Find the value of a double integral [closed]
Let $int_{-infty}^{infty}f(x)dx=1 $ and $int_{-infty}^{infty}g(x)dx=1 $,
Must it be the case that
$int_{-infty}^{infty}int_{-infty}^{infty}f(x)g(y-x)dxdy=1$?
integration
edited Nov 28 at 16:14
Mason
1,8971528
asked Nov 28 at 15:13
Anson NG
41
closed as off-topic by Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 0:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $int_{-infty}^{infty}f(x)dx=1 $ and $int_{-infty}^{infty}g(x)dx=1 $,
Must it be the case that
$int_{-infty}^{infty}int_{-infty}^{infty}f(x)g(y-x)dxdy=1$?
integration
edited Nov 28 at 16:14
Mason
1,8971528
asked Nov 28 at 15:13
Anson NG
41
closed as off-topic by Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 0:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
1
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29
add a comment |
Let $int_{-infty}^{infty}f(x)dx=1 $ and $int_{-infty}^{infty}g(x)dx=1 $,
Must it be the case that
$int_{-infty}^{infty}int_{-infty}^{infty}f(x)g(y-x)dxdy=1$?
integration
edited Nov 28 at 16:14
Mason
1,8971528
asked Nov 28 at 15:13
Anson NG
41
Let $int_{-infty}^{infty}f(x)dx=1 $ and $int_{-infty}^{infty}g(x)dx=1 $,
Must it be the case that
$int_{-infty}^{infty}int_{-infty}^{infty}f(x)g(y-x)dxdy=1$?
integration
integration
edited Nov 28 at 16:14
Mason
1,8971528
asked Nov 28 at 15:13
Anson NG
41
edited Nov 28 at 16:14
Mason
1,8971528
asked Nov 28 at 15:13
Anson NG
41
edited Nov 28 at 16:14
Mason
1,8971528
edited Nov 28 at 16:14
Mason
1,8971528
edited Nov 28 at 16:14
Mason
1,8971528
1,8971528
asked Nov 28 at 15:13
Anson NG
41
asked Nov 28 at 15:13
Anson NG
41
asked Nov 28 at 15:13
Anson NG
41
41
closed as off-topic by Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 0:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 0:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Gibbs, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
1
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29
add a comment |
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
1
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
1
1
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
Consider $f(x) = g(x) =sqrt{frac{2}{pi}} sin(x^2)$. This is an example where $int_{-infty}^infty f(x); dx =int_{-infty}^infty g(x); dx = 1$, converging as an improper integral, but not absolutely. I claim that the improper integral $int_{-infty}^infty f(x) g(y-x); dx$ does not exist for almost every $y$.
Note that $$sin(x^2) sin((y-x)^2) = frac{cos(2xy - y^2) - cos(2 x^2 - 2 y x + y^2)}{2}$$
The improper integral $int_{-infty}^infty cos(2 x^2 - 2 y x + y^2); dx$ exists (and can be expressed using Fresnel integrals), but $int_{-infty}^infty cos(2xy - y^2); dx$ does not for almost every $y$, as the antiderivative $sin(2xy-y^2)/(2y)$ oscillates between $pm 1/(2y)$.
answered Nov 28 at 16:30
Robert Israel
317k23206457
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider $f(x) = g(x) =sqrt{frac{2}{pi}} sin(x^2)$. This is an example where $int_{-infty}^infty f(x); dx =int_{-infty}^infty g(x); dx = 1$, converging as an improper integral, but not absolutely. I claim that the improper integral $int_{-infty}^infty f(x) g(y-x); dx$ does not exist for almost every $y$.
Note that $$sin(x^2) sin((y-x)^2) = frac{cos(2xy - y^2) - cos(2 x^2 - 2 y x + y^2)}{2}$$
The improper integral $int_{-infty}^infty cos(2 x^2 - 2 y x + y^2); dx$ exists (and can be expressed using Fresnel integrals), but $int_{-infty}^infty cos(2xy - y^2); dx$ does not for almost every $y$, as the antiderivative $sin(2xy-y^2)/(2y)$ oscillates between $pm 1/(2y)$.
answered Nov 28 at 16:30
Robert Israel
317k23206457
add a comment |
Consider $f(x) = g(x) =sqrt{frac{2}{pi}} sin(x^2)$. This is an example where $int_{-infty}^infty f(x); dx =int_{-infty}^infty g(x); dx = 1$, converging as an improper integral, but not absolutely. I claim that the improper integral $int_{-infty}^infty f(x) g(y-x); dx$ does not exist for almost every $y$.
Note that $$sin(x^2) sin((y-x)^2) = frac{cos(2xy - y^2) - cos(2 x^2 - 2 y x + y^2)}{2}$$
The improper integral $int_{-infty}^infty cos(2 x^2 - 2 y x + y^2); dx$ exists (and can be expressed using Fresnel integrals), but $int_{-infty}^infty cos(2xy - y^2); dx$ does not for almost every $y$, as the antiderivative $sin(2xy-y^2)/(2y)$ oscillates between $pm 1/(2y)$.
answered Nov 28 at 16:30
Robert Israel
317k23206457
add a comment |
Consider $f(x) = g(x) =sqrt{frac{2}{pi}} sin(x^2)$. This is an example where $int_{-infty}^infty f(x); dx =int_{-infty}^infty g(x); dx = 1$, converging as an improper integral, but not absolutely. I claim that the improper integral $int_{-infty}^infty f(x) g(y-x); dx$ does not exist for almost every $y$.
Note that $$sin(x^2) sin((y-x)^2) = frac{cos(2xy - y^2) - cos(2 x^2 - 2 y x + y^2)}{2}$$
The improper integral $int_{-infty}^infty cos(2 x^2 - 2 y x + y^2); dx$ exists (and can be expressed using Fresnel integrals), but $int_{-infty}^infty cos(2xy - y^2); dx$ does not for almost every $y$, as the antiderivative $sin(2xy-y^2)/(2y)$ oscillates between $pm 1/(2y)$.
answered Nov 28 at 16:30
Robert Israel
317k23206457
Consider $f(x) = g(x) =sqrt{frac{2}{pi}} sin(x^2)$. This is an example where $int_{-infty}^infty f(x); dx =int_{-infty}^infty g(x); dx = 1$, converging as an improper integral, but not absolutely. I claim that the improper integral $int_{-infty}^infty f(x) g(y-x); dx$ does not exist for almost every $y$.
Note that $$sin(x^2) sin((y-x)^2) = frac{cos(2xy - y^2) - cos(2 x^2 - 2 y x + y^2)}{2}$$
The improper integral $int_{-infty}^infty cos(2 x^2 - 2 y x + y^2); dx$ exists (and can be expressed using Fresnel integrals), but $int_{-infty}^infty cos(2xy - y^2); dx$ does not for almost every $y$, as the antiderivative $sin(2xy-y^2)/(2y)$ oscillates between $pm 1/(2y)$.
answered Nov 28 at 16:30
Robert Israel
317k23206457
edited Nov 28 at 16:53
answered Nov 28 at 16:30
Robert Israel
317k23206457
answered Nov 28 at 16:30
Robert Israel
317k23206457
answered Nov 28 at 16:30
Robert Israel
317k23206457
317k23206457
add a comment |
add a comment |
Have you actually tried to solve it yourself?
– RamenChef
Nov 28 at 15:21
Try a change of variable $z=y-x$, and keep the other as $x$
– Andrei
Nov 28 at 15:23
1
Are you assuming the improper integrals converge absolutely? If not, you have to be careful, as switching the order of integration might not work.
– Robert Israel
Nov 28 at 15:55
You really need to explain the context! The answer is yes or no, depending on exactly what we're given about $f$ and $g$.
– David C. Ullrich
Nov 28 at 18:29