Uniform convergence of iterated improper integrals on $(0,infty)$












1














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







share|cite|improve this question





























    1














    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







    share|cite|improve this question



























      1












      1








      1







      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







      share|cite|improve this question















      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






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      share|cite|improve this question













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      edited Dec 13 '18 at 19:02









      RRL

      49.4k42573




      49.4k42573










      asked Dec 12 '18 at 22:44









      WoodWorkerWoodWorker

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          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.






          share|cite|improve this answer





















          • 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.
            – WoodWorker
            Dec 13 '18 at 0:07












          • Also can you take a look at this: math.stackexchange.com/q/3020423/318852
            – WoodWorker
            Dec 13 '18 at 0:10






          • 1




            @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
            – RRL
            Dec 13 '18 at 5:51






          • 1




            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.
            – RRL
            Dec 13 '18 at 5:54













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          1














          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.






          share|cite|improve this answer





















          • 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.
            – WoodWorker
            Dec 13 '18 at 0:07












          • Also can you take a look at this: math.stackexchange.com/q/3020423/318852
            – WoodWorker
            Dec 13 '18 at 0:10






          • 1




            @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
            – RRL
            Dec 13 '18 at 5:51






          • 1




            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.
            – RRL
            Dec 13 '18 at 5:54


















          1














          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.






          share|cite|improve this answer





















          • 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.
            – WoodWorker
            Dec 13 '18 at 0:07












          • Also can you take a look at this: math.stackexchange.com/q/3020423/318852
            – WoodWorker
            Dec 13 '18 at 0:10






          • 1




            @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
            – RRL
            Dec 13 '18 at 5:51






          • 1




            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.
            – RRL
            Dec 13 '18 at 5:54
















          1












          1








          1






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 23:01









          RRLRRL

          49.4k42573




          49.4k42573












          • 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.
            – WoodWorker
            Dec 13 '18 at 0:07












          • Also can you take a look at this: math.stackexchange.com/q/3020423/318852
            – WoodWorker
            Dec 13 '18 at 0:10






          • 1




            @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
            – RRL
            Dec 13 '18 at 5:51






          • 1




            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.
            – RRL
            Dec 13 '18 at 5:54




















          • 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.
            – WoodWorker
            Dec 13 '18 at 0:07












          • Also can you take a look at this: math.stackexchange.com/q/3020423/318852
            – WoodWorker
            Dec 13 '18 at 0:10






          • 1




            @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
            – RRL
            Dec 13 '18 at 5:51






          • 1




            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.
            – RRL
            Dec 13 '18 at 5:54


















          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.
          – WoodWorker
          Dec 13 '18 at 0:07






          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.
          – WoodWorker
          Dec 13 '18 at 0:07














          Also can you take a look at this: math.stackexchange.com/q/3020423/318852
          – WoodWorker
          Dec 13 '18 at 0:10




          Also can you take a look at this: math.stackexchange.com/q/3020423/318852
          – WoodWorker
          Dec 13 '18 at 0:10




          1




          1




          @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
          – RRL
          Dec 13 '18 at 5:51




          @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 $left|int_{alpha_n}^{beta_n}f(x,y_n) , dx right| notto 0$.
          – RRL
          Dec 13 '18 at 5:51




          1




          1




          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.
          – RRL
          Dec 13 '18 at 5:54






          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.
          – RRL
          Dec 13 '18 at 5:54




















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          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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