Find the value of the given limit












2















Find the value of this limit:



$$lim_{xto infty}dfrac{1}{pi}int _0^infty cosleft(dfrac{t^3}{3}+txright) dt$$.




Now as $xto infty implies dfrac{t^3}{3}+txto infty $ for any fixed $t$.



then $cosleft(dfrac{t^3}{3}+txright)$ is oscillating



Now I dont understand how should I compute the limit in this case?



Can someone please help.






calculus integration limits







share|cite|improve this question




















  • 1




    @KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
    – Brevan Ellefsen
    Nov 23 '18 at 8:17








  • 1




    @BrevanEllefsen Thank you very much.
    – Kavi Rama Murthy
    Nov 23 '18 at 8:23






  • 1




    Having a look at Airy function may be useful.
    – Kemono Chen
    Nov 23 '18 at 8:26






  • 1




    @KemonoChen;I checked it but it does not give a proof why the integral converges to 0
    – Join_PhD
    Nov 23 '18 at 8:30






  • 3




    Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
    – Fabian
    Nov 23 '18 at 9:26


















2















Find the value of this limit:



$$lim_{xto infty}dfrac{1}{pi}int _0^infty cosleft(dfrac{t^3}{3}+txright) dt$$.




Now as $xto infty implies dfrac{t^3}{3}+txto infty $ for any fixed $t$.



then $cosleft(dfrac{t^3}{3}+txright)$ is oscillating



Now I dont understand how should I compute the limit in this case?



Can someone please help.






calculus integration limits







share|cite|improve this question




















  • 1




    @KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
    – Brevan Ellefsen
    Nov 23 '18 at 8:17








  • 1




    @BrevanEllefsen Thank you very much.
    – Kavi Rama Murthy
    Nov 23 '18 at 8:23






  • 1




    Having a look at Airy function may be useful.
    – Kemono Chen
    Nov 23 '18 at 8:26






  • 1




    @KemonoChen;I checked it but it does not give a proof why the integral converges to 0
    – Join_PhD
    Nov 23 '18 at 8:30






  • 3




    Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
    – Fabian
    Nov 23 '18 at 9:26
















2












2








2


1






Find the value of this limit:



$$lim_{xto infty}dfrac{1}{pi}int _0^infty cosleft(dfrac{t^3}{3}+txright) dt$$.




Now as $xto infty implies dfrac{t^3}{3}+txto infty $ for any fixed $t$.



then $cosleft(dfrac{t^3}{3}+txright)$ is oscillating



Now I dont understand how should I compute the limit in this case?



Can someone please help.






calculus integration limits







share|cite|improve this question
















Find the value of this limit:



$$lim_{xto infty}dfrac{1}{pi}int _0^infty cosleft(dfrac{t^3}{3}+txright) dt$$.




Now as $xto infty implies dfrac{t^3}{3}+txto infty $ for any fixed $t$.



then $cosleft(dfrac{t^3}{3}+txright)$ is oscillating



Now I dont understand how should I compute the limit in this case?



Can someone please help.






calculus integration limits




calculus integration limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 8:06









Rebellos

14.5k31246




14.5k31246










asked Nov 23 '18 at 7:56









Join_PhDJoin_PhD

3088




3088








  • 1




    @KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
    – Brevan Ellefsen
    Nov 23 '18 at 8:17








  • 1




    @BrevanEllefsen Thank you very much.
    – Kavi Rama Murthy
    Nov 23 '18 at 8:23






  • 1




    Having a look at Airy function may be useful.
    – Kemono Chen
    Nov 23 '18 at 8:26






  • 1




    @KemonoChen;I checked it but it does not give a proof why the integral converges to 0
    – Join_PhD
    Nov 23 '18 at 8:30






  • 3




    Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
    – Fabian
    Nov 23 '18 at 9:26
















  • 1




    @KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
    – Brevan Ellefsen
    Nov 23 '18 at 8:17








  • 1




    @BrevanEllefsen Thank you very much.
    – Kavi Rama Murthy
    Nov 23 '18 at 8:23






  • 1




    Having a look at Airy function may be useful.
    – Kemono Chen
    Nov 23 '18 at 8:26






  • 1




    @KemonoChen;I checked it but it does not give a proof why the integral converges to 0
    – Join_PhD
    Nov 23 '18 at 8:30






  • 3




    Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
    – Fabian
    Nov 23 '18 at 9:26










1




1




@KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
– Brevan Ellefsen
Nov 23 '18 at 8:17






@KaviRamaMurthy I believe finiteness should follow from the following argument, (though I don't have time to check my work right now): $$begin{align} left|int_0^{infty}cosleft(frac{t^3}{3}+txright)dtright| &= left|int_0^{infty}cos(t^3/3)cos(tx)dt - int_0^{infty}sin(t^3/3)sin(tx)dtright| \& le int_0^{infty}left|cos(t^3/3)cos(tx)right|dt + int_0^{infty}left|sin(t^3/3)sin(tx)right|dt \&le int_0^{infty} |cos(t^3/3)|dt + int_0^{infty}|sin(t^3/3)|dt \&= frac{Gamma(1/3)}{2sqrt{3}} + frac{Gamma(1/3)}{6} end{align}$$
– Brevan Ellefsen
Nov 23 '18 at 8:17






1




1




@BrevanEllefsen Thank you very much.
– Kavi Rama Murthy
Nov 23 '18 at 8:23




@BrevanEllefsen Thank you very much.
– Kavi Rama Murthy
Nov 23 '18 at 8:23




1




1




Having a look at Airy function may be useful.
– Kemono Chen
Nov 23 '18 at 8:26




Having a look at Airy function may be useful.
– Kemono Chen
Nov 23 '18 at 8:26




1




1




@KemonoChen;I checked it but it does not give a proof why the integral converges to 0
– Join_PhD
Nov 23 '18 at 8:30




@KemonoChen;I checked it but it does not give a proof why the integral converges to 0
– Join_PhD
Nov 23 '18 at 8:30




3




3




Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
– Fabian
Nov 23 '18 at 9:26






Take a look at the Riemann–Lebesgue lemma. (hint: the limit is 0)
– Fabian
Nov 23 '18 at 9:26












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