Uniform convergence of iterated improper integrals on $(0,infty)$
$begingroup$
I'm trying to get a better understanding of when it is permissible to swtich conditionally convergent improper integrals (when Fubini inapplicable) and I looked at a case where it works:
$$int_0^infty int_0^infty e^{-xy} sin x , dx, dy = int_0^infty int_0^infty e^{-xy} sin x , dy , dx $$
I know the inner iterated integrals are uniformly convergent by the Weierstrass test for $x , y in [c, infty)$ where $c > 0$. Since $|e^{-xy} sin x | leqslant e^{-cy}$ for $c leq x < infty$ , then $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $c leq x < infty$. Since $|e^{-xy} sin x | leqslant e^{-cx}$ for $c leq y < infty$ , then $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $c leq y < infty$.
The Weierstrass test is not helpful to consider uniform convergence on $(0,infty)$.
My question is how to determine if $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $0 < x < infty$ and $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $0 < y < infty$ and either prove it or disprove it.
real-analysis calculus improper-integrals uniform-convergence
edited Dec 13 '18 at 19:02
RRL
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
$endgroup$
add a comment |
$begingroup$
I'm trying to get a better understanding of when it is permissible to swtich conditionally convergent improper integrals (when Fubini inapplicable) and I looked at a case where it works:
$$int_0^infty int_0^infty e^{-xy} sin x , dx, dy = int_0^infty int_0^infty e^{-xy} sin x , dy , dx $$
I know the inner iterated integrals are uniformly convergent by the Weierstrass test for $x , y in [c, infty)$ where $c > 0$. Since $|e^{-xy} sin x | leqslant e^{-cy}$ for $c leq x < infty$ , then $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $c leq x < infty$. Since $|e^{-xy} sin x | leqslant e^{-cx}$ for $c leq y < infty$ , then $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $c leq y < infty$.
The Weierstrass test is not helpful to consider uniform convergence on $(0,infty)$.
My question is how to determine if $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $0 < x < infty$ and $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $0 < y < infty$ and either prove it or disprove it.
real-analysis calculus improper-integrals uniform-convergence
edited Dec 13 '18 at 19:02
RRL
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
$endgroup$
add a comment |
$begingroup$
I'm trying to get a better understanding of when it is permissible to swtich conditionally convergent improper integrals (when Fubini inapplicable) and I looked at a case where it works:
$$int_0^infty int_0^infty e^{-xy} sin x , dx, dy = int_0^infty int_0^infty e^{-xy} sin x , dy , dx $$
I know the inner iterated integrals are uniformly convergent by the Weierstrass test for $x , y in [c, infty)$ where $c > 0$. Since $|e^{-xy} sin x | leqslant e^{-cy}$ for $c leq x < infty$ , then $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $c leq x < infty$. Since $|e^{-xy} sin x | leqslant e^{-cx}$ for $c leq y < infty$ , then $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $c leq y < infty$.
The Weierstrass test is not helpful to consider uniform convergence on $(0,infty)$.
My question is how to determine if $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $0 < x < infty$ and $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $0 < y < infty$ and either prove it or disprove it.
real-analysis calculus improper-integrals uniform-convergence
edited Dec 13 '18 at 19:02
RRL
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
$endgroup$
I'm trying to get a better understanding of when it is permissible to swtich conditionally convergent improper integrals (when Fubini inapplicable) and I looked at a case where it works:
$$int_0^infty int_0^infty e^{-xy} sin x , dx, dy = int_0^infty int_0^infty e^{-xy} sin x , dy , dx $$
I know the inner iterated integrals are uniformly convergent by the Weierstrass test for $x , y in [c, infty)$ where $c > 0$. Since $|e^{-xy} sin x | leqslant e^{-cy}$ for $c leq x < infty$ , then $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $c leq x < infty$. Since $|e^{-xy} sin x | leqslant e^{-cx}$ for $c leq y < infty$ , then $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $c leq y < infty$.
The Weierstrass test is not helpful to consider uniform convergence on $(0,infty)$.
My question is how to determine if $int_0^infty e^{-xy} sin x , dy$ converges uniformly for $0 < x < infty$ and $int_0^infty e^{-xy} sin x , dx$ converges uniformly for $0 < y < infty$ and either prove it or disprove it.
real-analysis calculus improper-integrals uniform-convergence
real-analysis calculus improper-integrals uniform-convergence
edited Dec 13 '18 at 19:02
RRL
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
edited Dec 13 '18 at 19:02
RRL
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
edited Dec 13 '18 at 19:02
RRL
53.4k52574
edited Dec 13 '18 at 19:02
RRL
53.4k52574
edited Dec 13 '18 at 19:02
RRL
53.4k52574
53.4k52574
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
asked Dec 12 '18 at 22:44
WoodWorkerWoodWorker
470314
470314
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,infty)$.
For the first integral, with $y_n = (2npi + pi)^{-1} in (0,infty)$ we have
$$left|int_{2npi}^{2npi+pi} e^{-xy_n} sin x , dxright|geqslant e^{-(2npi+pi) y_n}int_{2npi}^{2npi+pi} sin x , dx = 2 e^{-(2npi+pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n to infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n in (0,infty)$ we have
$$left|int_n^infty e^{-x_ny} sin x_n , dyright| = left|frac{sin x_n}{x_n} right|e^{-nx_n} = frac{sin frac{1}{n}}{frac{1}{n}}e^{-1} ,,, xrightarrow[n to infty]{} ,,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
$endgroup$
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
add a comment |
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$begingroup$
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,infty)$.
For the first integral, with $y_n = (2npi + pi)^{-1} in (0,infty)$ we have
$$left|int_{2npi}^{2npi+pi} e^{-xy_n} sin x , dxright|geqslant e^{-(2npi+pi) y_n}int_{2npi}^{2npi+pi} sin x , dx = 2 e^{-(2npi+pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n to infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n in (0,infty)$ we have
$$left|int_n^infty e^{-x_ny} sin x_n , dyright| = left|frac{sin x_n}{x_n} right|e^{-nx_n} = frac{sin frac{1}{n}}{frac{1}{n}}e^{-1} ,,, xrightarrow[n to infty]{} ,,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
$endgroup$
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
add a comment |
$begingroup$
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,infty)$.
For the first integral, with $y_n = (2npi + pi)^{-1} in (0,infty)$ we have
$$left|int_{2npi}^{2npi+pi} e^{-xy_n} sin x , dxright|geqslant e^{-(2npi+pi) y_n}int_{2npi}^{2npi+pi} sin x , dx = 2 e^{-(2npi+pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n to infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n in (0,infty)$ we have
$$left|int_n^infty e^{-x_ny} sin x_n , dyright| = left|frac{sin x_n}{x_n} right|e^{-nx_n} = frac{sin frac{1}{n}}{frac{1}{n}}e^{-1} ,,, xrightarrow[n to infty]{} ,,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
$endgroup$
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
add a comment |
$begingroup$
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,infty)$.
For the first integral, with $y_n = (2npi + pi)^{-1} in (0,infty)$ we have
$$left|int_{2npi}^{2npi+pi} e^{-xy_n} sin x , dxright|geqslant e^{-(2npi+pi) y_n}int_{2npi}^{2npi+pi} sin x , dx = 2 e^{-(2npi+pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n to infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n in (0,infty)$ we have
$$left|int_n^infty e^{-x_ny} sin x_n , dyright| = left|frac{sin x_n}{x_n} right|e^{-nx_n} = frac{sin frac{1}{n}}{frac{1}{n}}e^{-1} ,,, xrightarrow[n to infty]{} ,,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
$endgroup$
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,infty)$.
For the first integral, with $y_n = (2npi + pi)^{-1} in (0,infty)$ we have
$$left|int_{2npi}^{2npi+pi} e^{-xy_n} sin x , dxright|geqslant e^{-(2npi+pi) y_n}int_{2npi}^{2npi+pi} sin x , dx = 2 e^{-(2npi+pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n to infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n in (0,infty)$ we have
$$left|int_n^infty e^{-x_ny} sin x_n , dyright| = left|frac{sin x_n}{x_n} right|e^{-nx_n} = frac{sin frac{1}{n}}{frac{1}{n}}e^{-1} ,,, xrightarrow[n to infty]{} ,,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
answered Dec 12 '18 at 23:01
RRLRRL
53.4k52574
53.4k52574
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
add a comment |
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals.
$endgroup$
– WoodWorker
Dec 13 '18 at 0:07
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
$begingroup$
Also can you take a look at this: math.stackexchange.com/q/3020423/318852
$endgroup$
– WoodWorker
Dec 13 '18 at 0:10
1
1
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
$begingroup$
@WoodWorker: Proving non-uniform convergence basically means we find $epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $left|int_{c_1}^{c_2}f(x,y_C) , dx right| > epsilon_0$ -- which follows if as I showed there are sequences $beta_n > alpha_n$ that diverge to $+infty$ and $y_n$ where $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
$endgroup$
– RRL
Dec 13 '18 at 5:51
1
1
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
$begingroup$
Uniform convergence of $int_0^infty f(x,y) , dx$ and $int_0^infty f(x,y) , dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = int_0^infty f(x,y) , dy$ is such that $int_0^infty F(x) , dx$ is uniformly convergent.
$endgroup$
– RRL
Dec 13 '18 at 5:54
add a comment |
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