1 TABLE OF CONTENT INTRODUCTION 2 CHAPTER I PARAMETER DEPENDENT INTEGRAL 3 1 DEFINITION 3 2 PROPERTIES OF PARAMETER DEPENDENT INTEGRALS 3 2 1 Continuity 3 2 2 Differentiability 4 2 3 Integrability 7 3.
TABLE OF CONTENT INTRODUCTION CHAPTER I: PARAMETER - DEPENDENT INTEGRAL DEFINITION PROPERTIES OF PARAMETER-DEPENDENT INTEGRALS 2.1 Continuity 2.2 Differentiability 2.3 Integrability THE PROPERTIES OF THE PARAMETER-DEPENDENT INTEGRAL WITH THE VARIABLE BOUND 3.1 Continuity 3.2 Differentiability EXERCISES 10 CHAPTER II: IMPROPER INTEGRALS DEPENDING ON A PARAMETER 13 CONCEPTION 13 UNIFORMLY CONVERGENCE AND UNIFORMLY CONVERGENCE CRITERION 15 CONTINUITY 20 DISABILITY 22 INTEGRABLE 25 EXERCISES 29 CHAPTER III: EULER INTEGRAL 33 GAMMA FUNCTION 33 BETA FUNCTION 34 CONCLUSION 38 REFERENCE 39 INTRODUCTION Calculus has been around since ancient times, and in its most basic form, it is used for counting Its meaning in the world of mathematics is to take up space in order to solve complex problems when simpler mathematics cannot provide an answer Many people are unaware that calculus is being taught outside of high school and college Calculus is all around us, from building design to loan payments Calculus is divided into two parts: differential calculus and integral calculus Differential calculus is concerned with derivatives and how they are used Integral calculus is a branch of mathematics that determines quantities, areas, and equation solutions Differential calculus is the study of functions and the rate at which they change as variables are changed Integral calculus is concerned with calculating mathematical answers such as total dimensions or values In this essay, we will study about integrals depending on parameter, as well as some related problems Due to limited time and capacity, there will inevitably be shortcomings I hope to receive constructive suggestions from you Thank you so much! CHAPTER I: PARAMETER - DEPENDENT INTEGRAL DEFINITION Let f be a function defined on the rectangle [a,b]x[α,β] ⊆ R2 and for each y ∈ [α,β], f is integrable with x on [a,b] Then the integral: a f ( x, y)dx (*) b is a function with the variable y We say that the integral (*) is a parameterdependent integral with the parameter y, denoted by a I ( y ) = f ( x, y )dx b Note that instead of y ∈ [α,β], we can consider y ∈ U ⊆ Rn and then I(y) will become a multi-variable function However, almost all the properties of a parameter-dependent integral with y ∈ Rn are similarly for y ∈ R, therefore we will just consider integrals depending on single parameter Moreover, since the integral (*) has two stable limits a and b, then we can say that (*) is a parameterdependent integral with stable domain If b = ψ(y) and a = φ(y) are functions ( y) depending on y, we say that f ( x, y )dx is a parameter-dependent integral with ( y) unstable domain PROPERTIES OF PARAMETER-DEPENDENT INTEGRALS 2.1 Continuity We still use the notation I(y) for parameter-dependent integral with unstable domain and assume that f is defined on the rectangle [a,b]x[α,β] ⊆ R2 and a ≤ ψ(y) ≤ b, a ≤ φ(y) ≤ b for all y ∈ [α,β] Theorem: Suppose that f is continuous on [a,b]x[α,β], ψ and φ are continuous on [α,β] Then: ( y) I ( y) = ( y) is continuous on [α,β] f ( x, y )dx Corollary: If f is continuous on [a,b]x[α,β], then the integral: a I ( y ) = f ( x, y )dx b is continuous on [α,β] and for all y0 ∈ [α,β], we have: b b a a b lim f ( x, y)dx = lim f ( x, y)dx = f ( x, y0 )dx y → y0 y → y0 a Example 1: Check the continuity of the integral I ( y) = yf ( x) dx , with f(x) is a positive x2 + y function and continuous on [0,1] Solution: The function g(x,y) = yf ( x) is continuous on each rectangle + y2 x [0,1]x[c,d] and [0,1]x[-d,-c] with any < d < c, then I(y) is continuous on [c,d], [-c,-d], or I(y) is continuous for all y ≠ Now we check the continuous of I(y) at y = Since f(x) is continuous on [0,1], there exists m = f ( x ) Then 𝑓(𝑥) ≥ 𝑚 > 0∀𝑥 ∈ [0,1] and with 0,1 𝜀 > 0, we have: f ( x) m x I ( ) = dx dx = m arctan 2 x + x + o 1 I (− ) = o − f ( x ) x + 2 dx − m x dx = −m arctan 2 x + x a Then we get I ( ) − I ( − ) 2m arctan → 2m when 𝜀 → 0, which means I ( ) − I ( − ) is not tend to when 𝜀 → 0, I(y) is discontinuous at 𝑦 = 2.2 Differentiability * Theorem : Assume that the function f is continuous and has partial derivative f 'y , continuous on the domain a, b , and functions , differentiable on , Then, the function I(y) is differentiable on , and: ( y) I '( y ) = f ' y ( x, y )dx + f ( ( y ), y ) '( y ) − f ( ( y ), y ) '( y ) ( y) Proof: We consider the function with three variables: v F ( y, u, v) = f ( x, y )dx , ( y, u, v) [ , ] [ , ] [ , ] u We need to show that the function is differentiable and continuous We must to show that F has partial derivatives and continuous Consider increments v y F ( y, u, v) = F ( y + y , u, v) − F ( y, u, v) = [f ( x, y + y ) − f ( x, y)]dx u Since f 'y is continuous, by mean value theorem: f ( x, y y ) − f ( x, y) = f '( x, y + y ) y Where [0,1] , depending on (x,y) Then: y F ( y, u, v) y Note that, the function v v u u − f ' y ( x, y ) dx = [f ' y ( x, y + y ) − f ' y ( x, y ) f 'y continuous on the domain a, b , , then it is continuous and for all δ > 0, y , we have: f ' y ( x, y + y ) − f ' y ( x, y ) b−a for all x,y Therefore, for all y , we have a remark: y F ( y, u , v) y v − f ' y ( x, y )dx u v −u b−a For some ε, we conclude: lim y →0 Show that f ' y ( x, y) y F ( y, u , v) y exists and continuous Moreover, we have: F 'u ( y, u, v) = − f (u, y ) F 'v ( y, u, v) = f (v, y ) v = f ' y ( x, y ) u Are continuous functions, then F is continuous and differentiable If φ and are differentiable, then by the union function theorem: I ( y ) = F ( y, ( y ), ( y )) Is differentiable and ( y) du dv I '( y ) = F ' y + F 'u + F 'v = f ' y ( x, y )dx + f ( ( y ), y ) '( y ) − f ( ( y ), y ) '( y ) dy dy ( y ) The theorem has been proven cos y Example 1: Given I ( y ) = e yx dx Applying the theorem 2.1, The function y I ( y) is differentiable and cos y I '( y ) = y e yx dx − e y cos y sin y − e y y Example 2: Calculate the following integrals a) I n ( ) = x ln n xdx , n ∈ Z Solution For each , the function f n ( x, ) = x ln n+1 x, n = 0,1, 2, is continuous with x on [0,1] f n ( x, ) = x ln n +1 x Since lim x ln n+1 x = , so x → 0+ It means that the function f n ( x, ) = x ln n x is continuous on [0,1] (0, +) satisfies conditions of the theorem, so: d d I 'n−1 ( ) = x ln n−1 xdx = ( x ln n−1 x)dx = x ln n xdx = I n ( ) dx dx 0 Similarly, we have I 'n−2 = I n−1 , , I '2 = I1 , I '1 = I , implies that Moreover, I ( ) = x dx = I n ( ) = +1 + (n) = I n ( ) = [I ( )]( n ) (−1) n n ! ( + 1) n +1 b) I(𝑦) = ln(1 + y sin x)dx , where 𝑦 > Solution Consider the function f ( x, y) = ln(1 + y sin x) satisfies the following conditions: +) f ( x, y) = ln(1 + y sin x) is defined on [0, ] (1, +) and for some y > -1, f ( x, y ) is continuous with respect to x on [0, ] +) There exists f '( x, y) = sin x defined, continuous on [0, ] (1, +) 2 + y sin x 2 sin x dx = 0 + y sin x By theorem 2.1, we have I '( y) = dx +y sin x Set t = tanx, then dx = dt , t + 1+ t + I '( y) = + t dt 1 1 = − 2 2 (t + 1)(1 + t + yt ) y (t + 1) + ( y + 1)t = = 1 + arctan t − y 1 − y + arctan(t y + 1) y +1 = y +1 1+ y 1+ 1+ y This implies that: I ( y ) = I '( y )dy = Since dy = ln(1 + + y ) + C 1+ y 1+ 1+ y I (0) = C = − ln Therefore, I ( y) = ln(1 + + y ) − ln 2.3 Integrability Theorem: Assume that f is continuous on [a,b]×[,] Then, the integrals 𝑏 ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥, ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 is integrable on [,], [a,b] (corresponding) and we have Fubini formula: 𝑏 𝑏 ∫ 𝑑𝑦 ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥 = ∫𝑦 𝑑𝑥 ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 Proof: At the end of section 4.3.1, we had Fubini formula from general theorem Here's another way to prove it Since f is continuous, the function I(y) = 𝑏 ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥 is continuous, it implies that I(y) is integrable on [,] Similarly, the function ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 is integrable on [a,b] Set 𝑡 𝑏 g(t) = ∫ 𝑑𝑦 ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥, h(t) =≤t≤ We will prove that g(t) = h(t) for all t ∈ [,] and there will be immediately the formula in the theorem when choosing t = Note that with t = , we have g() = h() = 0, so we only have to prove g’(t) = h’(t) Remark that the function 𝑏 I(y) = ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥 is continuous on [,], so 𝑏 g’(t) = I(t) = ∫𝑎 𝑓(𝑥, 𝑡)𝑑𝑥 for all t ∈ [,] Moreover, the two-variable function 𝑡 J(x,t) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥, (x,t) ∈ [a,b]×[,], is continuous and has a derivative in terms of a continuous variable t (since J’t(x,t) = f(x,t) ), so we can apply the theorem about the derivative of integrals that depend on a parameter h’(t) = 𝑑 𝑑𝑡 𝑏 𝑏 𝑏 (∫𝑎 𝐽(𝑥, 𝑡)𝑑𝑥) = ∫𝑎 𝐽′𝑡 (𝑥, 𝑡)𝑑𝑥 = ∫𝑎 𝑓(𝑥, 𝑡)𝑑𝑥 It implies that g’(t) = h’(t) and the theorem has been fully proven Note: In the above theorem, if f isn’t continuous, the formula to change the order of integrals is no longer correct For example, the function 𝑥2− 𝑦2 𝑤ℎ𝑒𝑛 (𝑥, 𝑦) ≠ (0,0) f(x,y) = { (𝑥 2+ 𝑦2)2 𝑤ℎ𝑒𝑛 (𝑥, 𝑦) = (0,0) isn’t continuous at the point (x,y) = (0,0) in the domain [0,1]×[0,1], and we have: 1 −𝑑𝑦 ∫0 𝑑𝑦 ∫0 𝑓(𝑥, 𝑦)𝑑𝑥 = ∫0 1+ 𝑦 𝜋 1 −𝑑𝑥 = − , ∫0 𝑑𝑥 ∫0 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫0 1+ 𝑥 𝜋 = 𝑥𝑏 − 𝑥𝑎 Example: Calculate ∫0 𝑙𝑛 𝑥 𝑑𝑥, (0 < a < b) Solution The function that integrates f(x) = undefined But lim+ 𝑥𝑏 − 𝑥𝑎 𝑙𝑛 𝑥 𝑥→0 𝑥 𝑏 −𝑥 𝑎 𝑙𝑛 𝑥 even though x = is = 0, we can put this integral in the category of definite integrals We have: 𝑥𝑏 − 𝑥𝑎 𝑙𝑛 𝑥 𝑏 𝑏 = F(x,b) – F(x,a) = ∫𝑎 𝐹𝑦′ (𝑥, 𝑦)𝑑𝑦 = ∫𝑎 𝑥 𝑦 𝑑𝑦; ( F(x,y) := 𝑥𝑦 𝑙𝑛 𝑥 ) So: 𝑥𝑏 − 𝑥𝑎 ∫0 𝑙𝑛 𝑥 𝑏 𝑏 𝑏 𝑑𝑥 = ∫0 (∫𝑎 𝑥 𝑦 𝑑𝑦) 𝑑𝑥 = ∫𝑎 (∫0 𝑥 𝑦 𝑑𝑥) 𝑑𝑦 = ∫𝑎 𝑦+1 𝑑𝑦 = 𝑙𝑛 𝑏+1 𝑎+1 THE PROPERTIES OF THE PARAMETER-DEPENDENT INTEGRAL WITH THE VARIABLE BOUND b Consider parameter-dependent integrals with variable bound ( y) = ' ( y)dy a b( y ) 𝐽 ( y) = f ( x, y )dx, with y [c, d ], a a ( y ), b( y ) by [c, d ] a( y) 3.1 Continuity Theorem: If i) The function f ( x, y ) is ii)The functions continuous on a( y ), b( y ) ) a a ( y ), b( y ) b y [c, d ] [a, b] [c, d ] are continuous on [c, d ] and satisfy the condition then 𝐽 ( y ) is a continuous function (with respect to y ) on [c, d ] 3.2 Differentiability Theorem (Leibniz Theorem): If i) The function f ( x, y ) is continuous on [a, b] [c, d ] ii) The functions f ' y ( x, y) are continuous on [a, b] [c, d ] iii) The functions a( y ), b( y ) are differentiable on [c, d ] and satisfy the conditions a a( y ), b( y ) b y [c, d ] then 𝐽 ( y ) is a differentiable function (with respect to y ) on and: [c, d ] b( y ) 𝐽 ( y ) = f (b( y ), y )by' ( y ) − f (a( y ), y )a 'y ( y ) + f y' ( x, y ) dx a( y) 1+ y Example: Find lim y →0 dx + y2 1+ x y 1+ y Solution It is easy to check that the function 𝐽 ( y ) = y 1+ y y = based on theorem 3.14, so lim y →0 y dx is continuous at + y2 1+ x dx dx = (0) = = 2 1+ x + y 1+ x 4 EXERCISES Type Calculating the parameter-dependent broad integral by double the order of integration b Suppose we need to calculate: ( y) = f ( x, y)dx a d Step 1: Express f ( x, y ) = F ( x, y )dy c Step Use the reorder property to get the counter product: b ( y ) = a d b d f ( x, y )dx = F ( x, y )dy dx = F ( x, y )dx dy ac ca b Type Calculating the integral by derivation over the integral sign b Suppose we need to calculate ( y) = f ( x, y)dx a b Step Calculate ( y) by ( y) = f y' ( x, y)dx ' ' a Step Use the Newton-Leibniz formula to recover I(𝑦) = ∫ 𝐼 ′ (𝑦) 𝑑𝑦 + 𝑐 Step Given a special value of y to determine C 10 ( y ) by +∞ But 𝐼0 (𝑦) = ∫0 𝑑𝑥 = 𝑥 +𝑦 𝜋 (2𝑛−1)‼ So, 𝐼𝑛 (𝑦) = ∙ (2𝑛)‼ ∙ arctan √𝑦 √𝑦 2𝑛+1 𝑥 √𝑦 | +∞ = 𝜋 2√𝑦 The remaining problem is the test of the condition for passing the derivative over the integral sign 1) The functions 𝑓(𝑥, 𝑦) = , 𝑓 ′ 𝑦 (𝑥, 𝑦) = 𝑥 +𝑦 −1 (𝑥 +𝑦)2 (𝑛) , … , 𝑓𝑦𝑛 (𝑥, 𝑦) = (−1)𝑛 (𝑥 +𝑦)𝑛+1 are continuous on [0, +) × [, +) for each given > 2) ≤ 𝑥 +𝑦 −1 𝑥 +𝜀 , |(𝑥 +𝑦) +∞ But the integrals ∫0 (𝑥 +𝜀)2 , … , |(𝑥 +∞ 𝑥 +𝜀 +∞ (−1)𝑛 |≤ 𝑑𝑥, … , ∫0 +𝑦)𝑛+1 +∞ ′ ∫ 𝑓(𝑥, 𝑦)𝑑𝑥, ∫ 0 (𝑥 +𝜀)𝑛+1 are converge, hence (𝑥 +𝜀)𝑛+1 +∞ |≤ 𝑓 𝑦 (𝑥, 𝑦)𝑑𝑥, … , ∫ (𝑛) 𝑓𝑦𝑛 (𝑥, 𝑦)𝑑𝑥 converge uniformly on [, +) for each given > INTEGRABLE Theorem: Assume that: i) The function f is continuous on [a, ) × [, ] ∞ ii) The integral 𝐼(𝑦) = ∫𝑎 𝑓(𝑥, 𝑦)𝑑𝑥 converges uniformly on [, ] Then I(y) is integrable on [, ] and 𝛽 ∞ ∞ 𝛽 ∫𝛼 𝑑𝑦 ∫𝛼 𝑓(𝑥, 𝑦)𝑑𝑥 = ∫𝛼 𝑑𝑥 ∫𝛼 𝑓(𝑥, 𝑦)𝑑𝑦 Proof: From the conditions of the theorem it follows that I(y) is a continuous function on [, ], therefore integrable To prove the above formula, just consider the sequence of functions 25 𝑎+𝑛 𝜑𝑛 (𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 and then apply the theorem on the integrability of a series of functions that converge to get 𝛽 𝛽 ∫ 𝐼(𝑦)𝑑𝑦 = lim ∫ 𝜑𝑛 𝑑𝑦 𝑛→∞ 𝛼 𝛼 𝛽 𝑎+𝑛 = lim ∫ 𝑑𝑦 ∫ 𝑛→∞ 𝛼 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 𝑎+𝑛 = lim ∫ 𝛽 ∞ 𝛽 𝑑𝑥 ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫ 𝑑𝑥 ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 𝑛→∞ 𝑎 𝛼 𝑎 𝛼 Note that: The above result can be extended to the case of the integral I(y) domain of infinite (eg [, )) Specifically: Assume that the function f is continuous and positive on the domain [, )×[, ) and integrals 𝛽 ∞ 𝐽(𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 , 𝐼(𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥, 𝛼 𝑎 converge to continuous functions Then if one of the integrals ∞ ∞ ∫𝑎 𝑑𝑥 ∫𝑎 𝑓𝑑𝑦, ∞ ∞ ∫𝑎 𝑑𝑥 ∫𝑎 𝑓𝑑𝑥 exists then the remaining integral also exists and they are equal +∞ 𝑒 −𝛼𝑥 −𝑒 −𝛽𝑥 Example 1: Find ∫0 𝑥 𝑑𝑥 , (𝛼, 𝛽 > 0) Solution: We have: 𝛼 𝛽 𝑒 −𝛼𝑥 − 𝑒 −𝛽𝑥 ′( ( ) ( ) ) 𝑑𝑥 = 𝐹 𝑥, 𝛼 − 𝐹 𝑥, 𝛽 = ∫ 𝐹𝑦 𝑥, 𝑦 = ∫ 𝑒 −𝑦𝑥 𝑑𝑦 𝑥 𝛽 𝛼 So, +∞ −𝛼𝑥 ∫ 𝑒 +∞ 𝛽 𝛽 +∞ 𝛽 − 𝑒 −𝛽𝑥 𝑑𝑦 𝛽 𝑑𝑥 = ∫ (∫ 𝑒 −𝑦𝑥 𝑑𝑦) 𝑑𝑥 = ∫ (∫ 𝑒 −𝑦𝑥 𝑑𝑥) 𝑑𝑦 = ∫ = 𝑙𝑛 𝑥 𝛼 𝛼 𝛼 𝛼 𝑦 Readers check for themselves the condition of exchanging the order of integration 26 Example 2: Integral Gauss ∞ √𝜋 2 𝐺 = ∫ 𝑒 −𝑥 𝑑𝑥 = Set x = ut, we have: +∞ 𝑒 −𝑢 𝐺 = 𝑢∫ 2𝑡 𝑑𝑡 We have: +∞ 𝐺 = 𝐺∫ 𝑒 −𝑢2 +∞ 𝑑𝑢 = ∫ (𝑢𝑒 −𝑢2 +∞ +∞ (∫ 0 ∞ 2𝑡 𝑑𝑡) 𝑑𝑢 +∞ =∫ 𝑒 −𝑢 ∫ 𝑢𝑒 −(1+𝑡 )𝑢2 That implies: 𝐺 = ∫0 𝑒 −𝑥 𝑑𝑥 = +∞ 𝑑𝑡 𝜋 𝑑𝑢) 𝑑𝑡 = ∫ = + 𝑡2 √𝜋 Readers check for themselves the condition of exchanging the order of integration Example 3: Integral Dirichlet ∞ 𝐷=∫ 𝑠𝑖𝑛𝑥 𝜋 𝑑𝑥 = 𝑥 Consider: ∞ = ∫ 𝑒 −𝑥𝑡 𝑑𝑡, 𝑥 We have: ∞ ∞ 𝐷 = ∫ 𝑠𝑖𝑛𝑥 (∫ 𝑒 0 ∞ −𝑥𝑡 ∞ ∞ 𝑑𝑡) 𝑑𝑥 = ∫ (∫ 𝑒 −𝑥𝑡 𝑠𝑖𝑛𝑥𝑑𝑥 ) 𝑑𝑡 = ∫ 0 𝑑𝑡 𝜋 = + 𝑡2 Readers check for themselves the condition of exchanging the order of integration Note that: Readers can compare a result in calculus III, this is ∑∞ 𝑛=1 sin 𝑛 𝑛 27 = 𝜋−1 Example 4: Applying the integral Dirichlet formula, prove that: +∞ 𝑎) ∫ +∞ − cos 𝑥 𝜋 𝑑𝑥 = 𝑥2 𝑏) ∫ +∞ 𝑐) ∫ 𝑠𝑖𝑛3 𝑥 𝜋 𝑑𝑥 = 𝑥 𝑠𝑖𝑛2 𝑥 𝜋 𝑑𝑥 = 𝑥2 [𝑠𝑢𝑔𝑔𝑒𝑠𝑡𝑖𝑜𝑛𝑠]: 4 a) Applying the descending order formula 𝑠𝑖𝑛3 𝑥 = 𝑠𝑖𝑛3𝑥 + 𝑠𝑖𝑛𝑥 b) Applying the formula for integration by parts, 𝑀 ∫ 𝜖 𝑀 𝑀 − 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝜖 − 𝑐𝑜𝑠𝑀 𝑠𝑖𝑛𝑥 𝑀 (1 𝑑𝑥 = − 𝑐𝑜𝑠𝑥)| + ∫ 𝑑𝑥 = − + ∫ 𝑑𝑥 𝜖 𝑥 𝑥 𝜖 𝑀 𝑥 𝜖 𝜖 𝑇𝑎𝑘𝑒 𝜖 → ∞ 𝑎𝑛𝑑 𝑀 → ∞, 𝑤𝑒 ℎ𝑎𝑣𝑒: − 𝑐𝑜𝑠𝜖 − 𝑐𝑜𝑠𝑀 = 𝑎𝑛𝑑 lim =0 𝜖→∞ 𝑀→∞ 𝜖 𝑀 lim So, +∞ ∫ 𝑥 +∞ 𝑠𝑖𝑛2 +∞ − 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛2 𝑢 𝑑𝑥 = ∫ =∫ 𝑑𝑢 𝑥2 𝑥2 𝑢2 0 Example 5: Integral Fresnel ∞ 𝑠𝑖𝑛𝑡 ∞ 𝑐𝑜𝑠𝑡 𝐼= ∫ 𝑑𝑡 𝑎𝑛𝑑 𝐽 = ∫ 𝑑𝑡 √𝑡 √𝑡 ∞ By Gauss integral formula, we have ∫0 𝑒 −𝑢 𝑑𝑢, we have: √𝑡 = √𝜋 ∞ ∫ 𝑒 −𝑡𝑢 𝑑𝑢 We have: ∞ ∫ 𝑠𝑖𝑛𝑡 √𝑡 𝑑𝑡 = √𝜋 ∞ ∞ ∫ 𝑠𝑖𝑛𝑡 ∫ 𝑒 −𝑡𝑢 𝑑𝑢 = 0 √𝜋 ∞ ∞ ∫ 𝑑𝑢 ∫ 𝑒 −𝑡𝑢 𝑠𝑖𝑛𝑡𝑑𝑡 = 0 √𝜋 ∞ ∫ 𝑑𝑢 𝜋 𝜋 = =√ 1+𝑢 √𝜋 2√2 𝜋 Therefore: 𝐼 = 𝐽 = √ 2 Readers can check the condition of changing the order of integration 28 EXERCISES Method 1: Calculate the parameter-dependent integral by changing the order of integration + Suppose you need to calculate I (y) = f ( x, y )dx a d Step 1: Performing f ( x, y) = F ( x, y)dy c Step 2: Use the reorder property to integrate: + I ( y) = a + d f ( x, y )dx = d + c a ( F ( x, y) dy) dx = ( F ( x, y)dx)dy a c Method 2: Calculate the parameter-dependent integral by derivation over the integral sign + Suppose that you need to calculate I (y) = f ( x, y )dx a + Step 1: Calculate I’ (y) by I '( y) = f ' y ( x, y)dx a Step 2: Use the Newton-Leibniz formula to recover I (y) by I ( y) = I '( y)dy + C Step 3: Take a special value of y to determine C Note: Must check the conditions for converting the order of integration in the Theorem of Integrability or turning the sign of the derivative over the integral in the theorem of differentiability 29 + Example 1: Calculate e− x − e− x , ( , 0) Solution 1: change the order of integration + 2: Derivative over integral sign e − x − e − x e− yx F ( x, y ) := x I ( ) = I '( ) = + Thus, + = e− x − e− x dx + e − yx = e dy dx − x −e − x dx = − sin bx − sin cx , ( 0) x Thus, I ( ) = I '( )d = − ln + C + = e − yx dx dy = f ' ( x, )dx + = F ' y ( x, y ) = e − yx dy + e− x − e− x dx x = F ( x , ) − F ( x, ) + On the other hand, we have: I ( ) = , dy = ln y so C = ln Conclusion: I = ln Check the condition to change derivative over integral sign With f ( x, ) = 1) f ( x, ) = e − x − e − x x e − x − e − x x , we have: is continuous with respect to x on 0, +; , 2) f ' ( x, ) = −e − x is continuous on + 3) f ' ( x, )dx = + − x −e dx = − [0, +) (0, +) is convergent to on open interval [ , +) according to standard Weierstrass, indeed, −e − x −e − x + , moreover −e convergent 30 − x dx = − is + e Exercise 2: Calculate − x sin bx − sin cx , ( 0) x Solution 1: change the order of integration 2: Derivative over integral sign We have: e− x sin bx − sin cx + Set I (b) = e− x x e − x sin yx = F ( x , b ) − F ( x , c ) F ( x, y ) = x + I 'b ( x, b) = e − x sin bx − sin cx dx x cos bx = b b = F ' y ( x, y)dy = e− x cos yxdx c Thus, I = c + b c +b so C = − arctan + − x = e cos yxdy dx c b = + y2 c = arctan b db = arctan On the other hand: Thus, I = e− x cos yxdy dx b a a + b2 b +C I (c ) = , c Conclusion: − arctan I = arctan c b − arctan c Check the condition to change derivative over integral sign sin bx − sin cx , we x With f ( x, b ) = e− x f ( x , b ) = e − x 1) sin bx − sin cx x have: is continuous with respect to x on [0, +), a, b, c 2) f 'b ( x, b ) = e− x cos bx is continuous on [0, +) ( 0, + ) 3) + + f 'b ( x, b ) dx = e + − x b cos bx = − 2 e − x cos bx + 2 e − x sin bx = 2 +b +b +b 0 Is convergent with respect to b on ( 0, + ) according to standard Weierstrass, indeed, e− x cos bx e− x is convergent, moreover + e− x dx is convergent In the above proof, we used the formula e − ax cosyxdx = − + Then e − ax a y e− ax cosyx + e− ax sin yx a +y a + y2 a cos yxdx = a + y2 + e − ax cos yxdx = 31 a a + y2 Exercise 3: In a similar way to Exercise 2, calculate + e − ax Answer: I = ln a2 + c2 a + b2 cos bx − cos cx ,a x Consequence 1: + + Exercise Calculate e cos bx − cos cx c = ln x b − x2 cos( yx)dx Solution: + Put + − x2 − x2 − xe sin yxdx = e sin yx |0+ − 0 2 f ' y ( x, y)dx = + ye− x cos yxdx = −y I ( y) We have: 1) f(x,y) is continuous on [0, +∞) x (-∞, +∞) 2) f '( x, y) = − xe− x sin yx continuous on [0, +∞) x (-∞, +∞) + + f ' y ( x, y)dx = − xe − x2 sin yxdx = e− x sin yx |0+ − 2 + ye − x2 cos yxdx = −y I ( y) converges uniformly according to the Weierstrass criterion, indeed: + xe− x dx = 2 | f ' y ( x, y) | xe− x that, + xe − x2 dx = convergence 2 y − I '( y) y = − I = Ce Therefore, according to the differentiability theorem, I ( y) − y4 Since I(0) = C= then I (y) = e 2 32 CHAPTER III: EULER INTEGRAL GAMMA FUNCTION +∞ Γ(𝑃) = ∫0 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥 defined on (0; +∞) Indeed +∞ Γ(𝑃) = ∫ 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥 + ∫ 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥 = 𝐼1 + 𝐼2 +∞ * With integral 𝐼2 we compare with ∫ 𝑙𝑖𝑚 (𝑥 𝑃−1 𝑥→+∞ +∞ where ∫ 𝑑𝑥 𝑥2 𝑑𝑥 We have: 𝑥2 𝑥 1+𝑃 − 𝑒 : ) = 𝑙𝑖𝑚 = 0, 𝑥→+∞ 𝑒 𝑥 𝑥 𝑥 converges, so 𝐼2 converges Even the integral 𝐼2 converges for all p ∈ R * With integral 𝐼1 , then if p ≥ we have 𝐼1 which is a definite integral If < p < 1, then we compare 𝐼1 with ∫ 𝑑𝑥 𝑥 1−𝑃 We have: 𝑙𝑖𝑚 (𝑥 𝑃−1 𝑒 −𝑥 : 1−𝑝) = 1, 𝑥 𝑥→0+ where ∫ 𝑑𝑥 𝑥 1−𝑃 converges, so 𝐼1 also converges If p < then ∫ 𝑑𝑥 𝑥 1−𝑃 diverges so 𝐼1 will diverge * Properties a) Lower order: Γ (p + 1) = pΓ (p) This formula can be easily proved by Integration by parts Indeed, +∞ +∞ +∞ 𝑝−1 −𝑥 Γ (p + 1) = ∫ 𝑥 𝑝 𝑒 −𝑥 𝑑𝑥 = − ∫ 𝑥 𝑝 𝑑 (𝑒 −𝑥 ) = −𝑥 𝑝 𝑒 −𝑥 |+∞ 𝑒 = 𝑝Γ + 𝑝∫ 𝑥 0 Meaning: To study Γ (p), we only need to study Γ (p) with < p ≤ 11, and for p > we will use the descending formula 33 b) In particular + Γ (1) = (direct calculation) so Γ (n) = (n − 1)! ∀𝑛 ∈ ℕ + From the Gaussian formula: +∞ ∫ 𝑒 −𝑥 = +∞ −𝑥 𝑒 √𝜋 ⇒𝛤( ) = ∫ 𝑑𝑥 = √𝜋 2 √𝑥 0 Therefore, Γ (𝑛 + ) = (2𝑛−1)‼ 2𝑛 √𝜋 +∞ c) Derivative of Gamma function: Γ (𝑘) (p)∫0 d) Γ (p).Γ (1 − p) = 𝜋 𝑠𝑖𝑛 𝑝𝜋 𝑥 𝑝−1 (ln𝑘 𝑥)𝑒 −𝑥 𝑑𝑥 ∀0 < p < BETA FUNCTION Form 1: B (p, q) =∫01 𝑥 𝑝−1(1 − 𝑥)𝑞−1 𝑑𝑥 By changing the variable x = +∞ Form 2: B (p, q) =∫ 𝑥 𝑝−1 (1+𝑥)𝑝+𝑞 𝑡 𝑡+1 we get: 𝑑𝑥 𝜋 Example 3.20 Indicate ∫0 𝑠𝑖𝑛𝑚 𝑥 ⋅ 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 through function B (m, n) The answer: Put sin x = √𝑡 ⇒ ≤ 𝑡 ≤ , cos 𝑥 𝑑𝑥 = 𝜋 𝜋 ∫ 𝑠𝑖𝑛𝑚 𝑥 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 = ∫ 𝑠𝑖𝑛𝑚 𝑥 (1 − sin2 𝑥) 2√𝑡 𝜋 𝑛−1 cos 𝑥 𝑑𝑥 𝑚 𝑛−1 1 𝑚+1 𝑛+1 = ∫ 𝑡 (1 − 𝑡) 𝑡 −2 𝑑𝑡 = 𝐵 ( ; ) 2 2 From this example we get: 𝜋 Trigonometric form: B (p, q) = ∫0 sin2𝑝−1 𝑡 cos 2𝑞−1 𝑡 𝑑𝑡 *Properties: a) Symmetry: B(p, q) = B(q, p) b) Downgrade: 34 𝐵(𝑝, 𝑞) = { 𝐵(𝑝, 𝑞) = 𝑝−1 𝑝+𝑞−1 𝑞−1 𝑝+𝑞−1 𝐵(𝑝 − 1, 𝑝), 𝑖𝑓 𝑝 > 𝐵(𝑝, 𝑞 − 1), 𝑖𝑓 𝑞 > The meaning of the above formula is that if we want to study the beta function, we only need to study it in (0, 1] × (0, 1] only c) In particular, B (1, 1) = so (𝑚 − 1)! (𝑛 − 1)! ∀𝑚, 𝑛 ∈ ℕ (𝑚 + 𝑛 − 1)! (𝑛 − 1)! 𝐵(𝑝, 𝑛) = 𝐵(𝑝, 1), ∀𝑛 ∈ ℕ (𝑝 + 𝑛 − 1)(𝑝 + 𝑛 − 2) … (𝑝 + 1)𝑝 { 𝐵(𝑚, 𝑛) = d) Relationship formula between Beta and Gamma: B(p,q) = Γ(p)+Γ(q) Γ(p+q) Prove: We have: +∞ 𝑝−1 −𝑡 Γ(p)Γ(q) = ∫0 𝑡 𝑒 +∞ 𝑞−1 −𝑆 𝑑𝑡 ∫0 𝑠 𝑒 +∞ 𝑑𝑠 = ∫ +∞ 𝑝−1 𝑞−1 −(𝑡+𝑠) ∫0 𝑡 𝑠 𝑒 Applying the transformation formula t = xy and s = x(1 − y), we get: { 𝐷 (𝑡, 𝑠) 𝑦 ≤ 𝑥 < +∞ ≤ 𝑡 < +∞ ⇒{ ,𝐽 = = |1 − 𝑦 0≤𝑦≤1 ≤ 𝑠 < +∞ 𝐷 (𝑥, 𝑦) 𝑥 −𝑥| = −𝑥 Therefore, +∞ 𝛤(𝑝)𝛤(𝑞) = ∫ ∫ 𝑒 −𝑥 𝑥 𝑝−1 𝑦 𝑝−1 𝑥 𝑞−1 (1 − 𝑦)𝑝−1 𝑥𝑑𝑥𝑑𝑦 0 +∞ = ∫ 𝑒 −𝑥 ⋅ 𝑥 𝑝+𝑞−1 𝑑𝑥 ∫ 𝑦 𝑝−1 (1 − 𝑦)𝑝−1 𝑑𝑦 = Γ(p + q)B (p, q) 0 B(p, − p) = Γ (p) Γ (1 − p) = 𝜋 𝑠𝑖𝑛 𝑝𝜋 Example: Prove the relation between Beta and Gamma function B (p, q) = Γ(p)+Γ(q) Γ(p+q) by double integral 35 𝑑𝑡 𝑑𝑠 +∞ Solution Starting from the formula 𝛤(𝑝) = ∫0 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥, performing the transformation of the variable x = 𝑡 , we get: +∞ +∞ 2 𝛤(𝑝) = ∫ 𝑒 −𝑡 𝑡 2𝑝−1 𝑑𝑡 = ∫ 𝑒 −𝑥 𝑥 2𝑝−1 𝑑𝑥 0 Therefore, +∞ +∞ 𝛤(𝑝)𝛤(𝑞) = ∫ ∫ 𝑒 −(𝑥 𝑀 𝑀 Put 𝐼𝑀 = ∫ ∫ 𝑒 −(𝑥 n 𝑙𝑖𝑚 = 𝛤(𝑝)𝛤(𝑞) 𝑀→+∞ +𝑦 ) +𝑦 ) 𝑥 2𝑝−1 𝑦 2𝑞−1 𝑑𝑥𝑑𝑦 𝑥 2𝑝−1 𝑦 2𝑞−1 𝑑𝑥𝑑𝑦 We have: Picture 2.16 We have: ∫ 𝑒 −(𝑥 +𝑦 ) 𝑥 2𝑝−1 𝑦 2𝑞−1 𝑑𝑥𝑑𝑦 ≤ 𝐼𝑀 ≤ ∫ 𝑒 −(𝑥 𝑂𝐴𝐸 +𝑦 ) 𝑥 2𝑝−1 𝑦 2𝑞−1 𝑑𝑥𝑑𝑦 𝑂𝐵𝐷 𝑥 = 𝑟 𝑐𝑜𝑠 𝜑 Perform the transformation of the variable in polar coordinates { 𝑦 = 𝑟 𝑠𝑖𝑛 𝜑 we have: 36 𝜋 𝑀 2𝑝−1 (𝑠𝑖𝑛 ∫(𝑐𝑜𝑠 𝜑) 2𝑞−1 𝜑) 𝑑𝜑 ∫ 𝑟 2𝑝+2𝑞−1 ⋅ 𝑒 −𝑟 𝑑𝑟 0 𝜋 𝑀√2 ≤ 𝐼𝑀 ≤ ∫(𝑐𝑜𝑠 𝜑)2𝑝−1 (𝑠𝑖𝑛 𝜑)2𝑞−1 d𝜑 ∫ 𝑟 2𝑝+2𝑞−1 𝑒 −𝑟 d𝑟 0 We have: 𝑀 +∞ 𝑙𝑖𝑚 ∫ 𝑟 2𝑝+2𝑞−1 ⋅ 𝑒 −𝑟 𝑀→+∞ 𝑑𝑟 = ∫ 𝑟 2(𝑝+𝑞)−1 ⋅ 𝑒 −𝑟 𝑑𝑟 = 0 Γ(p + q) And 𝑀√2 𝑙𝑖𝑚 ∫ 𝑟 +∞ 2𝑝+2𝑞−1 𝑀→+∞ ⋅𝑒 −𝑟 2 𝑑𝑟 = ∫ 𝑟 2(𝑝+𝑞)−1 ⋅ 𝑒 −𝑟 𝑑𝑟 = 0 Γ(p + q) Besides, 𝜋 ∫(𝑐𝑜𝑠 𝜑)2𝑝−1 (𝑠𝑖𝑛 𝜑)2𝑞−1 d𝜑 = 𝐵(𝑝, 𝑞) Let 𝑀 → +∞ in inequality (3.2), we get: 1 Γ(p)Γ(q) 1 𝐵(𝑝, 𝑞) Γ(p + q) ≤ ≤ 𝐵(𝑝, 𝑞) Γ(p + q) 2 2 Therefore, B(p,q) = Γ(p)+Γ(q) 37 Γ(p+q) (3.2) CONCLUSION The Calculus module is an important part of the current university's advanced math program and is taught beginning in the first semester In this context, integral refers to a critical component There is a partial parametric dependent integral in integral that cannot be preserved and aids in the solution of some types of related problems 38 REFERENCE Mathematical Analysis II - Bui Xuan Dieu - Hanoi University of Science and Technology Mathematical Analysis II – Claudio, Anita Tabacco Https://tailieutuoi.com/chu-de/giai-tich-2 39 ... cannot provide an answer Many people are unaware that calculus is being taught outside of high school and college Calculus is all around us, from building design to loan payments Calculus is divided... on parameter, as well as some related problems Due to limited time and capacity, there will inevitably be shortcomings I hope to receive constructive suggestions from you Thank you so much! CHAPTER... interval [0,1] but the order cannot be integrated in this case Please explain why Exercise 4: Proving the Bessel function