Prove that $tlongmapsto int_a^b f(x,t)dx$ continuous.

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Let $(t,x)longmapsto f(x,t)$ a continuous function on $[a,b]times mathbb R$. Prove that $$tlongmapsto int_a^b f(x,t)dx,$$
is continuous over $mathbb R$. Is this result stil true if we take $a=-infty $ and $b=+infty $ ?




integration continuity




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    Let $(t,x)longmapsto f(x,t)$ a continuous function on $[a,b]times mathbb R$. Prove that $$tlongmapsto int_a^b f(x,t)dx,$$
    is continuous over $mathbb R$. Is this result stil true if we take $a=-infty $ and $b=+infty $ ?




    integration continuity




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      Let $(t,x)longmapsto f(x,t)$ a continuous function on $[a,b]times mathbb R$. Prove that $$tlongmapsto int_a^b f(x,t)dx,$$
      is continuous over $mathbb R$. Is this result stil true if we take $a=-infty $ and $b=+infty $ ?




      integration continuity




      share|cite|improve this question











      Let $(t,x)longmapsto f(x,t)$ a continuous function on $[a,b]times mathbb R$. Prove that $$tlongmapsto int_a^b f(x,t)dx,$$
      is continuous over $mathbb R$. Is this result stil true if we take $a=-infty $ and $b=+infty $ ?





      integration continuity





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      share|cite|improve this question




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      asked Jul 16 at 13:02









      Peter

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          Let $|h|<1$ and fix $t$. Since $f$ is continuous on $[a,b]times [t-1,t+1]$, and thus uniformly continuous. Let $varepsilon>0$. There is $0<delta<1$ s.t. $$|f(x,t+h)-f(x,t)|<fracvarepsilonb-a,$$
          for all $|h|<1$ and all $xin [a,b]$. Therefore, if $|h|<delta$,
          $$left|int_a^bf(x,t+h)dx-int_a^b f(x,t)dxright|leq int_a^b |f(x,t+h)-f(x,t)|dx<fracvarepsilonb-a(b-a)=varepsilon. $$






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          • Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
            – Peter
            Jul 16 at 17:19











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          Let $|h|<1$ and fix $t$. Since $f$ is continuous on $[a,b]times [t-1,t+1]$, and thus uniformly continuous. Let $varepsilon>0$. There is $0<delta<1$ s.t. $$|f(x,t+h)-f(x,t)|<fracvarepsilonb-a,$$
          for all $|h|<1$ and all $xin [a,b]$. Therefore, if $|h|<delta$,
          $$left|int_a^bf(x,t+h)dx-int_a^b f(x,t)dxright|leq int_a^b |f(x,t+h)-f(x,t)|dx<fracvarepsilonb-a(b-a)=varepsilon. $$






          share|cite|improve this answer





















          • Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
            – Peter
            Jul 16 at 17:19















          up vote
          3
          down vote













          Let $|h|<1$ and fix $t$. Since $f$ is continuous on $[a,b]times [t-1,t+1]$, and thus uniformly continuous. Let $varepsilon>0$. There is $0<delta<1$ s.t. $$|f(x,t+h)-f(x,t)|<fracvarepsilonb-a,$$
          for all $|h|<1$ and all $xin [a,b]$. Therefore, if $|h|<delta$,
          $$left|int_a^bf(x,t+h)dx-int_a^b f(x,t)dxright|leq int_a^b |f(x,t+h)-f(x,t)|dx<fracvarepsilonb-a(b-a)=varepsilon. $$






          share|cite|improve this answer





















          • Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
            – Peter
            Jul 16 at 17:19













          up vote
          3
          down vote










          up vote
          3
          down vote









          Let $|h|<1$ and fix $t$. Since $f$ is continuous on $[a,b]times [t-1,t+1]$, and thus uniformly continuous. Let $varepsilon>0$. There is $0<delta<1$ s.t. $$|f(x,t+h)-f(x,t)|<fracvarepsilonb-a,$$
          for all $|h|<1$ and all $xin [a,b]$. Therefore, if $|h|<delta$,
          $$left|int_a^bf(x,t+h)dx-int_a^b f(x,t)dxright|leq int_a^b |f(x,t+h)-f(x,t)|dx<fracvarepsilonb-a(b-a)=varepsilon. $$






          share|cite|improve this answer













          Let $|h|<1$ and fix $t$. Since $f$ is continuous on $[a,b]times [t-1,t+1]$, and thus uniformly continuous. Let $varepsilon>0$. There is $0<delta<1$ s.t. $$|f(x,t+h)-f(x,t)|<fracvarepsilonb-a,$$
          for all $|h|<1$ and all $xin [a,b]$. Therefore, if $|h|<delta$,
          $$left|int_a^bf(x,t+h)dx-int_a^b f(x,t)dxright|leq int_a^b |f(x,t+h)-f(x,t)|dx<fracvarepsilonb-a(b-a)=varepsilon. $$







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          answered Jul 16 at 13:20









          Surb

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          36.3k84274











          • Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
            – Peter
            Jul 16 at 17:19

















          • Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
            – Peter
            Jul 16 at 17:19
















          Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
          – Peter
          Jul 16 at 17:19





          Thank you. And for $tlongmapsto int_-infty ^infty f(x,t)dx$ ?
          – Peter
          Jul 16 at 17:19













           

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