Prove that $lim_{n rightarrow infty} int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dx = 0$.












-2












$begingroup$


Prove that $lim_{n rightarrow infty} int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dx=0.$



I think I will use Riemann integrability, but how I do not know, could anyone help me in this?






real-analysis calculus integration riemann-integration







share|cite|improve this question











$endgroup$

















    -2












    $begingroup$


    Prove that $lim_{n rightarrow infty} int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dx=0.$



    I think I will use Riemann integrability, but how I do not know, could anyone help me in this?






    real-analysis calculus integration riemann-integration







    share|cite|improve this question











    $endgroup$















      -2












      -2








      -2





      $begingroup$


      Prove that $lim_{n rightarrow infty} int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dx=0.$



      I think I will use Riemann integrability, but how I do not know, could anyone help me in this?






      real-analysis calculus integration riemann-integration







      share|cite|improve this question











      $endgroup$




      Prove that $lim_{n rightarrow infty} int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dx=0.$



      I think I will use Riemann integrability, but how I do not know, could anyone help me in this?






      real-analysis calculus integration riemann-integration




      real-analysis calculus integration riemann-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 8:11









      gimusi

      92.9k94494




      92.9k94494










      asked Dec 4 '18 at 7:45









      hopefullyhopefully

      309113




      309113






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $|frac {sin, x} {x^{2}+n^{2}}| leq frac 1 {n^{2}}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I will not use the property that the function is Riemann integrable?
            $endgroup$
            – hopefully
            Dec 4 '18 at 7:54






          • 1




            $begingroup$
            @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 7:57












          • $begingroup$
            Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
            $endgroup$
            – hopefully
            Dec 4 '18 at 8:13






          • 1




            $begingroup$
            @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:21








          • 1




            $begingroup$
            Any partition is fine, but you can choose the uniform partition for definiteness.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:27



















          1












          $begingroup$

          HINT



          Recall that for $f$ continuous on $(a,b)$



          $$left|{int_a^b f left({t}right) mathrm dt}right| le int_a^b left|{f left({t}right)}right| mathrm dt$$



          therefore we have that



          $$0le left|int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dxright|le frac1{n^2}int_{0}^{2 pi} frac {1}{(x/n)^2 + 1} dx =frac{1}{n}arctanleft(frac{2pi}nright)$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $|frac {sin, x} {x^{2}+n^{2}}| leq frac 1 {n^{2}}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I will not use the property that the function is Riemann integrable?
              $endgroup$
              – hopefully
              Dec 4 '18 at 7:54






            • 1




              $begingroup$
              @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 7:57












            • $begingroup$
              Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
              $endgroup$
              – hopefully
              Dec 4 '18 at 8:13






            • 1




              $begingroup$
              @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:21








            • 1




              $begingroup$
              Any partition is fine, but you can choose the uniform partition for definiteness.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:27
















            1












            $begingroup$

            $|frac {sin, x} {x^{2}+n^{2}}| leq frac 1 {n^{2}}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I will not use the property that the function is Riemann integrable?
              $endgroup$
              – hopefully
              Dec 4 '18 at 7:54






            • 1




              $begingroup$
              @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 7:57












            • $begingroup$
              Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
              $endgroup$
              – hopefully
              Dec 4 '18 at 8:13






            • 1




              $begingroup$
              @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:21








            • 1




              $begingroup$
              Any partition is fine, but you can choose the uniform partition for definiteness.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:27














            1












            1








            1





            $begingroup$

            $|frac {sin, x} {x^{2}+n^{2}}| leq frac 1 {n^{2}}$.






            share|cite|improve this answer









            $endgroup$



            $|frac {sin, x} {x^{2}+n^{2}}| leq frac 1 {n^{2}}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 7:49









            Kavi Rama MurthyKavi Rama Murthy

            54.8k32055




            54.8k32055












            • $begingroup$
              I will not use the property that the function is Riemann integrable?
              $endgroup$
              – hopefully
              Dec 4 '18 at 7:54






            • 1




              $begingroup$
              @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 7:57












            • $begingroup$
              Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
              $endgroup$
              – hopefully
              Dec 4 '18 at 8:13






            • 1




              $begingroup$
              @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:21








            • 1




              $begingroup$
              Any partition is fine, but you can choose the uniform partition for definiteness.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:27


















            • $begingroup$
              I will not use the property that the function is Riemann integrable?
              $endgroup$
              – hopefully
              Dec 4 '18 at 7:54






            • 1




              $begingroup$
              @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 7:57












            • $begingroup$
              Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
              $endgroup$
              – hopefully
              Dec 4 '18 at 8:13






            • 1




              $begingroup$
              @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:21








            • 1




              $begingroup$
              Any partition is fine, but you can choose the uniform partition for definiteness.
              $endgroup$
              – Kavi Rama Murthy
              Dec 4 '18 at 8:27
















            $begingroup$
            I will not use the property that the function is Riemann integrable?
            $endgroup$
            – hopefully
            Dec 4 '18 at 7:54




            $begingroup$
            I will not use the property that the function is Riemann integrable?
            $endgroup$
            – hopefully
            Dec 4 '18 at 7:54




            1




            1




            $begingroup$
            @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 7:57






            $begingroup$
            @hopefully You have to use an elementary property of Riemann integrals which says that if $|f(x)| leq C$ then $|int_a^{b} f(x), dx| leq C(b-a)$. You can in fact prove this easily from definition of Riemann integral since all Riemann sums have the bound $C(b-a)$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 7:57














            $begingroup$
            Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
            $endgroup$
            – hopefully
            Dec 4 '18 at 8:13




            $begingroup$
            Could you provide a bit more details about the sequence of steps in proving so .... what should exactly I do?
            $endgroup$
            – hopefully
            Dec 4 '18 at 8:13




            1




            1




            $begingroup$
            @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:21






            $begingroup$
            @hopefully Yes. the integral is bounded by $frac {2pi} {n^{2}}$ so it tends to $0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:21






            1




            1




            $begingroup$
            Any partition is fine, but you can choose the uniform partition for definiteness.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:27




            $begingroup$
            Any partition is fine, but you can choose the uniform partition for definiteness.
            $endgroup$
            – Kavi Rama Murthy
            Dec 4 '18 at 8:27











            1












            $begingroup$

            HINT



            Recall that for $f$ continuous on $(a,b)$



            $$left|{int_a^b f left({t}right) mathrm dt}right| le int_a^b left|{f left({t}right)}right| mathrm dt$$



            therefore we have that



            $$0le left|int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dxright|le frac1{n^2}int_{0}^{2 pi} frac {1}{(x/n)^2 + 1} dx =frac{1}{n}arctanleft(frac{2pi}nright)$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              HINT



              Recall that for $f$ continuous on $(a,b)$



              $$left|{int_a^b f left({t}right) mathrm dt}right| le int_a^b left|{f left({t}right)}right| mathrm dt$$



              therefore we have that



              $$0le left|int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dxright|le frac1{n^2}int_{0}^{2 pi} frac {1}{(x/n)^2 + 1} dx =frac{1}{n}arctanleft(frac{2pi}nright)$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                HINT



                Recall that for $f$ continuous on $(a,b)$



                $$left|{int_a^b f left({t}right) mathrm dt}right| le int_a^b left|{f left({t}right)}right| mathrm dt$$



                therefore we have that



                $$0le left|int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dxright|le frac1{n^2}int_{0}^{2 pi} frac {1}{(x/n)^2 + 1} dx =frac{1}{n}arctanleft(frac{2pi}nright)$$






                share|cite|improve this answer











                $endgroup$



                HINT



                Recall that for $f$ continuous on $(a,b)$



                $$left|{int_a^b f left({t}right) mathrm dt}right| le int_a^b left|{f left({t}right)}right| mathrm dt$$



                therefore we have that



                $$0le left|int_{0}^{2 pi} frac {sin (nx)}{x^2 + n^2} dxright|le frac1{n^2}int_{0}^{2 pi} frac {1}{(x/n)^2 + 1} dx =frac{1}{n}arctanleft(frac{2pi}nright)$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 8:03

























                answered Dec 4 '18 at 7:47









                gimusigimusi

                92.9k94494




                92.9k94494






























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