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Trying to evaluate $int_0^pi/2x^scot xarctan(cot x) dx$
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$$G(s)= int_0^pi/2x^scot xarctan(cot x),mathrm dx$$
I am interested in evaluating $G(s)$ but I am unable to even make a start!
Any help? Thank you
I have try a substitution of $u=cot x$ but not right. I have try by parts but it is to complicate.
calculus integration definite-integrals trigonometric-integrals
edited Aug 6 at 10:24
Chappers
55k74191
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up vote
3
down vote
favorite
$$G(s)= int_0^pi/2x^scot xarctan(cot x),mathrm dx$$
I am interested in evaluating $G(s)$ but I am unable to even make a start!
Any help? Thank you
I have try a substitution of $u=cot x$ but not right. I have try by parts but it is to complicate.
calculus integration definite-integrals trigonometric-integrals
edited Aug 6 at 10:24
Chappers
55k74191
$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
1
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
1
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
6
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$G(s)= int_0^pi/2x^scot xarctan(cot x),mathrm dx$$
I am interested in evaluating $G(s)$ but I am unable to even make a start!
Any help? Thank you
I have try a substitution of $u=cot x$ but not right. I have try by parts but it is to complicate.
calculus integration definite-integrals trigonometric-integrals
edited Aug 6 at 10:24
Chappers
55k74191
$$G(s)= int_0^pi/2x^scot xarctan(cot x),mathrm dx$$
I am interested in evaluating $G(s)$ but I am unable to even make a start!
Any help? Thank you
I have try a substitution of $u=cot x$ but not right. I have try by parts but it is to complicate.
calculus integration definite-integrals trigonometric-integrals
edited Aug 6 at 10:24
Chappers
55k74191
edited Aug 6 at 10:24
Chappers
55k74191
edited Aug 6 at 10:24
Chappers
55k74191
edited Aug 6 at 10:24
Chappers
55k74191
55k74191
asked Aug 6 at 10:04
user565198
$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
1
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
1
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
6
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51
add a comment |Â
$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
1
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
1
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
6
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51
$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
1
1
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
1
1
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
6
6
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51
add a comment |Â
2 Answers
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Thanks to Milan Ivanov in the comments,
$$int x^scot xarctan(cot x),mathrm dx = fracpi2int x^scot x,mathrm dx - int x^s+1cot x,mathrm dx$$
Let's do rough calculation on that common form:
$$int x^ncot x,mathrm dx \= x^nsmallint cot x,mathrm dx - intleft(nx^n-1cdot smallint cot x,mathrm dxright)mathrm dx \= x^ncdotln|sin x,| - nint x^n-1cdot ln|sin x,|,mathrm dx$$
Mhh.. now, this $int x^n-1cdot ln|sin x,|,mathrm dx$ is something you should try yourself.
answered Aug 6 at 16:01
Nick
3,62163261
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The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series
$$-ln(sin(x)) = ln(2) + sum limits_k=1^infty fraccos(2kx)k , , x in (0,pi) , , tag1$$
and the Dirichlet eta function
$$ eta (s) = sum limits_k=1^infty frac(-1)^k-1k^s = left(1-2^1-sright) zeta(s) , , , operatornameRe(s) > 0 , , tag2$$
with the limit $eta(1) = ln(2)$ .
We can integrate by parts and use $(1)$ to obtain
beginalign
K_n &equiv int limits_0^pi/2 t^n cot(t) , mathrmd t = - n int limits_0^pi/2 t^n-1 ln(sin(t)) , mathrmd t\
&= left(fracpi2right)^n ln(2) + sum limits_k=1^infty frac1k n int limits_0^pi/2 t^n-1 cos(2kt), mathrmd t , .
endalign
It is now useful to distinguish between even and odd $n$ . For $nu in mathbbN$ we integrate by parts repeatedly to get
beginalign
K_2nu &= left(fracpi2right)^2 nu ln(2) + sum limits_k=1^infty frac1k frac(2nu)(2 nu -1)4 k^2left[ (-1)^k left( fracpi2right)^2nu-2 - (2 nu-2) int limits_0^pi/2 t^2 nu-3 cos(2kt) , mathrmd t right] \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) right] - frac(2 nu)!2^2 (2 nu -3)! sum limits_k=1^infty frac1k^3 int limits_0^pi/2 t^2nu-3 cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) + fracpi^2(nu-2)(2(nu-2))! eta(5) right] \
&phantom=+ frac(2 nu)!2^4 (2 nu -5)! sum limits_k=1^infty frac1k^5 int limits_0^pi/2 t^2nu-5 cos(2 k t) , mathrmd t \
&= dots \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu-1 frac(2 nu)!2^2(nu-1) sum limits_k=1^infty frac1k^2nu-1 int limits_0^pi/2 t cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu frac(2 nu)!2^2nu sum limits_k=1^infty frac1 + (-1)^k-1k^2nu+1 \
&= frac(2nu)!2^2 nu left[ sum limits_l=0^nu (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu zeta(2nu+1)right] , .
endalign
Similarly,
beginalign
K_2nu-1 &= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) + (-1)^nu-1 frac(2 nu-1)!2^2nu-3 sum limits_k=1^infty frac1k^2nu-2 int limits_0^pi/2 cos(2 k t) , mathrmd t \
&= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) , ,
endalign
but here the final integral vanishes and there is no extra term.
Milen Ivanov's hint from the comments now yields
$$G(n) = fracpi2 K_n - K_n+1$$
for $n in mathbbN$ . Note that the term containing $eta(1) = ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $pi$ times zeta values at the odd integers between $3$ and $2leftlfloorfracn+12rightrfloor+1$ with rational coefficients.
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Thanks to Milan Ivanov in the comments,
$$int x^scot xarctan(cot x),mathrm dx = fracpi2int x^scot x,mathrm dx - int x^s+1cot x,mathrm dx$$
Let's do rough calculation on that common form:
$$int x^ncot x,mathrm dx \= x^nsmallint cot x,mathrm dx - intleft(nx^n-1cdot smallint cot x,mathrm dxright)mathrm dx \= x^ncdotln|sin x,| - nint x^n-1cdot ln|sin x,|,mathrm dx$$
Mhh.. now, this $int x^n-1cdot ln|sin x,|,mathrm dx$ is something you should try yourself.
answered Aug 6 at 16:01
Nick
3,62163261
add a comment |Â
up vote
0
down vote
Thanks to Milan Ivanov in the comments,
$$int x^scot xarctan(cot x),mathrm dx = fracpi2int x^scot x,mathrm dx - int x^s+1cot x,mathrm dx$$
Let's do rough calculation on that common form:
$$int x^ncot x,mathrm dx \= x^nsmallint cot x,mathrm dx - intleft(nx^n-1cdot smallint cot x,mathrm dxright)mathrm dx \= x^ncdotln|sin x,| - nint x^n-1cdot ln|sin x,|,mathrm dx$$
Mhh.. now, this $int x^n-1cdot ln|sin x,|,mathrm dx$ is something you should try yourself.
answered Aug 6 at 16:01
Nick
3,62163261
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Thanks to Milan Ivanov in the comments,
$$int x^scot xarctan(cot x),mathrm dx = fracpi2int x^scot x,mathrm dx - int x^s+1cot x,mathrm dx$$
Let's do rough calculation on that common form:
$$int x^ncot x,mathrm dx \= x^nsmallint cot x,mathrm dx - intleft(nx^n-1cdot smallint cot x,mathrm dxright)mathrm dx \= x^ncdotln|sin x,| - nint x^n-1cdot ln|sin x,|,mathrm dx$$
Mhh.. now, this $int x^n-1cdot ln|sin x,|,mathrm dx$ is something you should try yourself.
answered Aug 6 at 16:01
Nick
3,62163261
Thanks to Milan Ivanov in the comments,
$$int x^scot xarctan(cot x),mathrm dx = fracpi2int x^scot x,mathrm dx - int x^s+1cot x,mathrm dx$$
Let's do rough calculation on that common form:
$$int x^ncot x,mathrm dx \= x^nsmallint cot x,mathrm dx - intleft(nx^n-1cdot smallint cot x,mathrm dxright)mathrm dx \= x^ncdotln|sin x,| - nint x^n-1cdot ln|sin x,|,mathrm dx$$
Mhh.. now, this $int x^n-1cdot ln|sin x,|,mathrm dx$ is something you should try yourself.
answered Aug 6 at 16:01
Nick
3,62163261
answered Aug 6 at 16:01
Nick
3,62163261
answered Aug 6 at 16:01
Nick
3,62163261
answered Aug 6 at 16:01
Nick
3,62163261
3,62163261
add a comment |Â
add a comment |Â
up vote
0
down vote
The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series
$$-ln(sin(x)) = ln(2) + sum limits_k=1^infty fraccos(2kx)k , , x in (0,pi) , , tag1$$
and the Dirichlet eta function
$$ eta (s) = sum limits_k=1^infty frac(-1)^k-1k^s = left(1-2^1-sright) zeta(s) , , , operatornameRe(s) > 0 , , tag2$$
with the limit $eta(1) = ln(2)$ .
We can integrate by parts and use $(1)$ to obtain
beginalign
K_n &equiv int limits_0^pi/2 t^n cot(t) , mathrmd t = - n int limits_0^pi/2 t^n-1 ln(sin(t)) , mathrmd t\
&= left(fracpi2right)^n ln(2) + sum limits_k=1^infty frac1k n int limits_0^pi/2 t^n-1 cos(2kt), mathrmd t , .
endalign
It is now useful to distinguish between even and odd $n$ . For $nu in mathbbN$ we integrate by parts repeatedly to get
beginalign
K_2nu &= left(fracpi2right)^2 nu ln(2) + sum limits_k=1^infty frac1k frac(2nu)(2 nu -1)4 k^2left[ (-1)^k left( fracpi2right)^2nu-2 - (2 nu-2) int limits_0^pi/2 t^2 nu-3 cos(2kt) , mathrmd t right] \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) right] - frac(2 nu)!2^2 (2 nu -3)! sum limits_k=1^infty frac1k^3 int limits_0^pi/2 t^2nu-3 cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) + fracpi^2(nu-2)(2(nu-2))! eta(5) right] \
&phantom=+ frac(2 nu)!2^4 (2 nu -5)! sum limits_k=1^infty frac1k^5 int limits_0^pi/2 t^2nu-5 cos(2 k t) , mathrmd t \
&= dots \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu-1 frac(2 nu)!2^2(nu-1) sum limits_k=1^infty frac1k^2nu-1 int limits_0^pi/2 t cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu frac(2 nu)!2^2nu sum limits_k=1^infty frac1 + (-1)^k-1k^2nu+1 \
&= frac(2nu)!2^2 nu left[ sum limits_l=0^nu (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu zeta(2nu+1)right] , .
endalign
Similarly,
beginalign
K_2nu-1 &= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) + (-1)^nu-1 frac(2 nu-1)!2^2nu-3 sum limits_k=1^infty frac1k^2nu-2 int limits_0^pi/2 cos(2 k t) , mathrmd t \
&= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) , ,
endalign
but here the final integral vanishes and there is no extra term.
Milen Ivanov's hint from the comments now yields
$$G(n) = fracpi2 K_n - K_n+1$$
for $n in mathbbN$ . Note that the term containing $eta(1) = ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $pi$ times zeta values at the odd integers between $3$ and $2leftlfloorfracn+12rightrfloor+1$ with rational coefficients.
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
add a comment |Â
up vote
0
down vote
The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series
$$-ln(sin(x)) = ln(2) + sum limits_k=1^infty fraccos(2kx)k , , x in (0,pi) , , tag1$$
and the Dirichlet eta function
$$ eta (s) = sum limits_k=1^infty frac(-1)^k-1k^s = left(1-2^1-sright) zeta(s) , , , operatornameRe(s) > 0 , , tag2$$
with the limit $eta(1) = ln(2)$ .
We can integrate by parts and use $(1)$ to obtain
beginalign
K_n &equiv int limits_0^pi/2 t^n cot(t) , mathrmd t = - n int limits_0^pi/2 t^n-1 ln(sin(t)) , mathrmd t\
&= left(fracpi2right)^n ln(2) + sum limits_k=1^infty frac1k n int limits_0^pi/2 t^n-1 cos(2kt), mathrmd t , .
endalign
It is now useful to distinguish between even and odd $n$ . For $nu in mathbbN$ we integrate by parts repeatedly to get
beginalign
K_2nu &= left(fracpi2right)^2 nu ln(2) + sum limits_k=1^infty frac1k frac(2nu)(2 nu -1)4 k^2left[ (-1)^k left( fracpi2right)^2nu-2 - (2 nu-2) int limits_0^pi/2 t^2 nu-3 cos(2kt) , mathrmd t right] \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) right] - frac(2 nu)!2^2 (2 nu -3)! sum limits_k=1^infty frac1k^3 int limits_0^pi/2 t^2nu-3 cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) + fracpi^2(nu-2)(2(nu-2))! eta(5) right] \
&phantom=+ frac(2 nu)!2^4 (2 nu -5)! sum limits_k=1^infty frac1k^5 int limits_0^pi/2 t^2nu-5 cos(2 k t) , mathrmd t \
&= dots \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu-1 frac(2 nu)!2^2(nu-1) sum limits_k=1^infty frac1k^2nu-1 int limits_0^pi/2 t cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu frac(2 nu)!2^2nu sum limits_k=1^infty frac1 + (-1)^k-1k^2nu+1 \
&= frac(2nu)!2^2 nu left[ sum limits_l=0^nu (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu zeta(2nu+1)right] , .
endalign
Similarly,
beginalign
K_2nu-1 &= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) + (-1)^nu-1 frac(2 nu-1)!2^2nu-3 sum limits_k=1^infty frac1k^2nu-2 int limits_0^pi/2 cos(2 k t) , mathrmd t \
&= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) , ,
endalign
but here the final integral vanishes and there is no extra term.
Milen Ivanov's hint from the comments now yields
$$G(n) = fracpi2 K_n - K_n+1$$
for $n in mathbbN$ . Note that the term containing $eta(1) = ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $pi$ times zeta values at the odd integers between $3$ and $2leftlfloorfracn+12rightrfloor+1$ with rational coefficients.
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series
$$-ln(sin(x)) = ln(2) + sum limits_k=1^infty fraccos(2kx)k , , x in (0,pi) , , tag1$$
and the Dirichlet eta function
$$ eta (s) = sum limits_k=1^infty frac(-1)^k-1k^s = left(1-2^1-sright) zeta(s) , , , operatornameRe(s) > 0 , , tag2$$
with the limit $eta(1) = ln(2)$ .
We can integrate by parts and use $(1)$ to obtain
beginalign
K_n &equiv int limits_0^pi/2 t^n cot(t) , mathrmd t = - n int limits_0^pi/2 t^n-1 ln(sin(t)) , mathrmd t\
&= left(fracpi2right)^n ln(2) + sum limits_k=1^infty frac1k n int limits_0^pi/2 t^n-1 cos(2kt), mathrmd t , .
endalign
It is now useful to distinguish between even and odd $n$ . For $nu in mathbbN$ we integrate by parts repeatedly to get
beginalign
K_2nu &= left(fracpi2right)^2 nu ln(2) + sum limits_k=1^infty frac1k frac(2nu)(2 nu -1)4 k^2left[ (-1)^k left( fracpi2right)^2nu-2 - (2 nu-2) int limits_0^pi/2 t^2 nu-3 cos(2kt) , mathrmd t right] \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) right] - frac(2 nu)!2^2 (2 nu -3)! sum limits_k=1^infty frac1k^3 int limits_0^pi/2 t^2nu-3 cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) + fracpi^2(nu-2)(2(nu-2))! eta(5) right] \
&phantom=+ frac(2 nu)!2^4 (2 nu -5)! sum limits_k=1^infty frac1k^5 int limits_0^pi/2 t^2nu-5 cos(2 k t) , mathrmd t \
&= dots \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu-1 frac(2 nu)!2^2(nu-1) sum limits_k=1^infty frac1k^2nu-1 int limits_0^pi/2 t cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu frac(2 nu)!2^2nu sum limits_k=1^infty frac1 + (-1)^k-1k^2nu+1 \
&= frac(2nu)!2^2 nu left[ sum limits_l=0^nu (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu zeta(2nu+1)right] , .
endalign
Similarly,
beginalign
K_2nu-1 &= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) + (-1)^nu-1 frac(2 nu-1)!2^2nu-3 sum limits_k=1^infty frac1k^2nu-2 int limits_0^pi/2 cos(2 k t) , mathrmd t \
&= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) , ,
endalign
but here the final integral vanishes and there is no extra term.
Milen Ivanov's hint from the comments now yields
$$G(n) = fracpi2 K_n - K_n+1$$
for $n in mathbbN$ . Note that the term containing $eta(1) = ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $pi$ times zeta values at the odd integers between $3$ and $2leftlfloorfracn+12rightrfloor+1$ with rational coefficients.
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series
$$-ln(sin(x)) = ln(2) + sum limits_k=1^infty fraccos(2kx)k , , x in (0,pi) , , tag1$$
and the Dirichlet eta function
$$ eta (s) = sum limits_k=1^infty frac(-1)^k-1k^s = left(1-2^1-sright) zeta(s) , , , operatornameRe(s) > 0 , , tag2$$
with the limit $eta(1) = ln(2)$ .
We can integrate by parts and use $(1)$ to obtain
beginalign
K_n &equiv int limits_0^pi/2 t^n cot(t) , mathrmd t = - n int limits_0^pi/2 t^n-1 ln(sin(t)) , mathrmd t\
&= left(fracpi2right)^n ln(2) + sum limits_k=1^infty frac1k n int limits_0^pi/2 t^n-1 cos(2kt), mathrmd t , .
endalign
It is now useful to distinguish between even and odd $n$ . For $nu in mathbbN$ we integrate by parts repeatedly to get
beginalign
K_2nu &= left(fracpi2right)^2 nu ln(2) + sum limits_k=1^infty frac1k frac(2nu)(2 nu -1)4 k^2left[ (-1)^k left( fracpi2right)^2nu-2 - (2 nu-2) int limits_0^pi/2 t^2 nu-3 cos(2kt) , mathrmd t right] \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) right] - frac(2 nu)!2^2 (2 nu -3)! sum limits_k=1^infty frac1k^3 int limits_0^pi/2 t^2nu-3 cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu left[fracpi^2nu(2nu)! eta(1) - fracpi^2(nu-1)(2(nu-1))! eta(3) + fracpi^2(nu-2)(2(nu-2))! eta(5) right] \
&phantom=+ frac(2 nu)!2^4 (2 nu -5)! sum limits_k=1^infty frac1k^5 int limits_0^pi/2 t^2nu-5 cos(2 k t) , mathrmd t \
&= dots \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu-1 frac(2 nu)!2^2(nu-1) sum limits_k=1^infty frac1k^2nu-1 int limits_0^pi/2 t cos(2 k t) , mathrmd t \
&= frac(2nu)!2^2 nu sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu frac(2 nu)!2^2nu sum limits_k=1^infty frac1 + (-1)^k-1k^2nu+1 \
&= frac(2nu)!2^2 nu left[ sum limits_l=0^nu (-1)^l fracpi^2(nu-l)(2(nu-l))! eta(2l+1) + (-1)^nu zeta(2nu+1)right] , .
endalign
Similarly,
beginalign
K_2nu-1 &= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) + (-1)^nu-1 frac(2 nu-1)!2^2nu-3 sum limits_k=1^infty frac1k^2nu-2 int limits_0^pi/2 cos(2 k t) , mathrmd t \
&= frac(2nu-1)!2^2 nu-1 sum limits_l=0^nu-1 (-1)^l fracpi^2(nu-l)-1(2(nu-l)-1)! eta(2l+1) , ,
endalign
but here the final integral vanishes and there is no extra term.
Milen Ivanov's hint from the comments now yields
$$G(n) = fracpi2 K_n - K_n+1$$
for $n in mathbbN$ . Note that the term containing $eta(1) = ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $pi$ times zeta values at the odd integers between $3$ and $2leftlfloorfracn+12rightrfloor+1$ with rational coefficients.
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
edited Aug 6 at 17:53
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
answered Aug 6 at 17:34
ComplexYetTrivial
2,817624
2,817624
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$s$ may be any real number, or whole numbers only?
– denklo
Aug 6 at 10:11
1
Try using $arctan(cot(x)) = pi/2 - arccot(cot(x)) = pi/2 - x$. This will reduce the problem to calculating $int_0^pi/2x^scot x dx$ for two different values of $s$.
– Milen Ivanov
Aug 6 at 10:22
let say it is whole numbers
– user565198
Aug 6 at 10:22
1
For $s$ a positive integer, Maple evaluates these to linear combinations of $zeta(3), zeta(5),...$ So perhaps a Taylor series is the way to go?
– B. Goddard
Aug 6 at 10:43
6
The integrals $int_0^pi/2 x^s cot(x) , mathrmd x$ can be done using the Fourier series of $log(sin)$ for $s in mathbbN$. I have given a general formula in this question.
– ComplexYetTrivial
Aug 6 at 10:51