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












1












$begingroup$


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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 19:02









      RRL

      53.4k52574




      53.4k52574










      asked Dec 12 '18 at 22:44









      WoodWorkerWoodWorker

      470314




      470314






















          1 Answer
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          active

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          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














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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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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




















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          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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