General formula for $int_0^{pi/2} tan^{alpha}(x) dx$?












3












$begingroup$


There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?



More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$






calculus integration improper-integrals gamma-function







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  • $begingroup$
    Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
    $endgroup$
    – robjohn♦
    Feb 1 at 4:03
















3












$begingroup$


There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?



More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$






calculus integration improper-integrals gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
    $endgroup$
    – robjohn♦
    Feb 1 at 4:03














3












3








3


1



$begingroup$


There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?



More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$






calculus integration improper-integrals gamma-function







share|cite|improve this question









$endgroup$




There are already questions about how to find $int tan^{1/2}(x) dx$. But how to derive a general formula for $int_0^{pi/2} tan^{alpha}(x) dx$ (which converges if $|alpha|<1$) ?



More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=tan^2(x)$ in $int_0^infty frac{v^beta}{1+v} dv$






calculus integration improper-integrals gamma-function




calculus integration improper-integrals gamma-function






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asked Feb 1 at 3:39









Holding ArthurHolding Arthur

1,555417




1,555417












  • $begingroup$
    Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
    $endgroup$
    – robjohn♦
    Feb 1 at 4:03


















  • $begingroup$
    Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
    $endgroup$
    – robjohn♦
    Feb 1 at 4:03
















$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03




$begingroup$
Euler's Reflection Formula is proven in this answer, which is referenced in my answer below.
$endgroup$
– robjohn♦
Feb 1 at 4:03










3 Answers
3






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4












$begingroup$

With $u=tan x$, we have
$$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
Now, according to this question:
$$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
So, plugging in $beta=frac pi 2$ yields
$$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Consider the integral
    $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
    Setting $t=sin(x)^2$:
    $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
    Then recall the definition of the beta function
    $$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
    We use this to see that
    $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
    Hence
    $$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
    Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
    It can be easily shown that
    $$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
    So
    $$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
    Which you know works for $|a|<1$.




    This can be used to show that
    $$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
    Or simply
    $$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
    To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      $$
      begin{align}
      int_0^{pi/2}tan^alpha(x),mathrm{d}x
      &=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
      &=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
      &=fracpi2secleft(frac{pialpha}2right)tag3
      end{align}
      $$
      Explanation:
      $(1)$: substitute $xmapstoarctan(x)$
      $(2)$: use this answer
      $(3)$: trigonometric identity






      share|cite|improve this answer











      $endgroup$














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        3 Answers
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        3 Answers
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        4












        $begingroup$

        With $u=tan x$, we have
        $$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
        Now, according to this question:
        $$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
        So, plugging in $beta=frac pi 2$ yields
        $$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          With $u=tan x$, we have
          $$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
          Now, according to this question:
          $$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
          So, plugging in $beta=frac pi 2$ yields
          $$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            With $u=tan x$, we have
            $$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
            Now, according to this question:
            $$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
            So, plugging in $beta=frac pi 2$ yields
            $$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$






            share|cite|improve this answer









            $endgroup$



            With $u=tan x$, we have
            $$int_0^{frac pi 2} tan^alpha (x)dx=int_0^{+infty}frac{u^alpha}{1+u^2}du$$
            Now, according to this question:
            $$int_0^infty frac{x^{alpha}dx}{1+2xcosbeta +x^{2}}=frac{pisin (alphabeta)}{sin (alphapi)sin beta }$$
            So, plugging in $beta=frac pi 2$ yields
            $$int_0^{frac pi 2} tan^alpha (x)dx=frac{pisin(frac {alpha pi}2)}{sin(alpha pi)}=frac{pi}{2cos(frac{alphapi}{2})}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 1 at 3:56









            Stefan LafonStefan Lafon

            3,005212




            3,005212























                2












                $begingroup$

                Consider the integral
                $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
                Setting $t=sin(x)^2$:
                $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                Then recall the definition of the beta function
                $$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
                We use this to see that
                $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                Hence
                $$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
                Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
                It can be easily shown that
                $$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
                So
                $$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
                Which you know works for $|a|<1$.




                This can be used to show that
                $$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
                Or simply
                $$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
                To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  Consider the integral
                  $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
                  Setting $t=sin(x)^2$:
                  $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                  Then recall the definition of the beta function
                  $$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
                  We use this to see that
                  $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                  Hence
                  $$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
                  Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
                  It can be easily shown that
                  $$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
                  So
                  $$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
                  Which you know works for $|a|<1$.




                  This can be used to show that
                  $$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
                  Or simply
                  $$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
                  To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Consider the integral
                    $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
                    Setting $t=sin(x)^2$:
                    $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                    Then recall the definition of the beta function
                    $$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
                    We use this to see that
                    $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                    Hence
                    $$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
                    Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
                    It can be easily shown that
                    $$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
                    So
                    $$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
                    Which you know works for $|a|<1$.




                    This can be used to show that
                    $$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
                    Or simply
                    $$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
                    To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.






                    share|cite|improve this answer











                    $endgroup$



                    Consider the integral
                    $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx,qquad a,b>-1$$
                    Setting $t=sin(x)^2$:
                    $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                    Then recall the definition of the beta function
                    $$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
                    We use this to see that
                    $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                    Hence
                    $$int_0^{pi/2}tan(x)^amathrm dx=frac12Gammaleft(frac{1+a}2right)Gammaleft(frac{1-a}2right)$$
                    Then using $$Gamma(1-s)Gamma(s)=frac{pi}{sinpi s}$$
                    It can be easily shown that
                    $$Gammaleft(frac{1+s}2right)Gammaleft(frac{1-s}2right)=pisecfrac{pi s}2$$
                    So
                    $$int_0^{pi/2}tan(x)^amathrm dx=fracpi2secfrac{pi a}2$$
                    Which you know works for $|a|<1$.




                    This can be used to show that
                    $$int_0^{pi/2} log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2,bigg|_{a=0}$$
                    Or simply
                    $$int_0^{pi/2} tan(x)^{a}log^{n}[tan x]mathrm dx=fracpi2left(frac{d}{da}right)^nsecfrac{pi a}2$$
                    To show this just take $left(frac{d}{da}right)^n$ on both sides for integer $n$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Feb 1 at 4:32

























                    answered Feb 1 at 4:25









                    clathratusclathratus

                    5,1141439




                    5,1141439























                        2












                        $begingroup$

                        $$
                        begin{align}
                        int_0^{pi/2}tan^alpha(x),mathrm{d}x
                        &=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
                        &=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
                        &=fracpi2secleft(frac{pialpha}2right)tag3
                        end{align}
                        $$
                        Explanation:
                        $(1)$: substitute $xmapstoarctan(x)$
                        $(2)$: use this answer
                        $(3)$: trigonometric identity






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          $$
                          begin{align}
                          int_0^{pi/2}tan^alpha(x),mathrm{d}x
                          &=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
                          &=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
                          &=fracpi2secleft(frac{pialpha}2right)tag3
                          end{align}
                          $$
                          Explanation:
                          $(1)$: substitute $xmapstoarctan(x)$
                          $(2)$: use this answer
                          $(3)$: trigonometric identity






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            $$
                            begin{align}
                            int_0^{pi/2}tan^alpha(x),mathrm{d}x
                            &=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
                            &=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
                            &=fracpi2secleft(frac{pialpha}2right)tag3
                            end{align}
                            $$
                            Explanation:
                            $(1)$: substitute $xmapstoarctan(x)$
                            $(2)$: use this answer
                            $(3)$: trigonometric identity






                            share|cite|improve this answer











                            $endgroup$



                            $$
                            begin{align}
                            int_0^{pi/2}tan^alpha(x),mathrm{d}x
                            &=int_0^inftyfrac{x^alpha}{1+x^2},mathrm{d}xtag1\
                            &=fracpi2cscleft(pifrac{alpha+1}2right)tag2\[3pt]
                            &=fracpi2secleft(frac{pialpha}2right)tag3
                            end{align}
                            $$
                            Explanation:
                            $(1)$: substitute $xmapstoarctan(x)$
                            $(2)$: use this answer
                            $(3)$: trigonometric identity







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Feb 1 at 4:56

























                            answered Feb 1 at 3:57









                            robjohn♦robjohn

                            270k27313642




                            270k27313642






























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