Improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable
up vote
3
down vote
favorite
Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.
I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.
real-analysis measure-theory improper-integrals lebesgue-integral
asked Nov 22 at 14:59
D. Brito
360111
add a comment |
up vote
3
down vote
favorite
Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.
I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.
real-analysis measure-theory improper-integrals lebesgue-integral
asked Nov 22 at 14:59
D. Brito
360111
2
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.
I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.
real-analysis measure-theory improper-integrals lebesgue-integral
asked Nov 22 at 14:59
D. Brito
360111
Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.
I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.
real-analysis measure-theory improper-integrals lebesgue-integral
real-analysis measure-theory improper-integrals lebesgue-integral
asked Nov 22 at 14:59
D. Brito
360111
asked Nov 22 at 14:59
D. Brito
360111
edited Nov 23 at 15:22
asked Nov 22 at 14:59
D. Brito
360111
asked Nov 22 at 14:59
D. Brito
360111
asked Nov 22 at 14:59
D. Brito
360111
360111
2
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22
add a comment |
2
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22
2
2
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Convergence of the improper integral.
This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$
See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity
See also the following questions:
- Definite integral of $cos (x)/ sqrt{x}$?
- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$
Lebesgue integrability.
Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$
answered Nov 22 at 15:36
user587192
1,473112
add a comment |
up vote
0
down vote
The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Convergence of the improper integral.
This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$
See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity
See also the following questions:
- Definite integral of $cos (x)/ sqrt{x}$?
- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$
Lebesgue integrability.
Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$
answered Nov 22 at 15:36
user587192
1,473112
add a comment |
up vote
4
down vote
accepted
Convergence of the improper integral.
This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$
See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity
See also the following questions:
- Definite integral of $cos (x)/ sqrt{x}$?
- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$
Lebesgue integrability.
Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$
answered Nov 22 at 15:36
user587192
1,473112
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Convergence of the improper integral.
This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$
See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity
See also the following questions:
- Definite integral of $cos (x)/ sqrt{x}$?
- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$
Lebesgue integrability.
Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$
answered Nov 22 at 15:36
user587192
1,473112
Convergence of the improper integral.
This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$
See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity
See also the following questions:
- Definite integral of $cos (x)/ sqrt{x}$?
- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$
Lebesgue integrability.
Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$
answered Nov 22 at 15:36
user587192
1,473112
edited Nov 22 at 15:52
answered Nov 22 at 15:36
user587192
1,473112
answered Nov 22 at 15:36
user587192
1,473112
answered Nov 22 at 15:36
user587192
1,473112
1,473112
add a comment |
add a comment |
up vote
0
down vote
The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
add a comment |
up vote
0
down vote
The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
add a comment |
up vote
0
down vote
up vote
0
down vote
The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
answered Nov 22 at 20:57
Jack D'Aurizio
285k33275654
285k33275654
add a comment |
add a comment |
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2
Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14
@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00
The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05
take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22