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.
PARAMETER - DEPENDENT INTEGRAL
DEFINITION
Let f be a function defined on the rectangle [a,b]x[α,β] ⊆ R 2 and for each y
∈ [α,β], f is integrable with x on [a,b] Then the integral:
(*) is a function with the variable y We say that the integral (*) is a parameter- dependent integral with the parameter y, denoted by
Instead of considering y within the interval [α,β], we can extend our analysis to y in U ⊆ R n, transforming I(y) into a multi-variable function However, most properties of parameter-dependent integrals for y in R n are analogous to those for y in R, leading us to focus primarily on integrals that depend on a single parameter Furthermore, since the integral has two stable limits, a and b, we classify it as a parameter-dependent integral with a stable domain If b = ψ(y) and a = φ(y) are functions that depend on y, this relationship is established.
is a parameter-dependent integral with unstable domain.
PROPERTIES OF PARAMETER-DEPENDENT INTEGRALS
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] ⊆ R 2 and a ≤ ψ(y)
Theorem: Suppose that f is continuous on [a,b]x[α,β], ψ and φ are continuous on [α,β] Then:
Corollary: If f is continuous on [a,b]x[α,β], then the integral:
I y = f x y dx is continuous on [α,β] and for all y 0 ∈ [α,β], we have:
0 0 lim ( , ) lim ( , ) ( , 0 ) b b b y y y a a a y f x y dx f x y dx f x y dx
Check the continuity of the integral 2
+ , with f(x) is a positive function and continuous on [0,1]
+ is continuous on each rectangle
[0,1]x[c,d] and [0,1]x[-d,-c] with any 0 < d < c, then I(y) is continuous on [c,d], [-c,-d], or I(y) is continuous for all y ≠ 0
Now we check the continuous of I(y) at y = 0 Since f(x) is continuous on
I − − I is not tend to 0 when 𝜀 → 0, I(y) is discontinuous at 𝑦 = 0
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:
Proof: We consider the function with three variables:
We need to show that the function is differentiable and continuous We must to show that F has partial derivatives and continuous Consider increments
Since f ' y is continuous, by mean value theorem:
Note that, the function f ' y continuous on the domain a b , , , then it is continuous and for all δ > 0, y , we have:
− for all x,y Therefore, for all y , we have a remark:
Show that f ' ( , ) y x y exists and continuous
Are continuous functions, then F is continuous and differentiable If φ and
are differentiable, then by the union function theorem:
The theorem has been proven
I y = e dx Applying the theorem 2.1, The function
Example 2: Calculate the following integrals a)
Solution For each 0, the function f n ( , ) x = x ln n + 1 x n , = 0,1, 2, is continuous with x on [0,1]
It means that the function f n ( , ) x = x ln n x satisfies conditions of the theorem, so:
I x xdx x x dx x xdx I dx dx
Similarly, we have I ' n − 2 = I n − 1 , , ' I 2 = I I 1 , ' 1 = I 0 , implies that I n ( ) = [ ( )] I 0 ( ) n
Solution Consider the function f x y ( , ) = ln(1 + y sin 2 x ) satisfies the following conditions:
+) f x y ( , ) = ln(1 + y sin 2 x ) is defined on [0, ] (1, )
( , ) f x y is continuous with respect to x on [0, ]
Theorem: Assume that f is continuous on [a,b]×[,] Then, the integrals
∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 𝑏 , ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 is integrable on [,], [a,b] (corresponding) and we have Fubini formula:
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) =
∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 𝑏 is continuous, it implies that I(y) is integrable on [,] Similarly, the function ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 is integrable on [a,b] Set g(t) = ∫ 𝑑𝑦 ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑡 𝑎 𝑏 , h(t) =≤t≤
We will demonstrate that g(t) equals h(t) for all t within the interval [α, β], leading to the theorem's formula when t is set to β It's important to note that at t = α, both g(α) and h(α) equal zero, which means we only need to establish that g’(t) is equal to h’(t).
I(y) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 𝑏 is continuous on [,], so g’(t) = I(t) = ∫ 𝑓(𝑥, 𝑡)𝑑𝑥 𝑎 𝑏 for all t ∈ [,]
Moreover, the two-variable function
J(x,t) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑡 , (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
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 f(x,y) = {
0 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:
Solution The function that integrates f(x) = 𝑥
𝑙𝑛 𝑥 even though x = 0 is undefined But lim
𝑙𝑛 𝑥 = 0, we can put this integral in the category of definite integrals
THE PROPERTIES OF THE PARAMETER-DEPENDENT INTEGRAL
Consider parameter-dependent integrals with variable bound ( ) ' ( ) b a y y dy
Theorem: If i) The function f x y ( , ) is continuous on [ , ] [ , ] a b c d ii)The functions a y b y ( ), ( ) ) are continuous on [ , ] c d and satisfy the condition
( ), ( ) [ , ] a a y b y b y c d then 𝐽( ) y is a continuous function (with respect to y ) on
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
10 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 [ , ] c d and:
Solution It is easy to check that the function 𝐽( ) y 1
EXERCISES
Type 1 Calculating the parameter-dependent broad integral by double the order of integration
Suppose we need to calculate: ( ) ( , ) b a y f x y dx
Step 2 Use the reorder property to get the counter product:
( ) ( , ) ( , ) ( , ) b b d d b a a c c a y f x y dx F x y dy dx F x y dx dy
Type 2 Calculating the integral by derivation over the integral sign
Suppose we need to calculate ( ) ( , ) b a y f x y dx
Step 2 Use the Newton-Leibniz formula to recover ( ) y by
Step 3 Given a special value of y to determine C
Note: Must check the condition to change the order of integration in Theorem
3.13 or convert the sign of the derivative over the integral in Theorem 3.12
Method 1: Exchange parameters for integral
Method 2: Derivative over the sign TP
Substituting the special value b = a into the integral expression 𝐼(b) we get
Exercise 2: Calculate the following integral: a) 𝐼(𝑦) = ∫ arctan 𝑥
0 b) I(𝑦)= ∫ ln(𝑥 0 1 2 + y 2 )dx [Suggestions] a) Step 1: Check that 𝐼(y) satisfies the conditions of the Theorem of scalability
Step 4: Substitute a special value y = y o to calculate C
For example, I(1) = ∫ arctan xdx 0 1 and calculate C = 0
12 b) Step 1: Check that I(𝑦) satisfies the conditions of the Theorem of Differentiability
𝑦 Step 4: Substitute a special value y = y 0 to calculate C
For example, I(0) = ∫ ln𝑥 0 1 2 dx, and calculate C = 0
That is the function I(y) = ∫ f(x + y)dx 0 1 on the interval [0,1] but the order cannot be integrated in this case Please explain why
Exercise 4: Proving the Bessel function
IMPROPER INTEGRALS DEPENDING ON A
CONCEPTION
Assume that f is a definite function on the domain [a,∞)×U, U ⊆ ℝ such that for each fixed y ∈ U, the function f(x,y) is integrable in terms of x on [a,b] for all b>a Integral:
𝑎 is called the improper integral depending on a parameter (with a neighborhood of +∞) This integral converges at y 0 ∈ U if the integral
∫ 𝑓(𝑥, 𝑦 𝑎 ∞ 0)𝑑𝑥 converges We say that the improper integrals depending on a parameter is converges on U if it converges at every point of U, i.e for all y ∈ U,
Similar to the above, we can define the improper integrals depending on a parameter with a neighborhood of −∞, or neighbourhoods of −∞ and +∞
The study of improper integrals that depend on a parameter, particularly those approaching +∞, is thoroughly explored through established definitions and theorems This section focuses specifically on these types of improper integrals, highlighting their significance in mathematical analysis.
+ − b) Prove that I(y) converges evenly to 1 over [ y 0 , +) for all 𝑦 0 > 0 c) Explain why I(y) does not converge evenly on (0, +∞)
The function I(y) is defined as I(y) = -e^(-yx) with the condition that I(y) approaches 1 for every y > 0 To demonstrate that I(y) converges uniformly to 1 on the interval [y0, +∞), it is essential to establish that for any ε > 0, there exists a number bε, which is solely dependent on ε and not on y.
I y − ye − yx dx then e − by b 1 y ln 1 But when y → 0 + Hence, it is not possible to choose a constant b that depends only on ε satisfying the requirement of uniformly convergence
+ with a > 0 and for every 𝑦 [Suggestion]
2 2 2 cos 2 cos sin ax a ax y ax e yxdx e yx e yx a y a y
UNIFORMLY CONVERGENCE AND UNIFORMLY CONVERGENCE
In the study of function series, the concept of uniform convergence is essential for establishing the continuous and differentiable properties of the resulting sum function This principle can also be applied to generalized integrals that depend on parameters and have upper bounds of +∞.
Assume that the generalized integral is parameter-dependent
I ( 𝑦 ) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converge on the domain U ⊆ ℝ We say that this integral converges uniformly on U if for all ε > 0 find a number 𝑏 0 such that:
| ∫ 𝑓(𝑥, 𝑦)𝑑𝑥| 𝑎 ∞ < ε, for all b > 𝑏 0 and ∀𝑦 ∈ U Note that the above definition is equivalent to the condition: lim sup | ( , y) dx | 0 b y U b f x
Theorem (Cauchy criterion): The integral I ( y ) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 0 ∞ converges uniformly on the set U if and only if for every ε > 0 there exists a number 𝑏 0 so that
The condition needs to be inferred from the definition because if
Sufficient conditions: With y fixed, the conditions of the inference theorem I(𝑦) converge Furthermore for b 2 → in the said condition b 1 b y 2 , U (ε) for all b 1 b y 2 , U
By definition, the integral converges uniformly on 𝑈
Theorem (Weierstrass Criterion): Suppose there exists a function F(x) ≥ 0 integrable and a number b ≥ a such that | f(x,y) | ≤ F(x) for all y ∈ U, x ≥ b and integral ( ) a
Proof: According to the Cauchy criterion for convergent integrals, for all 𝜀 > 0 exists b 0 so that
, for all b b 1 , 2 b 0 , 𝑦 ∈ U Applying Cauchy's theorem we conclude that the integral I(𝑦) converges on 𝑈
is convergent on any set U = [ t 0 , ) for some t 0 > 0
Solution: Note that e − yx 2 e − t x 0 2 for all y ∈U, x ≥ 0 Furthermore, the integral
− converges According to the Weierstrass theorem, the integral 2
To present some uniform convergence criterion for the integral of a product we need the following lemma, also known as Bonnet's theorem and is a form of the mean value theorem
Addendum (Bonnet’s theorem): If the function α(x) is monotone and the function g(x) integrable on [𝑎, 𝑏] then exists a point c ϵ [𝑎, 𝑏] such that
Proof: Consider the case α(x) does not increase and α(x) ≥ 0 ( The case α(x) no reduction is similar) Let P be any partition of [𝑎, 𝑏], given by the sequence point a = 𝑥 1 < 𝑥 2 < ⋯ < 𝑥 𝑛 = 𝑏 At that time
Note that α(x) is monotone and g(x) integrable should be bounded, i.e
|𝑔(𝑥)| < 𝛿 for all x ϵ [𝑎, 𝑏] and with for some 𝛿 > 0 The the second component on the right side of (*) can be evaluated as follows
Since α(x) is integrable, the subtraction in the above expression approaches the subtracted number unless the width of partitions ascending to 0 For the first
In the right-hand side of the equation, we observe that G(x) = ∫ g(x) dx, where g(x) is continuous on the interval [a, b] Consequently, G(x) attains a maximum value of M and a minimum value of m within this range Additionally, we can apply the following transformation.
Since 𝛼(𝑥 𝑖−1 ) − 𝛼(𝑥 𝑖 ) ≥ 0 and 𝛼(𝑥 𝑛 ) ≥ 0, the quantity 𝜎 𝑇 is clamped, namely m𝛼(𝑎) ≤ 𝜎 𝑇 ≤ 𝑀𝛼(𝑎) Through the limit when the width of the partition approaches 0 we have: m𝛼(𝑎) ≤ ∫ 𝑔(𝑥)𝛼(𝑥)𝑑𝑥 ≤ 𝑀𝛼(𝑎) 𝑎 𝑏 Since G(x) is continous funtion, there exists c ϵ [𝑎, 𝑏] such that
Now if 𝛼(𝑥) is not necessarily positive, for 𝛼(𝑥) − 𝛼(𝑏) ≥ 0 we consider the integral ∫ 𝑔(𝑥)( 𝑎 𝑏 𝛼(𝑥) − 𝛼(𝑏))𝑑𝑥 According to the above proof, we can find c ϵ [𝑎, 𝑏], so that
Is the necessary formula find
Theorem( Dirichlet criterion): It is assumed that: i) The integral ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 𝑏 is bounded by b and y, i.e there exits c > 0 so that
𝑖𝑖) 𝜑(𝑥, 𝑦) converges uniformly with respect to y ϵ U to 0 when x → ∞ and 𝜑(𝑥, 𝑦) are monotone with respect to x for each fixed y ϵ U
Then the integral ∫ 𝑓(𝑥, 𝑦) 𝑎 𝑏 𝜑(𝑥, 𝑦)𝑑𝑥 converges uniformly on U
Proof: Take 𝜀 > 0 arbitrary From ii) we can find 𝑏 0 to:
At that time, for every 𝑏 2 ≥ 𝑏 1 ≥ 𝑏 0 , combined with Bonnet’s theorem we have:
< 𝜀 4𝑐(2𝑐 + 2𝑐) = 𝜀, Where 𝜉 is a point in [𝑏 1 , 𝑏 2 ] According to Cauchy’s theorem, the integral
Abel's criterion states that if the integral ∫ f(x, y) dx from a to ∞ converges uniformly on U, and if the function ϕ(x, y) is uniformly bounded—meaning there exists a constant c > 0 such that |ϕ(x, y)| ≤ c for all x ≥ a and y in U—then for each fixed y in U, the function ϕ(., y) is monotone with respect to x.
Then the integral ∫ 𝑓(𝑥, 𝑦)𝜑(𝑥, 𝑦) 𝑎 ∞ converges uniformly on U
Prove Similar to the above theorem, apply Bonnet's theorem and Cauchy's theorem
Example: Investigate the uniform convergence of the integral ∫ sin (𝑦𝑥)
𝑥 𝑎𝑛𝑑 𝑓(𝑥, 𝑦) = 𝑠𝑖𝑛(𝑦𝑥) we immediately see that the conditions of the Dirichlet criterion is satisfied So this integral converges uniformly on U.
CONTINUITY
To investigate the properties of uniformly convergent integrals with close to infinity we establish the relationship of these integrals to the series of uniform convergence
Lemma: Assume that integral ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converges evenly on the set U and {𝑎 𝑛 } is a sequence of numbers up to +∞ where 𝑎 𝑛 > a Then the sequence of functions:
𝜑 𝑛 (y) = ∫ 𝑎 𝑎 𝑛 𝑓(𝑥, 𝑦)𝑑𝑥 converges evenly to the function I(y) on U
For each fixed value of \( y \) in the set \( U \), the integral \( \int f(x, y) \, dx \) converges, leading to the conclusion that the sequence of functions \( \{ \phi_n(y) \} \) converges to \( I(y) \) We will demonstrate that this series converges uniformly Given any \( \epsilon > 0 \), since \( I(y) \) converges uniformly, we can identify a value \( b_0 \) such that the uniform convergence criterion is satisfied.
Then there exists n 0 > 0 such that for every n ≥ n 0 , we have a n ≥ b (because{a n }an approaches ∞ ) So
| < 𝜀 for all n ≥ n 0 , 𝑦 ∈ 𝑈 Show that {φ n (y)} converges to I(y) on 𝑈
Theorem: Assume that the function f is deterministic and continuous over the domain [a,∞)×[α,β] and integral I(y) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converges evenly on [α,β] Then the function I(y) is continuous on [α,β]
Prove: Take the sequence {a n } an progressive to +∞, a n ≥ a and consider the sequence of functions
For any fixed value of n, the theorem regarding the continuity of parameter-dependent integrals with finite bounds ensures that the function φ n (y) remains continuous over the interval [α,β] By applying the relevant lemma, we find that the sequence {φ n (y)} converges uniformly to I(y) Consequently, based on the theorem concerning the continuity of uniformly convergent function series, we can deduce that the limit of the function is established.
𝑛→∞𝜑 𝑛 (𝑦) continuous on [𝛼, 𝛽] The theorem is proven
In the case of a positive function (f(x,y) ≥ 0), the converse of the previously mentioned theorem holds true, akin to Dini's theorem for series of functions Specifically, if f is continuous and positive over the domain [a,∞)×[α,β], and the integral ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 from a to ∞ converges to a continuous function I(y) on [α,β], then this integral converges uniformly To demonstrate this, we analyze the monotonicity of the sequence of continuous functions φ n (y) = ∫ 𝑎 𝑎+𝑛 𝑓(𝑥, 𝑦)𝑑𝑥, which converges to the continuous function I(y) on [α,β] By applying Dini's theorem, we conclude that this series of functions converges uniformly, ensuring that for every ε > 0, there exists an n₀ such that the convergence criteria are satisfied.
Prove the integral ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converges evenly on [α,β]
Example: Prove that I(y) =∫ 𝑎 +∞ 𝑦𝑒 −𝑦𝑥 𝑑𝑥 is not continuous at y = 0, that is,
Explain why the limit sign cannot be passed into the integral expression in this case.
DISABILITY
Theorem: Assuming that i) The function f is continuous and has partial 𝑓′ 𝑦 continuous on the domain
[a,∞)×[α,β] ii) Integral I(y) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converges on [α,β] iii) Integral ∫ 𝑓′ 𝑎 ∞ 𝑦 (𝑥, 𝑦)𝑑𝑥 converges evenly on [α,β]
Then the function I(y) is differentiable on [α,β] and the derivative is calculated by the formula:
Prove Consider the series of functions φ 𝑛 (y) = ∫ a 𝑎+𝑛 𝑓(𝑥, 𝑦)𝑑𝑥, y∈[α,β]
For each fixed n, by the theorem on the difference of integrals that depend on parameters with finite approach, the function φ n (y) is differentiable on [α,β] and
The sequence {φ′ 𝑛 (y)} converges within the interval [α,β], which, according to the previous lemma, indicates that the function I(y) is differentiable By applying the theorem on the differentiability of function sequences, we confirm the differentiability of I(y).
Example : Prove that the integral depends on the parameter
+ is a continuous function differentiable with respect to the variable 𝑦 Calculate I'(y) and then deduce the expression of I(y)
By the theorem 3.17, I(y) converges on [−∞, +∞]
∫ −∞ +∞ 𝑓 ′ 𝑦 (𝑥, 𝑦)𝑑𝑥 converges uniformly on [−∞, +∞] By the theorem 3.18, I(y) is differentiability on [−∞, +∞], and 𝐼 ′ (𝑦) = ∫ −∞ +∞ (1+𝑥 2 )[1+(𝑥+𝑦) 1 2 ] 𝑑𝑥
1+(𝑥+𝑦) 2 , using the coefficient homogenization method, we get:
The remaining problem is the test of the condition for passing the derivative over the integral sign
(𝑥 2 +𝑦) 𝑛+1 are continuous on [0, +) × [, +) for each given > 0
(𝑥 2 +𝜀) 𝑛+1 But the integrals ∫ 0 +∞ 𝑥 2 1 +𝜀 𝑑𝑥, … , ∫ 0 +∞ (𝑥 2 +𝜀) 1 𝑛+1 are converge, hence
0 converge uniformly on [, +) for each given > 0
INTEGRABLE
Theorem: Assume that: i) The function f is continuous on [a, ) × [, ] ii) The integral 𝐼(𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 𝑎 ∞ converges uniformly on [, ]
Then I(y) is integrable on [, ] and
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
𝑎 and then apply the theorem on the integrability of a series of functions that converge to get
The results can be extended to the integral I(y) over an infinite domain, such as [α, ∞) It is assumed that the function f remains continuous and positive within the domain [α, ∞) × [α, ∞), and the associated integrals are considered.
𝑎 converge to continuous functions Then if one of the integrals
∫ 𝑑𝑥 ∫ 𝑓𝑑𝑦, ∫ 𝑑𝑥 ∫ 𝑓𝑑𝑥 𝑎 ∞ 𝑎 ∞ 𝑎 ∞ 𝑎 ∞ exists then the remaining integral also exists and they are equal
Readers check for themselves the condition of exchanging the order of integration
Readers check for themselves the condition of exchanging the order of integration
2 Readers check for themselves the condition of exchanging the order of integration
Note that: Readers can compare a result in calculus III, this is
Example 4: Applying the integral Dirichlet formula, prove that:
[𝑠𝑢𝑔𝑔𝑒𝑠𝑡𝑖𝑜𝑛𝑠]: a) Applying the descending order formula 𝑠𝑖𝑛 3 𝑥 = 1
4𝑠𝑖𝑛𝑥 b) Applying the formula for integration by parts,
By Gauss integral formula, we have ∫ 𝑒 0 ∞ −𝑢 2 𝑑𝑢, we have:
2 Readers can check the condition of changing the order of integration
EXERCISES
Method 1: Calculate the parameter-dependent integral by changing the order of integration
Suppose you need to calculate (y) ( , ) a
Step 2: Use the reorder property to integrate:
Method 2: Calculate the parameter-dependent integral by derivation over the integral sign
Suppose that you need to calculate (y) ( , ) a
Step 2: Use the Newton-Leibniz formula to recover I (y) by
Step 3: Take a special value of y to determine C
When converting the order of integration in the Theorem of Integrability or changing the sign of the derivative in the Theorem of Differentiability, it is essential to verify the applicable conditions.
1: change the order of integration
On the other hand, we have:
Check the condition to change derivative over integral sign
= − − − is continuous with respect to x on 0, + ; , 0
is convergent to on open interval [ , + ) according to standard Weierstrass, indeed, − e − x − e − x , moreover
1: change the order of integration
We have: e x sin bx sin cx x
On the other hand: I c ( ) = 0, so arctan c
Check the condition to change derivative over integral sign
With f x b ( ) , e x sin bx sin cx x
= is continuous with respect to x on
2) f ' b ( ) x b , = e − x cos bx is continuous on [0, + ) ( 0, + )
' b , x cos x cos b x sin f x b dx e bx e bx e bx b b b
Is convergent with respect to b on ( 0, + ) according to standard Weierstrass, indeed, e − x cos bx e − x 2 is convergent, moreover
In the above proof, we used the formula
2 2 2 2 cosyx cosyx sin ax a ax y ax e dx e e yx a y a y
Exercise 3: In a similar way to Exercise 2, calculate
0 cos cos bx cx ln c x b
2 2 2 x x x y f x y dx xe yxdx e yx ye yxdx y I y
2) f '( , y) x = − xe − x 2 sin yx continuous on [0, +∞) x (-∞, +∞)
2 2 2 x x x y f x y dx xe yxdx e yx ye yxdx y I y
converges uniformly according to the Weierstrass criterion, indeed:
Therefore, according to the differentiability theorem,
EULER INTEGRAL
GAMMA FUNCTION
* With integral 𝐼 2 we compare with ∫ 𝑑𝑥
Even the integral 𝐼 2 converges for all p ∈ R
* With integral 𝐼 1 , then if p ≥ 1 we have 𝐼 1 which is a definite integral
If 0 < p < 1, then we compare 𝐼 1 with ∫ 𝑑𝑥
0 converges, so 𝐼 1 also converges If p < 0 then ∫ 𝑑𝑥
* Properties a) Lower order: Γ (p + 1) = pΓ (p) This formula can be easily proved by
Meaning: To study Γ (p), we only need to study Γ (p) with 0 < p ≤ 11, and for p > 1 we will use the descending formula
2 𝑛 √𝜋 c) Derivative of Gamma function: Γ (𝑘) (p)∫ 0 +∞ 𝑥 𝑝−1 (ln 𝑘 𝑥)𝑒 −𝑥 𝑑𝑥 d) Γ (p).Γ (1 − p) = 𝜋
BETA FUNCTION
Form 1: B (p, q) =∫ 𝑥 0 1 𝑝−1 (1 − 𝑥) 𝑞−1 𝑑𝑥 By changing the variable x = 𝑡
The answer: Put sin x = √𝑡 ⇒ 0 ≤ 𝑡 ≤ 1 , cos 𝑥 𝑑𝑥 = 1
From this example we get:
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) = 1 so
(𝑝 + 𝑛 − 1)(𝑝 + 𝑛 − 2) … (𝑝 + 1)𝑝 𝐵(𝑝, 1), ∀𝑛 ∈ ℕ d) Relationship formula between Beta and Gamma: B(p,q) = Γ(p)+Γ(q) Γ(p+q)
Applying the transformation formula t = xy and s = x(1 − y), we get:
Example: Prove the relation between Beta and Gamma function
Solution Starting from the formula 𝛤(𝑝) = ∫ 0 +∞ 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥, performing the transformation of the variable x = 𝑡 2 , we get:
Perform the transformation of the variable in polar coordinates {𝑥 = 𝑟 𝑐𝑜𝑠 𝜑
Let 𝑀 → +∞ in inequality (3.2), we get:
The Calculus module is a vital component of the university's advanced math curriculum, introduced in the first semester Integral calculus plays a crucial role, featuring a partial parametric dependent integral that is essential for solving various related mathematical problems.