Mixing
Showing that $sum_{n=-infty}^{infty}frac{(-1)^n}{(n-alpha)^2}=pi^2csc pi alpha cot pi alpha$
$begingroup$
I'm trying to show that $sum_{n=-infty}^{infty}frac{(-1)^n}{(n-alpha)^2}=pi^2csc pi alpha cot pi alpha$ for $0<alpha<1$. The method that I decided to use is contour (square) integrals and the residue theorem.
Progress
I noticed that $f(z)=frac{1}{z^2sinpi(z+alpha)}$ has poles at $z=0, k-alpha, kin mathbb{Z}$ with residues $$mbox{res}(f(z);0)=-pi cot (pi alpha) csc( pi alpha) \ mbox{res}(f(z);k-alpha)=frac{(-1)^k}{pi(k-alpha)^2}$$ which one can obtain by simply calculating. So then I considered a path $Gamma_N$ defined by a square with vertices at $N(1, i), N(1, -1), N(-1, i), N(-1, -i)$, $N in mathbb{N}$. So if we let $a_k$ be the list of poles of $f$ inside the region formed by $Gamma_N$, by the residue theorem, we get:
$$begin{align*} oint_{Gamma_N}f(z)dz &= 2pi isum_{a_i} mbox{res}(f(z);a_k) \ &= 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{n=-N+1}^N frac{(-1)^n}{pi(n-alpha)^2}right] \ &to 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}right] end{align*}$$
Now, supposing $oint_{Gamma_N}f(z)dz to 0$, we get $$0=-pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}$$ and then we're done.
But I cannot seem to prove that $oint_{Gamma_N}f(z)dz to 0$. I'm pretty sure the Estimation Lemma is used here, but I am not sure how to go about it. Is this even true? Are there any other ways of proving it?
sequences-and-series residue-calculus
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
$endgroup$
add a comment |
$begingroup$
I'm trying to show that $sum_{n=-infty}^{infty}frac{(-1)^n}{(n-alpha)^2}=pi^2csc pi alpha cot pi alpha$ for $0<alpha<1$. The method that I decided to use is contour (square) integrals and the residue theorem.
Progress
I noticed that $f(z)=frac{1}{z^2sinpi(z+alpha)}$ has poles at $z=0, k-alpha, kin mathbb{Z}$ with residues $$mbox{res}(f(z);0)=-pi cot (pi alpha) csc( pi alpha) \ mbox{res}(f(z);k-alpha)=frac{(-1)^k}{pi(k-alpha)^2}$$ which one can obtain by simply calculating. So then I considered a path $Gamma_N$ defined by a square with vertices at $N(1, i), N(1, -1), N(-1, i), N(-1, -i)$, $N in mathbb{N}$. So if we let $a_k$ be the list of poles of $f$ inside the region formed by $Gamma_N$, by the residue theorem, we get:
$$begin{align*} oint_{Gamma_N}f(z)dz &= 2pi isum_{a_i} mbox{res}(f(z);a_k) \ &= 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{n=-N+1}^N frac{(-1)^n}{pi(n-alpha)^2}right] \ &to 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}right] end{align*}$$
Now, supposing $oint_{Gamma_N}f(z)dz to 0$, we get $$0=-pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}$$ and then we're done.
But I cannot seem to prove that $oint_{Gamma_N}f(z)dz to 0$. I'm pretty sure the Estimation Lemma is used here, but I am not sure how to go about it. Is this even true? Are there any other ways of proving it?
sequences-and-series residue-calculus
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
$endgroup$
$begingroup$
Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
$endgroup$
– Professor Vector
Sep 29 '17 at 4:11
add a comment |
$begingroup$
I'm trying to show that $sum_{n=-infty}^{infty}frac{(-1)^n}{(n-alpha)^2}=pi^2csc pi alpha cot pi alpha$ for $0<alpha<1$. The method that I decided to use is contour (square) integrals and the residue theorem.
Progress
I noticed that $f(z)=frac{1}{z^2sinpi(z+alpha)}$ has poles at $z=0, k-alpha, kin mathbb{Z}$ with residues $$mbox{res}(f(z);0)=-pi cot (pi alpha) csc( pi alpha) \ mbox{res}(f(z);k-alpha)=frac{(-1)^k}{pi(k-alpha)^2}$$ which one can obtain by simply calculating. So then I considered a path $Gamma_N$ defined by a square with vertices at $N(1, i), N(1, -1), N(-1, i), N(-1, -i)$, $N in mathbb{N}$. So if we let $a_k$ be the list of poles of $f$ inside the region formed by $Gamma_N$, by the residue theorem, we get:
$$begin{align*} oint_{Gamma_N}f(z)dz &= 2pi isum_{a_i} mbox{res}(f(z);a_k) \ &= 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{n=-N+1}^N frac{(-1)^n}{pi(n-alpha)^2}right] \ &to 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}right] end{align*}$$
Now, supposing $oint_{Gamma_N}f(z)dz to 0$, we get $$0=-pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}$$ and then we're done.
But I cannot seem to prove that $oint_{Gamma_N}f(z)dz to 0$. I'm pretty sure the Estimation Lemma is used here, but I am not sure how to go about it. Is this even true? Are there any other ways of proving it?
sequences-and-series residue-calculus
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
$endgroup$
I'm trying to show that $sum_{n=-infty}^{infty}frac{(-1)^n}{(n-alpha)^2}=pi^2csc pi alpha cot pi alpha$ for $0<alpha<1$. The method that I decided to use is contour (square) integrals and the residue theorem.
Progress
I noticed that $f(z)=frac{1}{z^2sinpi(z+alpha)}$ has poles at $z=0, k-alpha, kin mathbb{Z}$ with residues $$mbox{res}(f(z);0)=-pi cot (pi alpha) csc( pi alpha) \ mbox{res}(f(z);k-alpha)=frac{(-1)^k}{pi(k-alpha)^2}$$ which one can obtain by simply calculating. So then I considered a path $Gamma_N$ defined by a square with vertices at $N(1, i), N(1, -1), N(-1, i), N(-1, -i)$, $N in mathbb{N}$. So if we let $a_k$ be the list of poles of $f$ inside the region formed by $Gamma_N$, by the residue theorem, we get:
$$begin{align*} oint_{Gamma_N}f(z)dz &= 2pi isum_{a_i} mbox{res}(f(z);a_k) \ &= 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{n=-N+1}^N frac{(-1)^n}{pi(n-alpha)^2}right] \ &to 2pi i left[ -pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}right] end{align*}$$
Now, supposing $oint_{Gamma_N}f(z)dz to 0$, we get $$0=-pi cot (pi alpha) csc( pi alpha)+sum_{-infty}^{infty} frac{(-1)^n}{pi(n-alpha)^2}$$ and then we're done.
But I cannot seem to prove that $oint_{Gamma_N}f(z)dz to 0$. I'm pretty sure the Estimation Lemma is used here, but I am not sure how to go about it. Is this even true? Are there any other ways of proving it?
sequences-and-series residue-calculus
sequences-and-series residue-calculus
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
edited Sep 29 '17 at 3:52
Jihoon Kang
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
asked Sep 29 '17 at 3:31
Jihoon KangJihoon Kang
48329
48329
$begingroup$
Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
$endgroup$
– Professor Vector
Sep 29 '17 at 4:11
add a comment |
$begingroup$
Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
$endgroup$
– Professor Vector
Sep 29 '17 at 4:11
$begingroup$
Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
$endgroup$
– Professor Vector
Sep 29 '17 at 4:11
$begingroup$
Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
$endgroup$
– Professor Vector
Sep 29 '17 at 4:11
add a comment |
3 Answers
3
active
oldest
votes
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Hint:
You can use an analogous function $$f(z) = frac{csc(pi z)}{(z-a)^2}$$ integrate around the square $Gamma_N$ with vertices $pm N pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:
$csc(z)$ is uniformly bounded on $Gamma_N$
Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.
The following result is also handy at residue summation, they're not difficult to prove either.
$csc(z), cot(z)$ are uniformly bounded on $Gamma_N$ with $N$ half integer.
$tan(z), sec(z)$ are uniformly bounded on $Gamma_N$ with $N$ integer.
answered Sep 29 '17 at 4:23
piscopisco
12k21743
$endgroup$
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
add a comment |
$begingroup$
In $(3)$ from this answer, it is shown that
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{k+z}=picsc(pi z)tag{1}
$$
Substituting $zmapsto-z$ in $(1)$ and taking the derivative yields
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{(k-z)^2}=pi^2csc(pi z)cot(pi z)tag{2}
$$
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
$endgroup$
add a comment |
$begingroup$
The OP had asked, "Are there any other ways of proving it?"
We begin by writing the Fourier series,
$$cos(alpha x)=a_0/2+sum_{n=1}^infty a_ncos(nx) tag1$$
for $xin [-pi/pi]$. The Fourier coefficients in $(1)$ are given by
$$begin{align}
a_n&=frac{2}{pi}int_0^pi cos(alpha x)cos(nx),dx\\
&=frac1pi (-1)^n sin(pi alpha)left(frac{1}{alpha +n}+frac{1}{alpha -n}right)tag2
end{align}$$
Substituting $2$ into $1$, dividing by $sin(pi y)$, and setting $x=0$ n $(2)$ reveals
$$begin{align}
pi csc(pi alpha)&=frac1alpha +sum_{n=1}^infty (-1)^nleft(frac{1}{alpha -n}+frac{1}{alpha +n}right)\\
&=sum_{n=-infty}^infty frac{(-1)^n}{alpha-n}tag3
end{align}$$
Now differentiating with respect to $alpha$, enforcing the substitution $alphato alpha/pi$, and dividing by $pi^2$ yields the coveted result
$$csc(alpha)cot(alpha)=sum_{n=-infty}^infty frac{(-1)^n}{(n-alpha)^2}$$
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
You can use an analogous function $$f(z) = frac{csc(pi z)}{(z-a)^2}$$ integrate around the square $Gamma_N$ with vertices $pm N pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:
$csc(z)$ is uniformly bounded on $Gamma_N$
Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.
The following result is also handy at residue summation, they're not difficult to prove either.
$csc(z), cot(z)$ are uniformly bounded on $Gamma_N$ with $N$ half integer.
$tan(z), sec(z)$ are uniformly bounded on $Gamma_N$ with $N$ integer.
answered Sep 29 '17 at 4:23
piscopisco
12k21743
$endgroup$
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
add a comment |
$begingroup$
Hint:
You can use an analogous function $$f(z) = frac{csc(pi z)}{(z-a)^2}$$ integrate around the square $Gamma_N$ with vertices $pm N pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:
$csc(z)$ is uniformly bounded on $Gamma_N$
Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.
The following result is also handy at residue summation, they're not difficult to prove either.
$csc(z), cot(z)$ are uniformly bounded on $Gamma_N$ with $N$ half integer.
$tan(z), sec(z)$ are uniformly bounded on $Gamma_N$ with $N$ integer.
answered Sep 29 '17 at 4:23
piscopisco
12k21743
$endgroup$
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
add a comment |
$begingroup$
Hint:
You can use an analogous function $$f(z) = frac{csc(pi z)}{(z-a)^2}$$ integrate around the square $Gamma_N$ with vertices $pm N pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:
$csc(z)$ is uniformly bounded on $Gamma_N$
Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.
The following result is also handy at residue summation, they're not difficult to prove either.
$csc(z), cot(z)$ are uniformly bounded on $Gamma_N$ with $N$ half integer.
$tan(z), sec(z)$ are uniformly bounded on $Gamma_N$ with $N$ integer.
answered Sep 29 '17 at 4:23
piscopisco
12k21743
$endgroup$
Hint:
You can use an analogous function $$f(z) = frac{csc(pi z)}{(z-a)^2}$$ integrate around the square $Gamma_N$ with vertices $pm N pm Ni$, where $N$ is a half integer. Regarding your question of vanishing of the integral, prove the following:
$csc(z)$ is uniformly bounded on $Gamma_N$
Since the denominator is of $O(1/z^2)$, this shows the integral tends to 0.
The following result is also handy at residue summation, they're not difficult to prove either.
$csc(z), cot(z)$ are uniformly bounded on $Gamma_N$ with $N$ half integer.
$tan(z), sec(z)$ are uniformly bounded on $Gamma_N$ with $N$ integer.
answered Sep 29 '17 at 4:23
piscopisco
12k21743
edited Sep 29 '17 at 4:31
answered Sep 29 '17 at 4:23
piscopisco
12k21743
answered Sep 29 '17 at 4:23
piscopisco
12k21743
answered Sep 29 '17 at 4:23
piscopisco
12k21743
12k21743
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
add a comment |
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
$begingroup$
Cheers got it ${}{}$
$endgroup$
– Jihoon Kang
Sep 29 '17 at 7:45
add a comment |
$begingroup$
In $(3)$ from this answer, it is shown that
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{k+z}=picsc(pi z)tag{1}
$$
Substituting $zmapsto-z$ in $(1)$ and taking the derivative yields
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{(k-z)^2}=pi^2csc(pi z)cot(pi z)tag{2}
$$
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
$endgroup$
add a comment |
$begingroup$
In $(3)$ from this answer, it is shown that
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{k+z}=picsc(pi z)tag{1}
$$
Substituting $zmapsto-z$ in $(1)$ and taking the derivative yields
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{(k-z)^2}=pi^2csc(pi z)cot(pi z)tag{2}
$$
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
$endgroup$
add a comment |
$begingroup$
In $(3)$ from this answer, it is shown that
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{k+z}=picsc(pi z)tag{1}
$$
Substituting $zmapsto-z$ in $(1)$ and taking the derivative yields
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{(k-z)^2}=pi^2csc(pi z)cot(pi z)tag{2}
$$
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
$endgroup$
In $(3)$ from this answer, it is shown that
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{k+z}=picsc(pi z)tag{1}
$$
Substituting $zmapsto-z$ in $(1)$ and taking the derivative yields
$$
sum_{kinmathbb{Z}}frac{(-1)^k}{(k-z)^2}=pi^2csc(pi z)cot(pi z)tag{2}
$$
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
answered Sep 29 '17 at 5:22
robjohn♦robjohn
270k27312640
270k27312640
add a comment |
add a comment |
$begingroup$
The OP had asked, "Are there any other ways of proving it?"
We begin by writing the Fourier series,
$$cos(alpha x)=a_0/2+sum_{n=1}^infty a_ncos(nx) tag1$$
for $xin [-pi/pi]$. The Fourier coefficients in $(1)$ are given by
$$begin{align}
a_n&=frac{2}{pi}int_0^pi cos(alpha x)cos(nx),dx\\
&=frac1pi (-1)^n sin(pi alpha)left(frac{1}{alpha +n}+frac{1}{alpha -n}right)tag2
end{align}$$
Substituting $2$ into $1$, dividing by $sin(pi y)$, and setting $x=0$ n $(2)$ reveals
$$begin{align}
pi csc(pi alpha)&=frac1alpha +sum_{n=1}^infty (-1)^nleft(frac{1}{alpha -n}+frac{1}{alpha +n}right)\\
&=sum_{n=-infty}^infty frac{(-1)^n}{alpha-n}tag3
end{align}$$
Now differentiating with respect to $alpha$, enforcing the substitution $alphato alpha/pi$, and dividing by $pi^2$ yields the coveted result
$$csc(alpha)cot(alpha)=sum_{n=-infty}^infty frac{(-1)^n}{(n-alpha)^2}$$
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
add a comment |
$begingroup$
The OP had asked, "Are there any other ways of proving it?"
We begin by writing the Fourier series,
$$cos(alpha x)=a_0/2+sum_{n=1}^infty a_ncos(nx) tag1$$
for $xin [-pi/pi]$. The Fourier coefficients in $(1)$ are given by
$$begin{align}
a_n&=frac{2}{pi}int_0^pi cos(alpha x)cos(nx),dx\\
&=frac1pi (-1)^n sin(pi alpha)left(frac{1}{alpha +n}+frac{1}{alpha -n}right)tag2
end{align}$$
Substituting $2$ into $1$, dividing by $sin(pi y)$, and setting $x=0$ n $(2)$ reveals
$$begin{align}
pi csc(pi alpha)&=frac1alpha +sum_{n=1}^infty (-1)^nleft(frac{1}{alpha -n}+frac{1}{alpha +n}right)\\
&=sum_{n=-infty}^infty frac{(-1)^n}{alpha-n}tag3
end{align}$$
Now differentiating with respect to $alpha$, enforcing the substitution $alphato alpha/pi$, and dividing by $pi^2$ yields the coveted result
$$csc(alpha)cot(alpha)=sum_{n=-infty}^infty frac{(-1)^n}{(n-alpha)^2}$$
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
add a comment |
$begingroup$
The OP had asked, "Are there any other ways of proving it?"
We begin by writing the Fourier series,
$$cos(alpha x)=a_0/2+sum_{n=1}^infty a_ncos(nx) tag1$$
for $xin [-pi/pi]$. The Fourier coefficients in $(1)$ are given by
$$begin{align}
a_n&=frac{2}{pi}int_0^pi cos(alpha x)cos(nx),dx\\
&=frac1pi (-1)^n sin(pi alpha)left(frac{1}{alpha +n}+frac{1}{alpha -n}right)tag2
end{align}$$
Substituting $2$ into $1$, dividing by $sin(pi y)$, and setting $x=0$ n $(2)$ reveals
$$begin{align}
pi csc(pi alpha)&=frac1alpha +sum_{n=1}^infty (-1)^nleft(frac{1}{alpha -n}+frac{1}{alpha +n}right)\\
&=sum_{n=-infty}^infty frac{(-1)^n}{alpha-n}tag3
end{align}$$
Now differentiating with respect to $alpha$, enforcing the substitution $alphato alpha/pi$, and dividing by $pi^2$ yields the coveted result
$$csc(alpha)cot(alpha)=sum_{n=-infty}^infty frac{(-1)^n}{(n-alpha)^2}$$
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
$endgroup$
The OP had asked, "Are there any other ways of proving it?"
We begin by writing the Fourier series,
$$cos(alpha x)=a_0/2+sum_{n=1}^infty a_ncos(nx) tag1$$
for $xin [-pi/pi]$. The Fourier coefficients in $(1)$ are given by
$$begin{align}
a_n&=frac{2}{pi}int_0^pi cos(alpha x)cos(nx),dx\\
&=frac1pi (-1)^n sin(pi alpha)left(frac{1}{alpha +n}+frac{1}{alpha -n}right)tag2
end{align}$$
Substituting $2$ into $1$, dividing by $sin(pi y)$, and setting $x=0$ n $(2)$ reveals
$$begin{align}
pi csc(pi alpha)&=frac1alpha +sum_{n=1}^infty (-1)^nleft(frac{1}{alpha -n}+frac{1}{alpha +n}right)\\
&=sum_{n=-infty}^infty frac{(-1)^n}{alpha-n}tag3
end{align}$$
Now differentiating with respect to $alpha$, enforcing the substitution $alphato alpha/pi$, and dividing by $pi^2$ yields the coveted result
$$csc(alpha)cot(alpha)=sum_{n=-infty}^infty frac{(-1)^n}{(n-alpha)^2}$$
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
answered Jan 29 at 23:39
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
add a comment |
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
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– Mark Viola
Jan 30 at 5:03
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:03
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Doing this rigorously with residues is tricky. It's easier to prove that starting from known expansions, especially 4.22.4 in the standard tables.
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– Professor Vector
Sep 29 '17 at 4:11