Definite integral depending on a parameter

3.1 Definite integral depending on a parameter

Definition 3.1.1 Assuming a function $f (x, y)$ where $x \in [a, b]$ and $y \in U \subseteq R^{n}$ . For any $y \in U$ , $f (\cdot, y) \in R [a, b]$ . Mark

∫ b F(y ) = f(x,y )dx, F : U → ℝ a

as the integral depending on a parameter, where $y$ is called the parameter of the integral.

What we are interested.

The limit/continuity of $F$ .
$∫ b ∫ b ∫ b lim f (x, y)dx −→? f (x,y )dx = lim f(x, y)dx y→y0 a a 0 a y→y0 ◟-------◝◜--------◞ f is continuous w.r.t. y at y0$
Derivability, $C^{r}$ and the computation of the derivative.
$∂ ∫ b ? ∫ b ∂ ---- f(x, y)dx −→ ---f (x, y)dx ∂yk a a ∂yk$
Integral.
$∫ ∫ ∫ ∫ d b ? b d dy f(x, y)dx −→ dx f(x,y)dy c a a c$

They are all questions like if the order of the original integral and some kind of the limit (i.e. limit, derivative and another integral) could be exchanged.

Limit. Assuming $f (x, y)$ is continuous(?) with respect to $y$ , i.e. for any $ϵ > 0$ , there exists $δ > 0$ , s.t. for any $y \in U$ , $∥ y - y_{0} ∥ < δ \Rightarrow | f (x, y) - f (x, y_{0}) ∥ < ϵ$ . Therefore,

∫ b ∫ b ∫ b ∫ b ∖left| f(x,y)dx − f(x, y )dx∖right| ≤ |f (x, y) − f(x,y )|dx ≤ 𝜖dx < 𝜖′ a a 0 a 0 a

The proof above is wrong! Since $δ$ might be related to $x$ ! So we should add additional conditions to $f$ .

Theorem 3.1.2 Assuming $f : [a, b] \times U \to R$ satisfies

$\forall y \in U, f (\cdot, y) \in R [a, b]$ .
$f (x, y)$ is continuous with respect to $y$ at $y_{0}$ , and is uniform with respect to $x$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ) > 0$ such that $\forall x \in [a, b], \forall y \in U$ , $∥ y - y_{0} ∥ < δ (ϵ) \Rightarrow | f (x, y) - f (y_{0}) | < ϵ$ .

Then $\int_{a}^{b} f (x, y) d x$ is continuous at $y_{0}$ .

Corollary 3.1.3 Assuming $U \in R^{n}$ is closed/open, $f : [a, b] \times U \to R$ is continuous with respect to $(x, y)$ , then $\int_{a}^{b} f (x, y) d x$ is continuous with respect to $y$ .

Proof If $U$ is open, $y_{0} \in U$ , then $\exists δ_{0} > 0$ such that $K = \overset{―}{B (y_{0}, δ_{0})} \subseteq U$ . If $U$ is closed, $y_{0} \in U$ , then arbitrarily take $δ_{0} > 0$ , $K = U \cap \overset{―}{B (y_{0}, δ_{0})}$ .

$f$ is uniformly continuous on bounded closed set $[a, b] \times K$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ) > 0$ such that $\forall (x_{1}, y_{1}) \in K, (x_{ϵ}, y_{ϵ}) \in K$ , $| x_{1} - x_{ϵ} | < δ (ϵ) \land ∥ y_{1} - y_{ϵ} ∥ < δ (ϵ) \Rightarrow | f (x_{1}, y_{1}) - f (x_{ϵ}, y_{ϵ}) | < ϵ$ . $◻$

Corollary 3.1.4 If $f$ satisfies

$\forall y \in U, f (\cdot, y) \in R [a, b]$ .
$\frac{\partial f}{\partial y_{1}}, . . ., \frac{\partial f}{\partial y_{n}}$ is bounded.

Then $\int_{a}^{b} f (x, y) d x$ is continuous at $y_{0}$ .

Proof Hint $y_{k + 1} = y_{k} + Δ y_{k}$ where $Δ y_{k} = (y_{k} - y_{k}^{0}) e_{k}$ .

Example 3.1.1 Define

∫ 1 F (y) = cos(xy)dx 0

$c o s (x y)$ is continuous with respect to $(x, y)$ . We have

∫ 1 ∫ 1 ∫ 1 lim F (y) = lim cos(xy )dx = lim cos(xy)dx = 1dx = 1 y→0 y→0 0 0 y→0 0

On the other hand, when $y \neq 0$ , we have

∫ y F (y) = 1-costdt = siny- 0 y y

Derivability.

Theorem 3.1.5 Assuming $\forall k = 1, . . ., n$ , $\frac{\partial f}{\partial y_{k}} (x, y)$ is continuous with respect to $(x, y)$ ¹, then $F$ is of $C^{1}$ and

∫ b ∫ b -∂-- f(x, y)dx = ∂F-(y) = ∂f--(x, y)dx ∂yk a ∂yk a ∂yk

Proof Only the last formula is needed to be proven. WLOG assuming $k = 1$ , i.e. to prove

∫ b ∂f α(t) = F (y + te1) − F (y) − t ---(x, y)dx = o(t), t → 0 a ∂y1

We have $\begin{aligned} α (t) & = | F (y + t e_{1}) - F (y) - t \int_{a}^{b} \frac{\partial f}{\partial y_{1}} (x, y) d x | \\ = | \int_{a}^{b} f (x, y + t e_{1}) d x - \int_{a}^{b} f (x, y) d x - t \int_{a}^{b} \frac{\partial f}{\partial y_{1}} (x, y) d x | \\ \leq \int_{a}^{b} \underset{= β (t) = o (t)}{\underset{⏟}{| f (x, y + t e_{1}) - f (x, y) - t \frac{\partial f}{\partial y_{1}} (x, y) |}} d x \\ \leq \int_{a}^{b} ϵ | t | d x = ϵ (b - a) | t | = o (t) \end{aligned}$

The proof above is wrong! Since $t$ might be related to $x$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ, x)$ such that $0 < | t | < δ \Rightarrow | β (t) | < ϵ$ . There are infinite $x$ ’s, possibly $inf_{x} δ (ϵ, x) = 0$ ! Therefore, $\begin{aligned} α (t) & \leq \int_{a}^{b} | \underset{⏟}{f (x, y + t e_{1}) - f (x, y)} - t \frac{\partial f}{\partial y_{1}} (x, y) | d x \\ = \int_{a}^{b} | t | | \frac{\partial f}{\partial y_{1}} (x, y + θ t e_{1}) - \frac{\partial f}{\partial y_{1}} (x, y) | d x, θ \in (0, 1) \end{aligned}$

Here $θ$ might also be related with $x$ ! Since $\frac{\partial f}{\partial y_{1}} (x, y)$ is continuous with respect to $(x, y)$ , take $δ_{1} > 0$ , the open ball at $(x, y_{0})$ , $B (y_{0}, δ_{1}) \subseteq U$ , $K = [a, b] \times \overset{―}{B (y_{0}, δ_{1})}$ is bounded closed, $\frac{\partial f}{\partial y_{1}}$ is uniformly continuous on $K$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ) > 0$ such that $\forall (x_{1}, y_{1}), (x_{2}, y_{2}) \in K$ ,

∂f ∂f |x1 − x2| < δ(𝜖) ∧ ∥y1 − y2 ∥ < δ(𝜖) ⇒ ∖left|---(x1, y1) − ---(x2,y2 )∖right | < 𝜖 ∂y1 ∂y1

Hence, when $| t | < δ (ϵ)$ , $\forall x \in [a, b]$ ,

∥(x,y0 + 𝜃te1) − (x,y0)∥ = ∥ 𝜃te1∥ = 𝜃|t| ≤ |t| < δ(𝜖)

Therefore

∂f-- -∂f- ∖left|∂y1(x,y + 𝜃te1 ) − ∂y1 (x,y)∖right | < 𝜖

Terminally, theorem is proved. $◻$

Corollary 3.1.6 Assuming all $k^{th}$ -order partial derivative of $f$ over $y$ is continuous with respect to $(x, y)$ , then $F$ is of $C^{k}$ with respect to $y$ and

∂kF ∫ b ∂kf -----------(y) = -----------(x, y)dx ∂yi1 ⋅⋅⋅∂yik a ∂yi1 ⋅⋅⋅∂yik

Integrability.

Theorem 3.1.7 Assuming $f : [a, b] \times [c, d] \to R$ is continuous, then $F (y) = \int_{a}^{b} f (x, y) d x$ satisfies

∫ d ∫ b ∫ b ∫ d dy f (x,y)dx = dx f (x,y)dy c a a c

Proof To prove

∫ t ∫ b ∫ b ∫ t c dy a f (x,y)dx = a dx c f(x, y)dy

holds for any $t \in [c, d]$ . Assuming

∫ ∫ ∫ ∫ t b b t G (t) = dy f (x, y)dx − dx f(x,y)dy c a a c

We have $G (c) = 0$ . Notice that

∫ b (?)∫ b ∫ t G ′(t) = f(x, t)dx − dx ⋅ d- f(x,y )dy = 0 a a dt c

So $G (d) = G (c) = 0$ .

Why (?) is true? Assuming $g (x, t) = \int_{c}^{t} f (x, y) d y$ , $\frac{\partial g}{\partial t} = f (x, t)$ is continuous with respect to $(x, y)$ , so

∂ ∫ b ∫ b ∂g --- g(x,t)dx = --(x, t)dx ∂t a a ∂t

i.e.

∂ ∫ b ∫ t ∫ b --- dx f (x, y)dy = f(x,t)dx ∂t a c a

$◻$

Example 3.1.2 When $b > a > 0$ , seek

∫ 1 b a x-−--x-dx 0 ln x

Assuming

∫ 1 b a F (b) = x-−--x-dx 0 ln x

Consider

xb − xa xb − xa lim --------= 0, lim --------= b − a x→0 ln x x→1 ln x

So we can define

f(x,b) =

then $f$ is continuous. Notice that $F (a) = 0$ , and

∫ 1 b a ∫ 1 F′(b) = -∂-∖left(x--−-x-∖right )dx = xbdx = --1-- 0 ∂b ln x 0 b + 1

We can do the derivative since $x^{b}$ is binarily continuous when $b > 0$ . Therefore,

∫ b b + 1 F (b) = F (a) + F ′(t)t = ln∖lef t(------∖right) a a + 1

Example 3.1.3 (Variation) Assuming $a, b \in R^{n}$ , $γ$ is a curve of at least $C^{2}$ from $a$ to $b$ . Assuming $x (t) : [0, 1] \to R^{n}$ is a regular representative of the curve, i.e. $x (0) = a$ , $x (1) = b$ , $x^{'} (t) \neq 0$ for any $t \in [0, 1]$ . Assuming

∫ 1 L (γ ) = ∥x ′(t)∥dt 0

What’s $γ$ when $L (γ)$ takes the minimum?

Let

y (t,s) : [0,1] × [− δ,δ] → ℝn

is of at least $C^{2}$ where $\begin{aligned} y (t, 0) = x (t), & \forall t \in [0, 1] \\ y (0, s) = a, y (1, s) = b, & \forall s \in [- δ, δ] \end{aligned}$

Then we have

∘ ------------------------------ ∫ 1 ∂y ∫ 1 ∂y ∂y L(s) = ∖lef t∥ ---(t,s )∖right ∥dt = ∖left⟨---(t,s ), ---(t,s)∖right⟩dt 0 ∂t 0 ∂t ∂t

Assuming $x (t) = y (t, 0)$ has the minimum length among all $y (t, s)$ , then $L^{'} (0) = 0$ , i.e. $\begin{aligned} L^{'} (s) & = \int_{0}^{1} \frac{\partial}{\partial s} \sqrt{⟨ \frac{\partial y}{\partial t} (t, s), \frac{\partial y}{\partial t} (t, s) ⟩} d t = \int_{0}^{1} \frac{⟨ \frac{\partial^{2} y}{\partial s \partial t} (t, s), \frac{\partial y}{\partial t} (t, s) ⟩}{\sqrt{⟨ \frac{\partial y}{\partial t} (t, s), \frac{\partial y}{\partial t} (t, s) ⟩}} d t \\ L^{'} (0) & = \int_{0}^{1} ⟨ \frac{\partial^{2} y}{\partial s \partial t} (t, 0), \frac{x^{'} (t)}{∥ x^{'} (t) ∥} ⟩ d t \\ \to_{by part}^{integral} {⟨ \frac{\partial y}{\partial s} (t, 0), \frac{x^{'} (t)}{∥ x^{'} (t) ∥} ⟩ |}_{t = 0}^{1} - \int_{0}^{1} ⟨ \frac{\partial y}{\partial s} (t, 0), \frac{d}{d t} (\frac{x^{'} (t)}{∥ x^{'} (t) ∥}) ⟩ d t \\ = - \int_{0}^{1} ⟨ \frac{\partial y}{\partial s} (t, 0), \frac{d}{d t} (\frac{x^{'} (t)}{∥ x^{'} (t) ∥}) ⟩ d t = 0 \end{aligned}$

Construct

′ -d- -x-(t)-- y (t,s) = x (t) + sdt∖lef t(∥x′(t)∥ ∖right)

Therefore

∫ 1 d x′(t) 0 = ∖left∥ --∖left(-------∖right)∖right∥2dt ≥ 0 0 dt ∥x′(t)∥

Since $x (t)$ is of $C^{2}$ , it takes the equality if and only if

d-- -x-′(t)- dt∖left(∥x ′(t)∥∖right) = 0

meaning $x (t)$ is a straight line.

$y (t, s)$ constructed above doesn’t satisfy the boundary conditions, so we can adjust it to

d x ′(t) y (t,s) = x (t) + s ⋅ h(t)-∖left(-′---∖right ) dt ∥x (t)∥

where $h (t)$ is of $C^{\infty}$ and

h(t) =

Therefore

∫ 1 d x ′(t) ∫ 1−𝜖 d x ′(t) 0 = h(t)∖lef t∥ --∖left(---′---∖right)∖right ∥2dt ≥ ∖left∥ --∖left(---′---∖right)∖right∥2dt ≥ 0 0 dt ∥x (t)∥ 𝜖 dt ∥x (t)∥

meaning

′ -x-(t)- = const vector ∥x ′(t)∥

i.e. $x (t)$ is a straight line.

Note: what we proved above is that if the curve of minimum length exists, then it can only be the straight line; yet we don’t prove that the curve of minimum length exists!

¹When it comes to integral depending on a parameter, the condition $\forall y \in U, f (\cdot, y) \in R [a, b]$ is always necessary. So later when we are illustrating theorems concerning integral depending on a parameter, we will neglect this condition.