Proving Asymptotic Comparisons of Integrals

[geoffrey159]

Homework Statement

Let $f$ be piecewise continuous from $[0,+\infty[$ into $V = \mathbb{R}$ or $\mathbb{C}$, such that $f(x) \longrightarrow_{ x\rightarrow +\infty} \ell$.

Show that $\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$

Homework Equations

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Integration of asymptotic comparisons

The Attempt at a Solution

[/B]
Can you tell me if this is correct please ?

Since $f - \ell = o_{+\infty}(1)$, and since $u: x \rightarrow 1$ is non-negative, piecewise continuous, and non-integrable on $[0,+\infty[$, then

$\int_0^x f(t) - \ell \ dt = o_{+\infty}(\int_0^x u(t) \ dt)$

which is the same as saying that $\int_0^x f(t) \ dt - x \ell = o_{+\infty}(x)$.

Multiplying left and right by $\frac{1}{x}$, I get that $\frac{1}{x}\ \int_0^x f(t) \ dt - \ell = o_{+\infty}(1)$ which proves that

$\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$.

Is this OK ?

 

[geoffrey159]

Never mind, I've had confirmation. Thanks !

 

[Mark44]

Homework Statement

Let $f$ be piecewise continuous from $[0,+\infty[$ into $V = \mathbb{R}$ or $\mathbb{C}$, such that $f(x) \longrightarrow_{ x\rightarrow +\infty} \ell$.

I think this is nicer notation: $\lim_{x \to \infty} f(x) = \ell$
My LaTeX script is ##\lim_{x \to \infty} f(x) = \ell##

Show that $\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$

Homework Equations

[/B]
Integration of asymptotic comparisons

The Attempt at a Solution

[/B]
Can you tell me if this is correct please ?

Since $f - \ell = o_{+\infty}(1)$, and since $u: x \rightarrow 1$ is non-negative, piecewise continuous, and non-integrable on $[0,+\infty[$, then

$\int_0^x f(t) - \ell \ dt = o_{+\infty}(\int_0^x u(t) \ dt)$

which is the same as saying that $\int_0^x f(t) \ dt - x \ell = o_{+\infty}(x)$.

Multiplying left and right by $\frac{1}{x}$, I get that $\frac{1}{x}\ \int_0^x f(t) \ dt - \ell = o_{+\infty}(1)$ which proves that

$\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$.

Is this OK ?

 

[HallsofIvy]

Homework Statement

Let $f$ be piecewise continuous from $[0,+\infty[$ into $V = \mathbb{R}$ or $\mathbb{C}$, such that $f(x) \longrightarrow_{ x\rightarrow +\infty} \ell$.

Show that $\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$

Homework Equations

[/B]
Integration of asymptotic comparisons

The Attempt at a Solution

[/B]
Can you tell me if this is correct please ?

Since $f - \ell = o_{+\infty}(1)$, and since $u: x \rightarrow 1$ is non-negative, piecewise continuous, and non-integrable on $[0,+\infty[$, thena

Surely you meant "integrable" not "non-integrable" here?

$\int_0^x f(t) - \ell \ dt = o_{+\infty}(\int_0^x u(t) \ dt)$

which is the same as saying that $\int_0^x f(t) \ dt - x \ell = o_{+\infty}(x)$.

Multiplying left and right by $\frac{1}{x}$, I get that $\frac{1}{x}\ \int_0^x f(t) \ dt - \ell = o_{+\infty}(1)$ which proves that

$\frac{1}{x}\ \int_0^x f(t) \ dt \longrightarrow_{ x\rightarrow +\infty} \ell$.

Is this OK ?

 

[geoffrey159]

I think this is nicer notation: $\lim_{x \to \infty} f(x) = \ell$
My LaTeX script is ##\lim_{x \to \infty} f(x) = \ell##

:-) Ok thanks, I'll try to follow that notation in the future

Surely you meant "integrable" not "non-integrable" here?

No, why do you say that? $u = 1$ is non-integrable on $[0,+\infty[$ since $\int_0^x u(t) \ dt$ does not have a finite limit as $x$ tends to infinity.