγ:[0,4] R 2 be defined by � � → (t,0) ift [0,1], ∈ (1,t 1) ift (1,2], γ(t):= − ∈ (3 t,1) ift (2,3], − ∈ (0,4 t) ift (3,4]. − ∈ Then the reader can check that the path is the unit square traversed counterclockwise. We can check that for example γ (t)=(1,t 1) and therefore (γ ) (t)=(0,1)=0 . It is good to |[1,2] − |[1,2] � � notice at this point that (γ ) (1) = (0,1), (γ ) (1) = (1,0), and γ (1) does not exist. That is, |[1,2] � |[0,1] � � at the corners γ is of course not differentiable, even though the restrictions are differentiable and the derivative depends on which restriction you take. Example 9.2.3: The condition that γ (t)=0 means that the image of γ has no “corners” where γ is � � smooth. For example, the function (t2,0) ift< 0, γ(t):= (0,t 2) ift 0. � ≥ ∗Note that the word “smooth” is used sometimes of continuously differentiable, sometimes for infinitely differen- tiable in the literature. 58 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES It is left for the reader to check that γ is continuously differentiable, yet the image γ(R) = (x,y) { ∈ R2 :(x,y) = ( s,0) or(x,y) = (0,s) for somes 0 has a “corner” at the origin. − ≥ } Example 9.2.4: A graph of a continuously differentiable function f:[a,b] R is a smooth path. → That is, defineγ:[a,b] R 2 by → γ(t):= t,f(t) . Thenγ �(t)= 1,f �(t) , which is never zero. � � There are other ways of parametrizing the path. That is, having a different path with the same � � image. For example, the function that takes t to (1 t)a+tb , takes the interval [0,1] to [a,b]. So let − α:[0,1] R 2 be defined by → α(t):= (1 t)a+tb,f((1 t)a+tb) .