Solving Riemann-Stieltjes integral:$int_{- pi/4}^{pi/4} f(x)dg(x)$












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I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







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  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14
















8












$begingroup$


I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14














8












8








8


4



$begingroup$


I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral







share|cite|improve this question











$endgroup$




I'm having trouble solving this Riemann-Stieltjes integral:



$$int_{- pi/4}^{pi/4} f(x)dg(x),$$ where $$f(x):= begin{cases} frac{sin^4x}{cos^2x}{} &text{if }xge0, \{}\ frac1{cos^3x} &text{if }x<0,end{cases}$$



and $$g(x)=begin{cases} phantom{-} 1+sin(x) &text{if }-pi/4 <x<pi/4, \ -1 &text{otherwise}.end{cases}$$



I believe the only jump discontinuities are at $-pi/4$ and $pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!






real-analysis calculus integration trigonometry stieltjes-integral




real-analysis calculus integration trigonometry stieltjes-integral






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edited Dec 14 '18 at 22:25









Batominovski

33.1k33293




33.1k33293










asked May 3 '13 at 0:53









DrakeDrake

4681515




4681515












  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14


















  • $begingroup$
    Check this theorem.
    $endgroup$
    – Mhenni Benghorbal
    May 3 '13 at 1:08










  • $begingroup$
    Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
    $endgroup$
    – Drake
    May 6 '13 at 13:51










  • $begingroup$
    CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
    $endgroup$
    – Mariusz Iwaniuk
    Oct 1 '18 at 22:14
















$begingroup$
Check this theorem.
$endgroup$
– Mhenni Benghorbal
May 3 '13 at 1:08




$begingroup$
Check this theorem.
$endgroup$
– Mhenni Benghorbal
May 3 '13 at 1:08












$begingroup$
Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
$endgroup$
– Drake
May 6 '13 at 13:51




$begingroup$
Can you elaborate how to solve the last integral in that theorem (related to my problem). I can see why I would use it since both f and g have no common discontinuities.
$endgroup$
– Drake
May 6 '13 at 13:51












$begingroup$
CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
$endgroup$
– Mariusz Iwaniuk
Oct 1 '18 at 22:14




$begingroup$
CAS says:$1-{frac {7,sqrt {2}}{12}}+ln left( 1+sqrt {2} right) $
$endgroup$
– Mariusz Iwaniuk
Oct 1 '18 at 22:14










3 Answers
3






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0












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Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






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  • $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50












  • $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55



















0












$begingroup$


Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
Therefore,
$$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



and
$$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
$$
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15










  • $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14










  • $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42










  • $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52










  • $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45



















0












$begingroup$

Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
$$I = int_{-T}^T f(x) dg(x) = \
lim_{epsilon downarrow 0} {
f(-T) (g(-T + epsilon)- g(-T)) + \
f(T) (g(T)- g(T - epsilon)) } + \
int_{-T}^T f(x) cos x ,dx = \
2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
It can be verified that
$$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
f(T) g(T) - f(-T) g(-T) - \
lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55
















    0












    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55














    0












    0








    0





    $begingroup$

    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$






    share|cite|improve this answer









    $endgroup$



    Since $g$ is differentiable on $-pi/4 <x<pi/4$ your integral changes to Riemann integral simply by the following theorem: $$int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx$$ so you will have $int_{- pi/4}^{pi/4} f(x)dg(x)= int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}d(1+sin x)=int_{- pi/4}^{pi/4}frac{sin^4x}{cos^2x}cos x dx$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 4 '15 at 10:01









    AliAli

    1,9812520




    1,9812520












    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55


















    • $begingroup$
      The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
      $endgroup$
      – New day rising
      Jan 7 '17 at 3:50












    • $begingroup$
      Psst... both of you need to take a closer look to what happens at the endpoints.
      $endgroup$
      – user361424
      Jun 28 '17 at 0:55
















    $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50






    $begingroup$
    The integral should really be $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)$$ or $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} cos x dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} cos x dx$$
    $endgroup$
    – New day rising
    Jan 7 '17 at 3:50














    $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55




    $begingroup$
    Psst... both of you need to take a closer look to what happens at the endpoints.
    $endgroup$
    – user361424
    Jun 28 '17 at 0:55











    0












    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45
















    0












    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45














    0












    0








    0





    $begingroup$


    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$






    share|cite|improve this answer











    $endgroup$




    Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'cdot f$




    Namely from here we have $$color{blue}{int_a^bf(x)dg(x)=int_a^bf(x)g'(x)dx}$$
    Therefore,
    $$int_{-pi/4}^{pi/4} f(x) dg(x) = int_{-pi/4}^0 frac 1{cos^3 x} d(1+sin x) + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(1+sin x)\=int_{-pi/4}^0 frac 1{cos^2 x} dx + int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x)$$



    But $$int_{-pi/4}^0 frac 1{cos^2 x} dx =int^{pi/4}_0 (tan x)' dx =1$$



    and
    $$int_0^{pi/4} frac{sin^4 x}{cos^2 x} d(sin x) = int_0^{pi/4} frac{sin^4 x}{1-sin^2 x} d(sin x) =int_0^{sqrt{2}/2} frac{t^4 }{1-t^2 } dt\=int_0^{sqrt{2}/2} -t^2+frac{t^2 }{1-t^2 } dt$$
    $$
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 2 '18 at 9:02

























    answered Dec 15 '17 at 14:27









    Guy FsoneGuy Fsone

    17.2k43074




    17.2k43074












    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45


















    • $begingroup$
      The first formula isn't applicable when $g$ has jump discontinuities.
      $endgroup$
      – Maxim
      May 28 '18 at 21:15










    • $begingroup$
      @Maxim It is possible the discontinuities are of first order.
      $endgroup$
      – Guy Fsone
      May 30 '18 at 7:14










    • $begingroup$
      There are zero order discontinuities at the endpoints. See my answer.
      $endgroup$
      – Maxim
      May 30 '18 at 9:42










    • $begingroup$
      @Maxim what do you mean by zero order discontnuity?
      $endgroup$
      – Guy Fsone
      May 31 '18 at 1:52










    • $begingroup$
      I mean that the function itself (the derivative of zero order) is discontinuous.
      $endgroup$
      – Maxim
      May 31 '18 at 9:45
















    $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15




    $begingroup$
    The first formula isn't applicable when $g$ has jump discontinuities.
    $endgroup$
    – Maxim
    May 28 '18 at 21:15












    $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14




    $begingroup$
    @Maxim It is possible the discontinuities are of first order.
    $endgroup$
    – Guy Fsone
    May 30 '18 at 7:14












    $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42




    $begingroup$
    There are zero order discontinuities at the endpoints. See my answer.
    $endgroup$
    – Maxim
    May 30 '18 at 9:42












    $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52




    $begingroup$
    @Maxim what do you mean by zero order discontnuity?
    $endgroup$
    – Guy Fsone
    May 31 '18 at 1:52












    $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45




    $begingroup$
    I mean that the function itself (the derivative of zero order) is discontinuous.
    $endgroup$
    – Maxim
    May 31 '18 at 9:45











    0












    $begingroup$

    Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
    $$I = int_{-T}^T f(x) dg(x) = \
    lim_{epsilon downarrow 0} {
    f(-T) (g(-T + epsilon)- g(-T)) + \
    f(T) (g(T)- g(T - epsilon)) } + \
    int_{-T}^T f(x) cos x ,dx = \
    2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
    It can be verified that
    $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
    f(T) g(T) - f(-T) g(-T) - \
    lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
    int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
    int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
      $$I = int_{-T}^T f(x) dg(x) = \
      lim_{epsilon downarrow 0} {
      f(-T) (g(-T + epsilon)- g(-T)) + \
      f(T) (g(T)- g(T - epsilon)) } + \
      int_{-T}^T f(x) cos x ,dx = \
      2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
      It can be verified that
      $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
      f(T) g(T) - f(-T) g(-T) - \
      lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
      int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
      int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
        $$I = int_{-T}^T f(x) dg(x) = \
        lim_{epsilon downarrow 0} {
        f(-T) (g(-T + epsilon)- g(-T)) + \
        f(T) (g(T)- g(T - epsilon)) } + \
        int_{-T}^T f(x) cos x ,dx = \
        2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
        It can be verified that
        $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
        f(T) g(T) - f(-T) g(-T) - \
        lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
        int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
        int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$






        share|cite|improve this answer











        $endgroup$



        Let $T = pi/4$. The first term in the Riemann-Stieltjes sum, $f(xi_1)(g(x_1) - g(-T))$, does not tend to zero when $x_1 downarrow -T$ and $xi_1 downarrow -T$, and similarly for the last term. The integral is
        $$I = int_{-T}^T f(x) dg(x) = \
        lim_{epsilon downarrow 0} {
        f(-T) (g(-T + epsilon)- g(-T)) + \
        f(T) (g(T)- g(T - epsilon)) } + \
        int_{-T}^T f(x) cos x ,dx = \
        2 operatorname{arctanh}(sqrt 2 - 1) + frac {19 sqrt 2} 6 - 2.$$
        It can be verified that
        $$I = f(T) g(T) - f(-T) g(-T) - int_{-T}^T g(x) df(x) = \
        f(T) g(T) - f(-T) g(-T) - \
        lim_{epsilon downarrow 0} g(0) (f(epsilon) - f(-epsilon)) - \
        int_{-T}^0 g(x) left( frac 1 {cos^3 x} right)' dx -
        int_0^T g(x) left( frac {sin^4 x} {cos^2 x} right)' dx.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 28 '18 at 21:35

























        answered May 28 '18 at 20:47









        MaximMaxim

        6,1681221




        6,1681221






























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