Mixing
Conjecture:$int_0^{pi/2} left(frac{sin(2n+1)x}{sin x}right)^beta dx$ is integer multiple of $pi/2$,for...
I was solving the integral $$I_n=int_0^{frac {pi}{2}} left(frac {sin ((2n+1)x)}{sin x}right)^2 dx$$
With $nge 0$ And $nin mathbb{N}$
On solving, I got $$I_n =frac {(2n+1)pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as
$$I_n(beta) =int_0^{frac {pi}{2}} left(frac {sin (2n+1)x}{sin x}right)^{beta} dx$$
On trying various values of $betagt 2$ and $betain mathbb{N}$, I conjectured that
$$I_n(beta) =c_{beta} frac{pi}{2}$$
where $c_{beta}$ denotes "Number of arrays of $beta$ integers in $-n$ to $n$ with sum $0$"
But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
calculus integration trigonometry definite-integrals
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
|
show 5 more comments
I was solving the integral $$I_n=int_0^{frac {pi}{2}} left(frac {sin ((2n+1)x)}{sin x}right)^2 dx$$
With $nge 0$ And $nin mathbb{N}$
On solving, I got $$I_n =frac {(2n+1)pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as
$$I_n(beta) =int_0^{frac {pi}{2}} left(frac {sin (2n+1)x}{sin x}right)^{beta} dx$$
On trying various values of $betagt 2$ and $betain mathbb{N}$, I conjectured that
$$I_n(beta) =c_{beta} frac{pi}{2}$$
where $c_{beta}$ denotes "Number of arrays of $beta$ integers in $-n$ to $n$ with sum $0$"
But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
calculus integration trigonometry definite-integrals
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
1
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
2
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01
|
show 5 more comments
I was solving the integral $$I_n=int_0^{frac {pi}{2}} left(frac {sin ((2n+1)x)}{sin x}right)^2 dx$$
With $nge 0$ And $nin mathbb{N}$
On solving, I got $$I_n =frac {(2n+1)pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as
$$I_n(beta) =int_0^{frac {pi}{2}} left(frac {sin (2n+1)x}{sin x}right)^{beta} dx$$
On trying various values of $betagt 2$ and $betain mathbb{N}$, I conjectured that
$$I_n(beta) =c_{beta} frac{pi}{2}$$
where $c_{beta}$ denotes "Number of arrays of $beta$ integers in $-n$ to $n$ with sum $0$"
But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
calculus integration trigonometry definite-integrals
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
I was solving the integral $$I_n=int_0^{frac {pi}{2}} left(frac {sin ((2n+1)x)}{sin x}right)^2 dx$$
With $nge 0$ And $nin mathbb{N}$
On solving, I got $$I_n =frac {(2n+1)pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as
$$I_n(beta) =int_0^{frac {pi}{2}} left(frac {sin (2n+1)x}{sin x}right)^{beta} dx$$
On trying various values of $betagt 2$ and $betain mathbb{N}$, I conjectured that
$$I_n(beta) =c_{beta} frac{pi}{2}$$
where $c_{beta}$ denotes "Number of arrays of $beta$ integers in $-n$ to $n$ with sum $0$"
But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
calculus integration trigonometry definite-integrals
calculus integration trigonometry definite-integrals
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
edited Jan 3 at 3:12
Digamma
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
asked Dec 29 '18 at 6:53
DigammaDigamma
6,1691439
6,1691439
1
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
2
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01
|
show 5 more comments
1
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
2
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01
1
1
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
2
2
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01
|
show 5 more comments
2 Answers
2
active
oldest
votes
$defb{beta}$begin{align*}
newcommandcmt[1]{{smalltextrm{#1}}}
I_n(b) &= int_0^{pi/2}
left(frac{sin (2n+1)x}{sin x}right)^b dx \
&= frac 1 4 int_0^{2pi}
left(frac{sin (2n+1)x}{sin x}right)^b dx
& cmt{begin similar to user630708} \
&= frac{1}{4i} oint_gamma
left(frac{z^{4n+2}-1}{z^2-1}right)^b
frac{dz}{z^{2nb+1}}
& cmt{let $z=e^{ix}$} \
&= frac{1}{4i} oint_gamma
left(sum_{k=0}^{2n}z^{2k}right)^b
frac{dz}{z^{2nb+1}}
& cmt{partial sum of geometric series} \
&= left.frac{1}{4i} frac{2pi i}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
left(sum_{k=0}^{2n}z^{2k}right)^b right|_{z=0}
& cmt{Cauchy integral formula} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
prod (z^{2k})^{x_k}
right|_{z=0}
& cmt{multinomial expansion, $k=0,1,ldots,2n$} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
z^{2sum k x_k}
right|_{z=0} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum k x_k = nb}
frac{b!}{prod x_k!}
& cmt{only surviving terms have $sum k x_k = nb$} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum (n-k) x_k = 0}
frac{b!}{prod x_k!}
end{align*}
In the last line note that
$sum_{k=0}^{2n} n x_k=nb$ and so
$sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$sum_{sum_{k=0}^{2n} x_k=b atop sum_{k=0}^{2n} (n-k) x_k = 0}
frac{b!}{prod x_k!}
= textrm{number of arrays of $b$ integers in $-n,ldots,n$ with sum equal to 0,}$$
i.e.,
$$I_n(b) = frac{pi}{2} T(b,n),$$
where $T(b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(b,n)$ which, after a simple substitution, gives $I_n(b) = frac{pi}{2} T(b,n)$.)
answered Jan 1 at 23:41
user26872user26872
14.9k22773
add a comment |
This is easy using Residue Theory:
Note that by symmetry $int_{0}^{pi/2}...dx=1/4int_{-pi}^{pi}…dx$ (use parity and a sub $y=pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,beta}=oint_C left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{dz}{i z^{2beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=pm i$ are removable singularities). We have
$$
4 I_{n,beta}=2pi text{Res}(left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{1}{z^{2beta n+1}}
,z=0)
$$
Using $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-beta}=frac{1}{2^{beta-1}}sum_{mgeq0}binom{m+beta-1}{beta-1}z^{2m}\
(1-z^{4n+2})^{beta}=z^{2beta}sum_{kgeq0}(-1)^kbinom{beta}{k}z^{4k}
$$
which means that we have the condition $4k+2(m+beta)-2beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,beta}=frac{pi}{2^{beta}}sum_{mgeq0}(-1)^{beta n /2-(beta+m)/2}binom{m+beta-1}{beta-1}binom{beta}{beta n /2-(beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> beta (n-1)$)
answered Dec 31 '18 at 14:41
user630708user630708
492
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
add a comment |
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2 Answers
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$defb{beta}$begin{align*}
newcommandcmt[1]{{smalltextrm{#1}}}
I_n(b) &= int_0^{pi/2}
left(frac{sin (2n+1)x}{sin x}right)^b dx \
&= frac 1 4 int_0^{2pi}
left(frac{sin (2n+1)x}{sin x}right)^b dx
& cmt{begin similar to user630708} \
&= frac{1}{4i} oint_gamma
left(frac{z^{4n+2}-1}{z^2-1}right)^b
frac{dz}{z^{2nb+1}}
& cmt{let $z=e^{ix}$} \
&= frac{1}{4i} oint_gamma
left(sum_{k=0}^{2n}z^{2k}right)^b
frac{dz}{z^{2nb+1}}
& cmt{partial sum of geometric series} \
&= left.frac{1}{4i} frac{2pi i}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
left(sum_{k=0}^{2n}z^{2k}right)^b right|_{z=0}
& cmt{Cauchy integral formula} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
prod (z^{2k})^{x_k}
right|_{z=0}
& cmt{multinomial expansion, $k=0,1,ldots,2n$} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
z^{2sum k x_k}
right|_{z=0} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum k x_k = nb}
frac{b!}{prod x_k!}
& cmt{only surviving terms have $sum k x_k = nb$} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum (n-k) x_k = 0}
frac{b!}{prod x_k!}
end{align*}
In the last line note that
$sum_{k=0}^{2n} n x_k=nb$ and so
$sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$sum_{sum_{k=0}^{2n} x_k=b atop sum_{k=0}^{2n} (n-k) x_k = 0}
frac{b!}{prod x_k!}
= textrm{number of arrays of $b$ integers in $-n,ldots,n$ with sum equal to 0,}$$
i.e.,
$$I_n(b) = frac{pi}{2} T(b,n),$$
where $T(b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(b,n)$ which, after a simple substitution, gives $I_n(b) = frac{pi}{2} T(b,n)$.)
answered Jan 1 at 23:41
user26872user26872
14.9k22773
add a comment |
$defb{beta}$begin{align*}
newcommandcmt[1]{{smalltextrm{#1}}}
I_n(b) &= int_0^{pi/2}
left(frac{sin (2n+1)x}{sin x}right)^b dx \
&= frac 1 4 int_0^{2pi}
left(frac{sin (2n+1)x}{sin x}right)^b dx
& cmt{begin similar to user630708} \
&= frac{1}{4i} oint_gamma
left(frac{z^{4n+2}-1}{z^2-1}right)^b
frac{dz}{z^{2nb+1}}
& cmt{let $z=e^{ix}$} \
&= frac{1}{4i} oint_gamma
left(sum_{k=0}^{2n}z^{2k}right)^b
frac{dz}{z^{2nb+1}}
& cmt{partial sum of geometric series} \
&= left.frac{1}{4i} frac{2pi i}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
left(sum_{k=0}^{2n}z^{2k}right)^b right|_{z=0}
& cmt{Cauchy integral formula} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
prod (z^{2k})^{x_k}
right|_{z=0}
& cmt{multinomial expansion, $k=0,1,ldots,2n$} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
z^{2sum k x_k}
right|_{z=0} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum k x_k = nb}
frac{b!}{prod x_k!}
& cmt{only surviving terms have $sum k x_k = nb$} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum (n-k) x_k = 0}
frac{b!}{prod x_k!}
end{align*}
In the last line note that
$sum_{k=0}^{2n} n x_k=nb$ and so
$sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$sum_{sum_{k=0}^{2n} x_k=b atop sum_{k=0}^{2n} (n-k) x_k = 0}
frac{b!}{prod x_k!}
= textrm{number of arrays of $b$ integers in $-n,ldots,n$ with sum equal to 0,}$$
i.e.,
$$I_n(b) = frac{pi}{2} T(b,n),$$
where $T(b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(b,n)$ which, after a simple substitution, gives $I_n(b) = frac{pi}{2} T(b,n)$.)
answered Jan 1 at 23:41
user26872user26872
14.9k22773
add a comment |
$defb{beta}$begin{align*}
newcommandcmt[1]{{smalltextrm{#1}}}
I_n(b) &= int_0^{pi/2}
left(frac{sin (2n+1)x}{sin x}right)^b dx \
&= frac 1 4 int_0^{2pi}
left(frac{sin (2n+1)x}{sin x}right)^b dx
& cmt{begin similar to user630708} \
&= frac{1}{4i} oint_gamma
left(frac{z^{4n+2}-1}{z^2-1}right)^b
frac{dz}{z^{2nb+1}}
& cmt{let $z=e^{ix}$} \
&= frac{1}{4i} oint_gamma
left(sum_{k=0}^{2n}z^{2k}right)^b
frac{dz}{z^{2nb+1}}
& cmt{partial sum of geometric series} \
&= left.frac{1}{4i} frac{2pi i}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
left(sum_{k=0}^{2n}z^{2k}right)^b right|_{z=0}
& cmt{Cauchy integral formula} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
prod (z^{2k})^{x_k}
right|_{z=0}
& cmt{multinomial expansion, $k=0,1,ldots,2n$} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
z^{2sum k x_k}
right|_{z=0} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum k x_k = nb}
frac{b!}{prod x_k!}
& cmt{only surviving terms have $sum k x_k = nb$} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum (n-k) x_k = 0}
frac{b!}{prod x_k!}
end{align*}
In the last line note that
$sum_{k=0}^{2n} n x_k=nb$ and so
$sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$sum_{sum_{k=0}^{2n} x_k=b atop sum_{k=0}^{2n} (n-k) x_k = 0}
frac{b!}{prod x_k!}
= textrm{number of arrays of $b$ integers in $-n,ldots,n$ with sum equal to 0,}$$
i.e.,
$$I_n(b) = frac{pi}{2} T(b,n),$$
where $T(b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(b,n)$ which, after a simple substitution, gives $I_n(b) = frac{pi}{2} T(b,n)$.)
answered Jan 1 at 23:41
user26872user26872
14.9k22773
$defb{beta}$begin{align*}
newcommandcmt[1]{{smalltextrm{#1}}}
I_n(b) &= int_0^{pi/2}
left(frac{sin (2n+1)x}{sin x}right)^b dx \
&= frac 1 4 int_0^{2pi}
left(frac{sin (2n+1)x}{sin x}right)^b dx
& cmt{begin similar to user630708} \
&= frac{1}{4i} oint_gamma
left(frac{z^{4n+2}-1}{z^2-1}right)^b
frac{dz}{z^{2nb+1}}
& cmt{let $z=e^{ix}$} \
&= frac{1}{4i} oint_gamma
left(sum_{k=0}^{2n}z^{2k}right)^b
frac{dz}{z^{2nb+1}}
& cmt{partial sum of geometric series} \
&= left.frac{1}{4i} frac{2pi i}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
left(sum_{k=0}^{2n}z^{2k}right)^b right|_{z=0}
& cmt{Cauchy integral formula} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
prod (z^{2k})^{x_k}
right|_{z=0}
& cmt{multinomial expansion, $k=0,1,ldots,2n$} \
&= left.frac{pi}{2} frac{1}{(2nb)!}
left(frac{d}{dz}right)^{2nb}
sum_{sum x_k=b} frac{b!}{prod x_k!}
z^{2sum k x_k}
right|_{z=0} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum k x_k = nb}
frac{b!}{prod x_k!}
& cmt{only surviving terms have $sum k x_k = nb$} \
&= frac{pi}{2}
sum_{sum x_k=b atop sum (n-k) x_k = 0}
frac{b!}{prod x_k!}
end{align*}
In the last line note that
$sum_{k=0}^{2n} n x_k=nb$ and so
$sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$sum_{sum_{k=0}^{2n} x_k=b atop sum_{k=0}^{2n} (n-k) x_k = 0}
frac{b!}{prod x_k!}
= textrm{number of arrays of $b$ integers in $-n,ldots,n$ with sum equal to 0,}$$
i.e.,
$$I_n(b) = frac{pi}{2} T(b,n),$$
where $T(b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(b,n)$ which, after a simple substitution, gives $I_n(b) = frac{pi}{2} T(b,n)$.)
answered Jan 1 at 23:41
user26872user26872
14.9k22773
edited Jan 5 at 20:21
answered Jan 1 at 23:41
user26872user26872
14.9k22773
answered Jan 1 at 23:41
user26872user26872
14.9k22773
answered Jan 1 at 23:41
user26872user26872
14.9k22773
14.9k22773
add a comment |
add a comment |
This is easy using Residue Theory:
Note that by symmetry $int_{0}^{pi/2}...dx=1/4int_{-pi}^{pi}…dx$ (use parity and a sub $y=pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,beta}=oint_C left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{dz}{i z^{2beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=pm i$ are removable singularities). We have
$$
4 I_{n,beta}=2pi text{Res}(left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{1}{z^{2beta n+1}}
,z=0)
$$
Using $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-beta}=frac{1}{2^{beta-1}}sum_{mgeq0}binom{m+beta-1}{beta-1}z^{2m}\
(1-z^{4n+2})^{beta}=z^{2beta}sum_{kgeq0}(-1)^kbinom{beta}{k}z^{4k}
$$
which means that we have the condition $4k+2(m+beta)-2beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,beta}=frac{pi}{2^{beta}}sum_{mgeq0}(-1)^{beta n /2-(beta+m)/2}binom{m+beta-1}{beta-1}binom{beta}{beta n /2-(beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> beta (n-1)$)
answered Dec 31 '18 at 14:41
user630708user630708
492
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
add a comment |
This is easy using Residue Theory:
Note that by symmetry $int_{0}^{pi/2}...dx=1/4int_{-pi}^{pi}…dx$ (use parity and a sub $y=pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,beta}=oint_C left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{dz}{i z^{2beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=pm i$ are removable singularities). We have
$$
4 I_{n,beta}=2pi text{Res}(left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{1}{z^{2beta n+1}}
,z=0)
$$
Using $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-beta}=frac{1}{2^{beta-1}}sum_{mgeq0}binom{m+beta-1}{beta-1}z^{2m}\
(1-z^{4n+2})^{beta}=z^{2beta}sum_{kgeq0}(-1)^kbinom{beta}{k}z^{4k}
$$
which means that we have the condition $4k+2(m+beta)-2beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,beta}=frac{pi}{2^{beta}}sum_{mgeq0}(-1)^{beta n /2-(beta+m)/2}binom{m+beta-1}{beta-1}binom{beta}{beta n /2-(beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> beta (n-1)$)
answered Dec 31 '18 at 14:41
user630708user630708
492
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
add a comment |
This is easy using Residue Theory:
Note that by symmetry $int_{0}^{pi/2}...dx=1/4int_{-pi}^{pi}…dx$ (use parity and a sub $y=pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,beta}=oint_C left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{dz}{i z^{2beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=pm i$ are removable singularities). We have
$$
4 I_{n,beta}=2pi text{Res}(left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{1}{z^{2beta n+1}}
,z=0)
$$
Using $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-beta}=frac{1}{2^{beta-1}}sum_{mgeq0}binom{m+beta-1}{beta-1}z^{2m}\
(1-z^{4n+2})^{beta}=z^{2beta}sum_{kgeq0}(-1)^kbinom{beta}{k}z^{4k}
$$
which means that we have the condition $4k+2(m+beta)-2beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,beta}=frac{pi}{2^{beta}}sum_{mgeq0}(-1)^{beta n /2-(beta+m)/2}binom{m+beta-1}{beta-1}binom{beta}{beta n /2-(beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> beta (n-1)$)
answered Dec 31 '18 at 14:41
user630708user630708
492
This is easy using Residue Theory:
Note that by symmetry $int_{0}^{pi/2}...dx=1/4int_{-pi}^{pi}…dx$ (use parity and a sub $y=pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,beta}=oint_C left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{dz}{i z^{2beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=pm i$ are removable singularities). We have
$$
4 I_{n,beta}=2pi text{Res}(left[frac{z^{4n+2}-1}{z^2-1}right]^{beta}frac{1}{z^{2beta n+1}}
,z=0)
$$
Using $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-beta}=frac{1}{2^{beta-1}}sum_{mgeq0}binom{m+beta-1}{beta-1}z^{2m}\
(1-z^{4n+2})^{beta}=z^{2beta}sum_{kgeq0}(-1)^kbinom{beta}{k}z^{4k}
$$
which means that we have the condition $4k+2(m+beta)-2beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,beta}=frac{pi}{2^{beta}}sum_{mgeq0}(-1)^{beta n /2-(beta+m)/2}binom{m+beta-1}{beta-1}binom{beta}{beta n /2-(beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> beta (n-1)$)
answered Dec 31 '18 at 14:41
user630708user630708
492
edited Dec 31 '18 at 14:50
answered Dec 31 '18 at 14:41
user630708user630708
492
answered Dec 31 '18 at 14:41
user630708user630708
492
answered Dec 31 '18 at 14:41
user630708user630708
492
492
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
add a comment |
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
This seems cool, but where does the $beta neq 4$ exception come from? Is it from that line about how we "kill one of the sums"?
– goblin
Jan 1 at 23:46
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Notice that the final sum is not necessarily real.
– user26872
Jan 2 at 0:37
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
Things seem to go off the rails with $beta!(z^2-1)^{-beta}=((2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$, which is false for $beta>1$. It is true that $(beta-1)!(z^2-1)^{-beta}=((-2z)^{-1}partial_z)^{beta-1}(z^2-1)^{-1}$.
– user26872
Jan 5 at 20:45
add a comment |
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1
@Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $sin((2n+1)x)$.
– xbh
Dec 29 '18 at 7:08
@Masacroso Edited!!!
– Digamma
Dec 29 '18 at 7:10
oeis.org/A201552
– James Arathoon
Dec 29 '18 at 12:21
2
See here: math.stackexchange.com/a/2885887/515527
– Zacky
Dec 29 '18 at 12:36
@James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs.
– Digamma
Dec 29 '18 at 13:01