Evaluate $int_{0}^{2pi} lfloorcot^{-1}xrfloor,dx$ from $0$ to $2 pi$.












0












$begingroup$


Evaluate $int_{0}^{2pi} lfloorcot^{-1}xrfloor,dx$ from $0$ to $2 pi$. Where $lfloor{}cdot{}rfloor$ denotes the greatest integer function.



Here $x$ varies from $0$ to $2 pi$, so



$cot^{-1}0 > cot^{-1}x > cot^{-1}2 pi$



$infty > cot^{-1}x > cot^{-1}2 pi$



Now
$int_{cot1}^{ 2pi} lfloorcot^{-1}xrfloor,dx + int_{0}^{cot1}lfloorcot^{-1}xrfloor,dx$



The first part of integral is clearly zero, how to evaluate second part?






calculus







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  • $begingroup$
    @egreg you edited the question correctly.
    $endgroup$
    – Mathsaddict
    Jan 17 at 14:19










  • $begingroup$
    @Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
    $endgroup$
    – egreg
    Jan 17 at 14:21










  • $begingroup$
    math.stackexchange.com/questions/3076606/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 14:26
















0












$begingroup$


Evaluate $int_{0}^{2pi} lfloorcot^{-1}xrfloor,dx$ from $0$ to $2 pi$. Where $lfloor{}cdot{}rfloor$ denotes the greatest integer function.



Here $x$ varies from $0$ to $2 pi$, so



$cot^{-1}0 > cot^{-1}x > cot^{-1}2 pi$



$infty > cot^{-1}x > cot^{-1}2 pi$



Now
$int_{cot1}^{ 2pi} lfloorcot^{-1}xrfloor,dx + int_{0}^{cot1}lfloorcot^{-1}xrfloor,dx$



The first part of integral is clearly zero, how to evaluate second part?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    @egreg you edited the question correctly.
    $endgroup$
    – Mathsaddict
    Jan 17 at 14:19










  • $begingroup$
    @Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
    $endgroup$
    – egreg
    Jan 17 at 14:21










  • $begingroup$
    math.stackexchange.com/questions/3076606/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 14:26














0












0








0





$begingroup$


Evaluate $int_{0}^{2pi} lfloorcot^{-1}xrfloor,dx$ from $0$ to $2 pi$. Where $lfloor{}cdot{}rfloor$ denotes the greatest integer function.



Here $x$ varies from $0$ to $2 pi$, so



$cot^{-1}0 > cot^{-1}x > cot^{-1}2 pi$



$infty > cot^{-1}x > cot^{-1}2 pi$



Now
$int_{cot1}^{ 2pi} lfloorcot^{-1}xrfloor,dx + int_{0}^{cot1}lfloorcot^{-1}xrfloor,dx$



The first part of integral is clearly zero, how to evaluate second part?






calculus







share|cite|improve this question











$endgroup$




Evaluate $int_{0}^{2pi} lfloorcot^{-1}xrfloor,dx$ from $0$ to $2 pi$. Where $lfloor{}cdot{}rfloor$ denotes the greatest integer function.



Here $x$ varies from $0$ to $2 pi$, so



$cot^{-1}0 > cot^{-1}x > cot^{-1}2 pi$



$infty > cot^{-1}x > cot^{-1}2 pi$



Now
$int_{cot1}^{ 2pi} lfloorcot^{-1}xrfloor,dx + int_{0}^{cot1}lfloorcot^{-1}xrfloor,dx$



The first part of integral is clearly zero, how to evaluate second part?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 14:20









egreg

183k1486204




183k1486204










asked Jan 17 at 14:11









MathsaddictMathsaddict

3619




3619












  • $begingroup$
    @egreg you edited the question correctly.
    $endgroup$
    – Mathsaddict
    Jan 17 at 14:19










  • $begingroup$
    @Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
    $endgroup$
    – egreg
    Jan 17 at 14:21










  • $begingroup$
    math.stackexchange.com/questions/3076606/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 14:26


















  • $begingroup$
    @egreg you edited the question correctly.
    $endgroup$
    – Mathsaddict
    Jan 17 at 14:19










  • $begingroup$
    @Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
    $endgroup$
    – egreg
    Jan 17 at 14:21










  • $begingroup$
    math.stackexchange.com/questions/3076606/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 14:26
















$begingroup$
@egreg you edited the question correctly.
$endgroup$
– Mathsaddict
Jan 17 at 14:19




$begingroup$
@egreg you edited the question correctly.
$endgroup$
– Mathsaddict
Jan 17 at 14:19












$begingroup$
@Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
$endgroup$
– egreg
Jan 17 at 14:21




$begingroup$
@Mathsaddict The symbol $lfloor{}cdot{}rfloor$ is more commonly used than $[{}cdot{}]$.
$endgroup$
– egreg
Jan 17 at 14:21












$begingroup$
math.stackexchange.com/questions/3076606/…
$endgroup$
– lab bhattacharjee
Jan 17 at 14:26




$begingroup$
math.stackexchange.com/questions/3076606/…
$endgroup$
– lab bhattacharjee
Jan 17 at 14:26










1 Answer
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$begingroup$

$$0le xlecot1$$



$$dfracpi2gecot^{-1}xge1impliesleftlfloorcot^{-1}xrightrfloor=1$$






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    $begingroup$

    $$0le xlecot1$$



    $$dfracpi2gecot^{-1}xge1impliesleftlfloorcot^{-1}xrightrfloor=1$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$0le xlecot1$$



      $$dfracpi2gecot^{-1}xge1impliesleftlfloorcot^{-1}xrightrfloor=1$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$0le xlecot1$$



        $$dfracpi2gecot^{-1}xge1impliesleftlfloorcot^{-1}xrightrfloor=1$$






        share|cite|improve this answer









        $endgroup$



        $$0le xlecot1$$



        $$dfracpi2gecot^{-1}xge1impliesleftlfloorcot^{-1}xrightrfloor=1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 14:18









        lab bhattacharjeelab bhattacharjee

        226k15157275




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