If $f(x)=o(g(x))$ prove that $int_a^x f(t)dt=C+oleft(int_a^x g(t)dtright)$

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I have a big doubt. @Did gave me a doubts in his comment in this post.

Is it true that (if $f$ and $g$ are integrable), that $$f(x)=o(g(x))implies int_a^t f(x)dx=oleft(int_a^t g(x)dxright) ?$$

For me it was always true since for example

$$frac11+x=1-x+o(x)implies ln(1+t)=int_0^tfrac11+xdx=t-fract^22+o(t^2).$$

Now I tried to prove it as :

---

Suppose $f(x)=o_x=0(g(x))$. Let $varepsilon>0$. There is $delta>0$ s.t. $$|x|<deltaimplies |f(x)|leq varepsilon|g(x)|,$$
and thus, $$|t|<deltaimplies int_0^t|f(x)|dxleq varepsilon int_0^t |g(x)|dx.$$

Unfortunately, I only get $$int_0^t|f(x)|dx=oleft(int_0^t |g(x)|dxright),$$
but not
$$int_0^t f(x)dx=oleft(int_0^t g(x)dxright),$$
so maybe I'm wrong ?

real-analysis asymptotics

share|cite|improve this question

edited Jul 18 at 16:28

asked Jul 18 at 15:20

Peter

348112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
  – Did
  Jul 18 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
  – Peter
  Jul 18 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
  – Did
  Jul 18 at 15:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did : so it doesn't work. Thank you, that really help :)
  – Peter
  Jul 18 at 15:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No it does not. BTW, why the completely illogical $C$?
  – Did
  Jul 18 at 15:59
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

1

I have a big doubt. @Did gave me a doubts in his comment in this post.

Is it true that (if $f$ and $g$ are integrable), that $$f(x)=o(g(x))implies int_a^t f(x)dx=oleft(int_a^t g(x)dxright) ?$$

For me it was always true since for example

$$frac11+x=1-x+o(x)implies ln(1+t)=int_0^tfrac11+xdx=t-fract^22+o(t^2).$$

Now I tried to prove it as :

---

Suppose $f(x)=o_x=0(g(x))$. Let $varepsilon>0$. There is $delta>0$ s.t. $$|x|<deltaimplies |f(x)|leq varepsilon|g(x)|,$$
and thus, $$|t|<deltaimplies int_0^t|f(x)|dxleq varepsilon int_0^t |g(x)|dx.$$

Unfortunately, I only get $$int_0^t|f(x)|dx=oleft(int_0^t |g(x)|dxright),$$
but not
$$int_0^t f(x)dx=oleft(int_0^t g(x)dxright),$$
so maybe I'm wrong ?

real-analysis asymptotics

share|cite|improve this question

edited Jul 18 at 16:28

asked Jul 18 at 15:20

Peter

348112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
  – Did
  Jul 18 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
  – Peter
  Jul 18 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
  – Did
  Jul 18 at 15:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did : so it doesn't work. Thank you, that really help :)
  – Peter
  Jul 18 at 15:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No it does not. BTW, why the completely illogical $C$?
  – Did
  Jul 18 at 15:59
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

I have a big doubt. @Did gave me a doubts in his comment in this post.

Is it true that (if $f$ and $g$ are integrable), that $$f(x)=o(g(x))implies int_a^t f(x)dx=oleft(int_a^t g(x)dxright) ?$$

For me it was always true since for example

$$frac11+x=1-x+o(x)implies ln(1+t)=int_0^tfrac11+xdx=t-fract^22+o(t^2).$$

Now I tried to prove it as :

---

Suppose $f(x)=o_x=0(g(x))$. Let $varepsilon>0$. There is $delta>0$ s.t. $$|x|<deltaimplies |f(x)|leq varepsilon|g(x)|,$$
and thus, $$|t|<deltaimplies int_0^t|f(x)|dxleq varepsilon int_0^t |g(x)|dx.$$

Unfortunately, I only get $$int_0^t|f(x)|dx=oleft(int_0^t |g(x)|dxright),$$
but not
$$int_0^t f(x)dx=oleft(int_0^t g(x)dxright),$$
so maybe I'm wrong ?

real-analysis asymptotics

share|cite|improve this question

edited Jul 18 at 16:28

asked Jul 18 at 15:20

Peter

348112

I have a big doubt. @Did gave me a doubts in his comment in this post.

Is it true that (if $f$ and $g$ are integrable), that $$f(x)=o(g(x))implies int_a^t f(x)dx=oleft(int_a^t g(x)dxright) ?$$

For me it was always true since for example

$$frac11+x=1-x+o(x)implies ln(1+t)=int_0^tfrac11+xdx=t-fract^22+o(t^2).$$

Now I tried to prove it as :

---

Suppose $f(x)=o_x=0(g(x))$. Let $varepsilon>0$. There is $delta>0$ s.t. $$|x|<deltaimplies |f(x)|leq varepsilon|g(x)|,$$
and thus, $$|t|<deltaimplies int_0^t|f(x)|dxleq varepsilon int_0^t |g(x)|dx.$$

Unfortunately, I only get $$int_0^t|f(x)|dx=oleft(int_0^t |g(x)|dxright),$$
but not
$$int_0^t f(x)dx=oleft(int_0^t g(x)dxright),$$
so maybe I'm wrong ?

real-analysis asymptotics

share|cite|improve this question

edited Jul 18 at 16:28

asked Jul 18 at 15:20

Peter

348112

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 16:28

edited Jul 18 at 16:28

edited Jul 18 at 16:28

asked Jul 18 at 15:20

Peter

348112

asked Jul 18 at 15:20

Peter

348112

asked Jul 18 at 15:20

Peter

348112

348112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
  – Did
  Jul 18 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
  – Peter
  Jul 18 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
  – Did
  Jul 18 at 15:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did : so it doesn't work. Thank you, that really help :)
  – Peter
  Jul 18 at 15:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No it does not. BTW, why the completely illogical $C$?
  – Did
  Jul 18 at 15:59
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
  – Did
  Jul 18 at 15:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
  – Peter
  Jul 18 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
  – Did
  Jul 18 at 15:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did : so it doesn't work. Thank you, that really help :)
  – Peter
  Jul 18 at 15:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No it does not. BTW, why the completely illogical $C$?
  – Did
  Jul 18 at 15:59
  

  
  
  
  

Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
– Did
Jul 18 at 15:49

Are you assuming that $g$ is nonnegative? Because, as you know, $f=o(g)$ is strictly equivalent to $|f|=o(|g|)$...
– Did
Jul 18 at 15:49

@Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
– Peter
Jul 18 at 15:50

@Did: thank you for all your answers :) In fact, I don't know that $g$ is non negative.
– Peter
Jul 18 at 15:50

Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
– Did
Jul 18 at 15:52

Then the integrals of $g$ might be too small, due to some cancellations between parts where $g>0$ and parts where $g<0$.
– Did
Jul 18 at 15:52

@Did : so it doesn't work. Thank you, that really help :)
– Peter
Jul 18 at 15:53

@Did : so it doesn't work. Thank you, that really help :)
– Peter
Jul 18 at 15:53

No it does not. BTW, why the completely illogical $C$?
– Did
Jul 18 at 15:59

No it does not. BTW, why the completely illogical $C$?
– Did
Jul 18 at 15:59

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

If $g$ is non negative then as $left| int_0^t f right| leq int_0^t |f|$ then you obtain your result.

If it is not the case consider for example:

• $f(x)=x^1+alpha left| sinleft( frac1x right) right|$


• $g(x)=x sinleft( frac1x right)$

Then for $alpha >0$, $f(x)=o(g(x))$.

More over:
$$int_0^t g(x) dx =int_frac1t^infty fracsin(u)u^3 du$$
$$int_0^t f(x) dx =int_frac1t^infty fracsin(u)u^3+alpha du$$
to estimate the integrals notice that:
$$int_k pi^(k+1) pi fracsin(u)u^3=(-1)^kint_0^pi fracsin(s)(k pi+s)^3 du$$
so:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac(-1)^k(k pi)^3$$

Similarly, as:
$$int_k pi^(k+1) pi fracsin(u)u^3+alpha=int_0^pi fracsin(s)(k pi+s)^3+alpha du$$
we have:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac1(k pi)^3+alpha$$

Now as the first series is an alternating series and the second one a Riemann series you can show that:
$$left|int_0^t g(x) dx right| leq C t^3$$
$$left|int_0^t f(x) dx right| geq c t^2+alpha$$
so wuth any $alpha < 1$ you obtain a counterexample.

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

If $g$ is non negative then as $left| int_0^t f right| leq int_0^t |f|$ then you obtain your result.

If it is not the case consider for example:

• $f(x)=x^1+alpha left| sinleft( frac1x right) right|$


• $g(x)=x sinleft( frac1x right)$

Then for $alpha >0$, $f(x)=o(g(x))$.

More over:
$$int_0^t g(x) dx =int_frac1t^infty fracsin(u)u^3 du$$
$$int_0^t f(x) dx =int_frac1t^infty fracsin(u)u^3+alpha du$$
to estimate the integrals notice that:
$$int_k pi^(k+1) pi fracsin(u)u^3=(-1)^kint_0^pi fracsin(s)(k pi+s)^3 du$$
so:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac(-1)^k(k pi)^3$$

Similarly, as:
$$int_k pi^(k+1) pi fracsin(u)u^3+alpha=int_0^pi fracsin(s)(k pi+s)^3+alpha du$$
we have:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac1(k pi)^3+alpha$$

Now as the first series is an alternating series and the second one a Riemann series you can show that:
$$left|int_0^t g(x) dx right| leq C t^3$$
$$left|int_0^t f(x) dx right| geq c t^2+alpha$$
so wuth any $alpha < 1$ you obtain a counterexample.

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

add a comment | 

up vote
1
down vote



accepted

If $g$ is non negative then as $left| int_0^t f right| leq int_0^t |f|$ then you obtain your result.

If it is not the case consider for example:

• $f(x)=x^1+alpha left| sinleft( frac1x right) right|$


• $g(x)=x sinleft( frac1x right)$

Then for $alpha >0$, $f(x)=o(g(x))$.

More over:
$$int_0^t g(x) dx =int_frac1t^infty fracsin(u)u^3 du$$
$$int_0^t f(x) dx =int_frac1t^infty fracsin(u)u^3+alpha du$$
to estimate the integrals notice that:
$$int_k pi^(k+1) pi fracsin(u)u^3=(-1)^kint_0^pi fracsin(s)(k pi+s)^3 du$$
so:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac(-1)^k(k pi)^3$$

Similarly, as:
$$int_k pi^(k+1) pi fracsin(u)u^3+alpha=int_0^pi fracsin(s)(k pi+s)^3+alpha du$$
we have:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac1(k pi)^3+alpha$$

Now as the first series is an alternating series and the second one a Riemann series you can show that:
$$left|int_0^t g(x) dx right| leq C t^3$$
$$left|int_0^t f(x) dx right| geq c t^2+alpha$$
so wuth any $alpha < 1$ you obtain a counterexample.

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

If $g$ is non negative then as $left| int_0^t f right| leq int_0^t |f|$ then you obtain your result.

If it is not the case consider for example:

• $f(x)=x^1+alpha left| sinleft( frac1x right) right|$


• $g(x)=x sinleft( frac1x right)$

Then for $alpha >0$, $f(x)=o(g(x))$.

More over:
$$int_0^t g(x) dx =int_frac1t^infty fracsin(u)u^3 du$$
$$int_0^t f(x) dx =int_frac1t^infty fracsin(u)u^3+alpha du$$
to estimate the integrals notice that:
$$int_k pi^(k+1) pi fracsin(u)u^3=(-1)^kint_0^pi fracsin(s)(k pi+s)^3 du$$
so:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac(-1)^k(k pi)^3$$

Similarly, as:
$$int_k pi^(k+1) pi fracsin(u)u^3+alpha=int_0^pi fracsin(s)(k pi+s)^3+alpha du$$
we have:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac1(k pi)^3+alpha$$

Now as the first series is an alternating series and the second one a Riemann series you can show that:
$$left|int_0^t g(x) dx right| leq C t^3$$
$$left|int_0^t f(x) dx right| geq c t^2+alpha$$
so wuth any $alpha < 1$ you obtain a counterexample.

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

If $g$ is non negative then as $left| int_0^t f right| leq int_0^t |f|$ then you obtain your result.

If it is not the case consider for example:

• $f(x)=x^1+alpha left| sinleft( frac1x right) right|$


• $g(x)=x sinleft( frac1x right)$

Then for $alpha >0$, $f(x)=o(g(x))$.

More over:
$$int_0^t g(x) dx =int_frac1t^infty fracsin(u)u^3 du$$
$$int_0^t f(x) dx =int_frac1t^infty fracsin(u)u^3+alpha du$$
to estimate the integrals notice that:
$$int_k pi^(k+1) pi fracsin(u)u^3=(-1)^kint_0^pi fracsin(s)(k pi+s)^3 du$$
so:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac(-1)^k(k pi)^3$$

Similarly, as:
$$int_k pi^(k+1) pi fracsin(u)u^3+alpha=int_0^pi fracsin(s)(k pi+s)^3+alpha du$$
we have:
$$int_frac1t^infty fracsin(u)u^3 sim 2sum_k =frac1pi t^infty frac1(k pi)^3+alpha$$

Now as the first series is an alternating series and the second one a Riemann series you can show that:
$$left|int_0^t g(x) dx right| leq C t^3$$
$$left|int_0^t f(x) dx right| geq c t^2+alpha$$
so wuth any $alpha < 1$ you obtain a counterexample.

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 16:20

Delta-u

4,742518

answered Jul 18 at 16:20

Delta-u

4,742518

answered Jul 18 at 16:20

Delta-u

4,742518

4,742518

add a comment | 

add a comment | 

 

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