Mixing
Does the improper integral $int_{R^2} e^{-xy}sin(x)$ converge?
$begingroup$
Does the improper integral $int_{R^2} e^{-xy}sin(x)$ converge?
I know the when the integral is only on $R^2 +$ then it converges to $frac{pi}{2}$.
However here the boundaries are different. Also it is not an positive function so I need to show it on the absolute value. It means I need to show that $$int_{R^2} e^{-xy}|sin(x)|$$ converges.
I'm pretty sure it doesn't converge, since I can't bound it from the right by a converge-bale function (I tried with $e^{-xy}$, also, Wolfram Alpha said so too).
However I can't think of a function that won't converge and will bound it from the left.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 9 at 0:19
Gabi GGabi G
39819
$endgroup$
add a comment |
$begingroup$
Does the improper integral $int_{R^2} e^{-xy}sin(x)$ converge?
I know the when the integral is only on $R^2 +$ then it converges to $frac{pi}{2}$.
However here the boundaries are different. Also it is not an positive function so I need to show it on the absolute value. It means I need to show that $$int_{R^2} e^{-xy}|sin(x)|$$ converges.
I'm pretty sure it doesn't converge, since I can't bound it from the right by a converge-bale function (I tried with $e^{-xy}$, also, Wolfram Alpha said so too).
However I can't think of a function that won't converge and will bound it from the left.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 9 at 0:19
Gabi GGabi G
39819
$endgroup$
$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43
add a comment |
$begingroup$
Does the improper integral $int_{R^2} e^{-xy}sin(x)$ converge?
I know the when the integral is only on $R^2 +$ then it converges to $frac{pi}{2}$.
However here the boundaries are different. Also it is not an positive function so I need to show it on the absolute value. It means I need to show that $$int_{R^2} e^{-xy}|sin(x)|$$ converges.
I'm pretty sure it doesn't converge, since I can't bound it from the right by a converge-bale function (I tried with $e^{-xy}$, also, Wolfram Alpha said so too).
However I can't think of a function that won't converge and will bound it from the left.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 9 at 0:19
Gabi GGabi G
39819
$endgroup$
Does the improper integral $int_{R^2} e^{-xy}sin(x)$ converge?
I know the when the integral is only on $R^2 +$ then it converges to $frac{pi}{2}$.
However here the boundaries are different. Also it is not an positive function so I need to show it on the absolute value. It means I need to show that $$int_{R^2} e^{-xy}|sin(x)|$$ converges.
I'm pretty sure it doesn't converge, since I can't bound it from the right by a converge-bale function (I tried with $e^{-xy}$, also, Wolfram Alpha said so too).
However I can't think of a function that won't converge and will bound it from the left.
Help would be appreciated.
integration multivariable-calculus improper-integrals
integration multivariable-calculus improper-integrals
asked Jan 9 at 0:19
Gabi GGabi G
39819
asked Jan 9 at 0:19
Gabi GGabi G
39819
asked Jan 9 at 0:19
Gabi GGabi G
39819
asked Jan 9 at 0:19
Gabi GGabi G
39819
asked Jan 9 at 0:19
Gabi GGabi G
39819
39819
$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43
add a comment |
$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43
$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43
add a comment |
1 Answer
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$begingroup$
To begin, the function $(x,y) mapsto e^{-xy} sin x$ is not absolutely integrable. It is enough to show this for the region $[0,infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$int_0^infty int_0^infty e^{-xy} | sin x| , dx , dy = int_0^infty int_0^infty e^{-xy} | sin x| , dy , dx $$
However,
$$int_0^infty e^{-xy} | sin x| , dy= frac{|sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,infty]$.
The question seems to be is the improper Riemann integral $int_{mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} sin x$.
For integrals over $mathbb{R}$ the definition of an improper integral as $lim_{a to -infty, b to +infty}int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$tag{*}int_{mathbb{R}^2}f = lim_{n to infty}int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n subset A_{n+1}$ for all $n$ and $cup_{n=1}^infty A_n = mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $int_{mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $int_{mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $int_{mathbb{R}^2} e^{-xy} |sin x|$ does not converge it follows that the improper integral $int_{mathbb{R}^2} e^{-xy} sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
answered Jan 9 at 5:08
RRLRRL
50.6k42573
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
To begin, the function $(x,y) mapsto e^{-xy} sin x$ is not absolutely integrable. It is enough to show this for the region $[0,infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$int_0^infty int_0^infty e^{-xy} | sin x| , dx , dy = int_0^infty int_0^infty e^{-xy} | sin x| , dy , dx $$
However,
$$int_0^infty e^{-xy} | sin x| , dy= frac{|sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,infty]$.
The question seems to be is the improper Riemann integral $int_{mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} sin x$.
For integrals over $mathbb{R}$ the definition of an improper integral as $lim_{a to -infty, b to +infty}int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$tag{*}int_{mathbb{R}^2}f = lim_{n to infty}int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n subset A_{n+1}$ for all $n$ and $cup_{n=1}^infty A_n = mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $int_{mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $int_{mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $int_{mathbb{R}^2} e^{-xy} |sin x|$ does not converge it follows that the improper integral $int_{mathbb{R}^2} e^{-xy} sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
answered Jan 9 at 5:08
RRLRRL
50.6k42573
$endgroup$
add a comment |
$begingroup$
To begin, the function $(x,y) mapsto e^{-xy} sin x$ is not absolutely integrable. It is enough to show this for the region $[0,infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$int_0^infty int_0^infty e^{-xy} | sin x| , dx , dy = int_0^infty int_0^infty e^{-xy} | sin x| , dy , dx $$
However,
$$int_0^infty e^{-xy} | sin x| , dy= frac{|sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,infty]$.
The question seems to be is the improper Riemann integral $int_{mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} sin x$.
For integrals over $mathbb{R}$ the definition of an improper integral as $lim_{a to -infty, b to +infty}int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$tag{*}int_{mathbb{R}^2}f = lim_{n to infty}int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n subset A_{n+1}$ for all $n$ and $cup_{n=1}^infty A_n = mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $int_{mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $int_{mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $int_{mathbb{R}^2} e^{-xy} |sin x|$ does not converge it follows that the improper integral $int_{mathbb{R}^2} e^{-xy} sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
answered Jan 9 at 5:08
RRLRRL
50.6k42573
$endgroup$
add a comment |
$begingroup$
To begin, the function $(x,y) mapsto e^{-xy} sin x$ is not absolutely integrable. It is enough to show this for the region $[0,infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$int_0^infty int_0^infty e^{-xy} | sin x| , dx , dy = int_0^infty int_0^infty e^{-xy} | sin x| , dy , dx $$
However,
$$int_0^infty e^{-xy} | sin x| , dy= frac{|sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,infty]$.
The question seems to be is the improper Riemann integral $int_{mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} sin x$.
For integrals over $mathbb{R}$ the definition of an improper integral as $lim_{a to -infty, b to +infty}int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$tag{*}int_{mathbb{R}^2}f = lim_{n to infty}int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n subset A_{n+1}$ for all $n$ and $cup_{n=1}^infty A_n = mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $int_{mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $int_{mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $int_{mathbb{R}^2} e^{-xy} |sin x|$ does not converge it follows that the improper integral $int_{mathbb{R}^2} e^{-xy} sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
answered Jan 9 at 5:08
RRLRRL
50.6k42573
$endgroup$
To begin, the function $(x,y) mapsto e^{-xy} sin x$ is not absolutely integrable. It is enough to show this for the region $[0,infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$int_0^infty int_0^infty e^{-xy} | sin x| , dx , dy = int_0^infty int_0^infty e^{-xy} | sin x| , dy , dx $$
However,
$$int_0^infty e^{-xy} | sin x| , dy= frac{|sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,infty]$.
The question seems to be is the improper Riemann integral $int_{mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} sin x$.
For integrals over $mathbb{R}$ the definition of an improper integral as $lim_{a to -infty, b to +infty}int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$tag{*}int_{mathbb{R}^2}f = lim_{n to infty}int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n subset A_{n+1}$ for all $n$ and $cup_{n=1}^infty A_n = mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $int_{mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $int_{mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $int_{mathbb{R}^2} e^{-xy} |sin x|$ does not converge it follows that the improper integral $int_{mathbb{R}^2} e^{-xy} sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
answered Jan 9 at 5:08
RRLRRL
50.6k42573
edited Jan 9 at 22:25
answered Jan 9 at 5:08
RRLRRL
50.6k42573
answered Jan 9 at 5:08
RRLRRL
50.6k42573
answered Jan 9 at 5:08
RRLRRL
50.6k42573
50.6k42573
add a comment |
add a comment |
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$begingroup$
No it does not. However, it does converge on ${mathbb{R}_+}^2times {mathbb{R}_-}^2$.
$endgroup$
– Zachary
Jan 9 at 1:34
$begingroup$
Can you think of a function to bound it so I can formally prove it?
$endgroup$
– Gabi G
Jan 9 at 2:05
$begingroup$
look at the part $e^{-xy}$ for $ytoinfty$
$endgroup$
– Henry Lee
Jan 9 at 11:43
$begingroup$
Are you asking if $int_{-infty}^{infty}int_{-infty}^{infty} e^{-xy} sin x , dx , dy$ converges as iterated improper integrals or if $int_{mathbb{R}^2} e^{-xy} sin x$ converges as an improper integral in the sense of a limit of Riemann integrals over compact rectifiable sets $A_n$ where $A_n subset A_{n+1}$ and $cup_n A_N = mathbb{R}^2$? You have to be careful about how an improper multiple integral is defined.
$endgroup$
– RRL
Jan 9 at 21:40
$begingroup$
As far as iterated integrals are concerned -- how could $int_0^infty e^{-xy} sin x , dx$ possibly converge if $y < 0$?
$endgroup$
– RRL
Jan 9 at 21:43