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A two-variable recurrence relation for $int_0^pi/2 t^n cot^m (t) , mathrmd t$
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I have found a general expression for the integrals $I_n$ discussed in this question in terms of $K_l^(l) , , , 0 leq l leq n , , $ where
$$ K_n^(m) equiv int limits_0^pi/2 t^n cot^m (t) , mathrmd t , , , n in mathbbN_0 , , , 0 leq m leq n , .$$
I would now like to obtain an explicit formula for these integrals in order to complete the solution of the original problem.
The simplest case is trivial:
$$ K_n^(0) = int limits_0^pi/2 t^n , mathrmd t = frac1n+1 left(fracpi2right)^n+1 , , , n in mathbbN_0 , . tag1 $$
With some more effort it is possible to compute
beginalign
K_n^(1) &= 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 \
&= fracn!2^n left[sum limits_l=0^left lfloor fracn-12 right rfloor (-1)^l fracpi^n-2l(n-2l)! eta (2l+1) + operatorname1_2 mathbbN (n) (-1)^left lfloor fracn+12 right rfloor [zeta (n+1) + eta (n+1)]right] tag2
endalign
for $n in mathbbN$ using the Fourier series of $ln(sin)$ and repeated integration by parts. $zeta$ is the Riemann zeta function, $eta$ is the Dirichlet eta function and $1_2mathbbN$ is the indicator function of the even positive integers, i. e. $1_2mathbbN (2k) = 1$ and $1_2mathbbN(2k-1) = 0$ for $k in mathbbN$ .
For $n geq m geq 2$ we obtain the recurrence relation
$$ K_n^(m) = fracnm-1 K_n-1^(m-1) - K_n^(m-2) tag3 $$
from the definition. It enables us to calculate any $K_n^(m)$ in terms of the integrals $(1)$ and $(2)$ , but I would prefer a closed-form expression.
I have tried to solve the recurrence using generating functions. The functions
$$F^(0)(x) equiv sum limits_n=0^infty K_n^(0) x^n = - fraclnleft(1-fracpi x2right)x$$
and
$$ F^(1)(x) equiv sum limits_n=1^infty K_n^(1) x^n $$
are known (at least in principle). If we define $K_n^(m) = 0$ for $m>n$ , we find (if I did not make a mistake) the following PDE for the function
$$ F(x,y) equiv sum limits_n=0^infty sum limits_m=0^infty K_n^(m) x^n y^m - F^(0) (x)$$
from $(3)$ :
$$ x fracpartialpartial x left[x F(x,y)right] = y fracpartialpartial y left[y left(F(x,y)+F^(0)(x) - fracpi2right) + frac1y F(x,y)right] , . tag4 $$
It is subject to the initial conditions $F(x,0) = 0$ and $ partial_2 F(x,0) = F^(1)(x) $ .
However, I am not sure how to proceed from here and could really use a hint or two regarding the following questions:
Is the method of generating functions the simplest way to solve the recurrence relation given by $(1)$, $(2)$ and $(3)$? If so, how can we find a solution to the PDE $(4)$? If not, what are possible alternative approaches ?
integration pde definite-integrals recurrence-relations generating-functions
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
 |Â
show 1 more comment
up vote
4
down vote
favorite
I have found a general expression for the integrals $I_n$ discussed in this question in terms of $K_l^(l) , , , 0 leq l leq n , , $ where
$$ K_n^(m) equiv int limits_0^pi/2 t^n cot^m (t) , mathrmd t , , , n in mathbbN_0 , , , 0 leq m leq n , .$$
I would now like to obtain an explicit formula for these integrals in order to complete the solution of the original problem.
The simplest case is trivial:
$$ K_n^(0) = int limits_0^pi/2 t^n , mathrmd t = frac1n+1 left(fracpi2right)^n+1 , , , n in mathbbN_0 , . tag1 $$
With some more effort it is possible to compute
beginalign
K_n^(1) &= 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 \
&= fracn!2^n left[sum limits_l=0^left lfloor fracn-12 right rfloor (-1)^l fracpi^n-2l(n-2l)! eta (2l+1) + operatorname1_2 mathbbN (n) (-1)^left lfloor fracn+12 right rfloor [zeta (n+1) + eta (n+1)]right] tag2
endalign
for $n in mathbbN$ using the Fourier series of $ln(sin)$ and repeated integration by parts. $zeta$ is the Riemann zeta function, $eta$ is the Dirichlet eta function and $1_2mathbbN$ is the indicator function of the even positive integers, i. e. $1_2mathbbN (2k) = 1$ and $1_2mathbbN(2k-1) = 0$ for $k in mathbbN$ .
For $n geq m geq 2$ we obtain the recurrence relation
$$ K_n^(m) = fracnm-1 K_n-1^(m-1) - K_n^(m-2) tag3 $$
from the definition. It enables us to calculate any $K_n^(m)$ in terms of the integrals $(1)$ and $(2)$ , but I would prefer a closed-form expression.
I have tried to solve the recurrence using generating functions. The functions
$$F^(0)(x) equiv sum limits_n=0^infty K_n^(0) x^n = - fraclnleft(1-fracpi x2right)x$$
and
$$ F^(1)(x) equiv sum limits_n=1^infty K_n^(1) x^n $$
are known (at least in principle). If we define $K_n^(m) = 0$ for $m>n$ , we find (if I did not make a mistake) the following PDE for the function
$$ F(x,y) equiv sum limits_n=0^infty sum limits_m=0^infty K_n^(m) x^n y^m - F^(0) (x)$$
from $(3)$ :
$$ x fracpartialpartial x left[x F(x,y)right] = y fracpartialpartial y left[y left(F(x,y)+F^(0)(x) - fracpi2right) + frac1y F(x,y)right] , . tag4 $$
It is subject to the initial conditions $F(x,0) = 0$ and $ partial_2 F(x,0) = F^(1)(x) $ .
However, I am not sure how to proceed from here and could really use a hint or two regarding the following questions:
Is the method of generating functions the simplest way to solve the recurrence relation given by $(1)$, $(2)$ and $(3)$? If so, how can we find a solution to the PDE $(4)$? If not, what are possible alternative approaches ?
integration pde definite-integrals recurrence-relations generating-functions
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
Formula dosen't work forn=4,6,8,10...unless I'm wrong ?
– Mariusz Iwaniuk
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
I checked 5 times for:n=4the indicator function must be equal1/2to agree with formula ? I'm confused.
– Mariusz Iwaniuk
13 hours ago
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have found a general expression for the integrals $I_n$ discussed in this question in terms of $K_l^(l) , , , 0 leq l leq n , , $ where
$$ K_n^(m) equiv int limits_0^pi/2 t^n cot^m (t) , mathrmd t , , , n in mathbbN_0 , , , 0 leq m leq n , .$$
I would now like to obtain an explicit formula for these integrals in order to complete the solution of the original problem.
The simplest case is trivial:
$$ K_n^(0) = int limits_0^pi/2 t^n , mathrmd t = frac1n+1 left(fracpi2right)^n+1 , , , n in mathbbN_0 , . tag1 $$
With some more effort it is possible to compute
beginalign
K_n^(1) &= 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 \
&= fracn!2^n left[sum limits_l=0^left lfloor fracn-12 right rfloor (-1)^l fracpi^n-2l(n-2l)! eta (2l+1) + operatorname1_2 mathbbN (n) (-1)^left lfloor fracn+12 right rfloor [zeta (n+1) + eta (n+1)]right] tag2
endalign
for $n in mathbbN$ using the Fourier series of $ln(sin)$ and repeated integration by parts. $zeta$ is the Riemann zeta function, $eta$ is the Dirichlet eta function and $1_2mathbbN$ is the indicator function of the even positive integers, i. e. $1_2mathbbN (2k) = 1$ and $1_2mathbbN(2k-1) = 0$ for $k in mathbbN$ .
For $n geq m geq 2$ we obtain the recurrence relation
$$ K_n^(m) = fracnm-1 K_n-1^(m-1) - K_n^(m-2) tag3 $$
from the definition. It enables us to calculate any $K_n^(m)$ in terms of the integrals $(1)$ and $(2)$ , but I would prefer a closed-form expression.
I have tried to solve the recurrence using generating functions. The functions
$$F^(0)(x) equiv sum limits_n=0^infty K_n^(0) x^n = - fraclnleft(1-fracpi x2right)x$$
and
$$ F^(1)(x) equiv sum limits_n=1^infty K_n^(1) x^n $$
are known (at least in principle). If we define $K_n^(m) = 0$ for $m>n$ , we find (if I did not make a mistake) the following PDE for the function
$$ F(x,y) equiv sum limits_n=0^infty sum limits_m=0^infty K_n^(m) x^n y^m - F^(0) (x)$$
from $(3)$ :
$$ x fracpartialpartial x left[x F(x,y)right] = y fracpartialpartial y left[y left(F(x,y)+F^(0)(x) - fracpi2right) + frac1y F(x,y)right] , . tag4 $$
It is subject to the initial conditions $F(x,0) = 0$ and $ partial_2 F(x,0) = F^(1)(x) $ .
However, I am not sure how to proceed from here and could really use a hint or two regarding the following questions:
Is the method of generating functions the simplest way to solve the recurrence relation given by $(1)$, $(2)$ and $(3)$? If so, how can we find a solution to the PDE $(4)$? If not, what are possible alternative approaches ?
integration pde definite-integrals recurrence-relations generating-functions
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
I have found a general expression for the integrals $I_n$ discussed in this question in terms of $K_l^(l) , , , 0 leq l leq n , , $ where
$$ K_n^(m) equiv int limits_0^pi/2 t^n cot^m (t) , mathrmd t , , , n in mathbbN_0 , , , 0 leq m leq n , .$$
I would now like to obtain an explicit formula for these integrals in order to complete the solution of the original problem.
The simplest case is trivial:
$$ K_n^(0) = int limits_0^pi/2 t^n , mathrmd t = frac1n+1 left(fracpi2right)^n+1 , , , n in mathbbN_0 , . tag1 $$
With some more effort it is possible to compute
beginalign
K_n^(1) &= 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 \
&= fracn!2^n left[sum limits_l=0^left lfloor fracn-12 right rfloor (-1)^l fracpi^n-2l(n-2l)! eta (2l+1) + operatorname1_2 mathbbN (n) (-1)^left lfloor fracn+12 right rfloor [zeta (n+1) + eta (n+1)]right] tag2
endalign
for $n in mathbbN$ using the Fourier series of $ln(sin)$ and repeated integration by parts. $zeta$ is the Riemann zeta function, $eta$ is the Dirichlet eta function and $1_2mathbbN$ is the indicator function of the even positive integers, i. e. $1_2mathbbN (2k) = 1$ and $1_2mathbbN(2k-1) = 0$ for $k in mathbbN$ .
For $n geq m geq 2$ we obtain the recurrence relation
$$ K_n^(m) = fracnm-1 K_n-1^(m-1) - K_n^(m-2) tag3 $$
from the definition. It enables us to calculate any $K_n^(m)$ in terms of the integrals $(1)$ and $(2)$ , but I would prefer a closed-form expression.
I have tried to solve the recurrence using generating functions. The functions
$$F^(0)(x) equiv sum limits_n=0^infty K_n^(0) x^n = - fraclnleft(1-fracpi x2right)x$$
and
$$ F^(1)(x) equiv sum limits_n=1^infty K_n^(1) x^n $$
are known (at least in principle). If we define $K_n^(m) = 0$ for $m>n$ , we find (if I did not make a mistake) the following PDE for the function
$$ F(x,y) equiv sum limits_n=0^infty sum limits_m=0^infty K_n^(m) x^n y^m - F^(0) (x)$$
from $(3)$ :
$$ x fracpartialpartial x left[x F(x,y)right] = y fracpartialpartial y left[y left(F(x,y)+F^(0)(x) - fracpi2right) + frac1y F(x,y)right] , . tag4 $$
It is subject to the initial conditions $F(x,0) = 0$ and $ partial_2 F(x,0) = F^(1)(x) $ .
However, I am not sure how to proceed from here and could really use a hint or two regarding the following questions:
Is the method of generating functions the simplest way to solve the recurrence relation given by $(1)$, $(2)$ and $(3)$? If so, how can we find a solution to the PDE $(4)$? If not, what are possible alternative approaches ?
integration pde definite-integrals recurrence-relations generating-functions
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
edited 15 hours ago
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
asked Aug 3 at 20:25
ComplexYetTrivial
2,592624
2,592624
What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
Formula dosen't work forn=4,6,8,10...unless I'm wrong ?
– Mariusz Iwaniuk
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
I checked 5 times for:n=4the indicator function must be equal1/2to agree with formula ? I'm confused.
– Mariusz Iwaniuk
13 hours ago
 |Â
show 1 more comment
What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
Formula dosen't work forn=4,6,8,10...unless I'm wrong ?
– Mariusz Iwaniuk
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
I checked 5 times for:n=4the indicator function must be equal1/2to agree with formula ? I'm confused.
– Mariusz Iwaniuk
13 hours ago
What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
Formula dosen't work for
n=4,6,8,10... unless I'm wrong ?– Mariusz Iwaniuk
14 hours ago
Formula dosen't work for
n=4,6,8,10... unless I'm wrong ?– Mariusz Iwaniuk
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
I checked 5 times for:
n=4the indicator function must be equal 1/2 to agree with formula ? I'm confused.– Mariusz Iwaniuk
13 hours ago
I checked 5 times for:
n=4the indicator function must be equal 1/2 to agree with formula ? I'm confused.– Mariusz Iwaniuk
13 hours ago
 |Â
show 1 more comment
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What is :$eta(n)$ and $operatorname1_2 mathbbN (n)$ ?
– Mariusz Iwaniuk
17 hours ago
@MariuszIwaniuk Sorry, I forgot to define these functions. I have added the explanations.
– ComplexYetTrivial
15 hours ago
Formula dosen't work for
n=4,6,8,10...unless I'm wrong ?– Mariusz Iwaniuk
14 hours ago
@MariuszIwaniuk It seems to work for me. It yields $K_4^(1) = frac2416 left(fracpi^424 ln(2) - fracpi^22 frac34 zeta(3) + frac3116 zeta(5) right)$ , which is equal to the value given by WolframAlpha. The results appear to agree for larger $n$ as well.
– ComplexYetTrivial
14 hours ago
I checked 5 times for:
n=4the indicator function must be equal1/2to agree with formula ? I'm confused.– Mariusz Iwaniuk
13 hours ago