Mixing
A certain series expansion for cosecant
$begingroup$
Has anyone encountered the summation
$$pi csc(pi alpha) stackrel{?}{=} sum_{k = 0}^infty binom{alpha}{k} dfrac{(-1)^k}{alpha-k}, quad alpha neq mathbb{Z}$$
somewhere, and if yes, could you give me an idea how to derive such expansion? I got the series expansion after integrating some stuff and tried Mathematica to give me hints as to how this series behaves, and it quickly gave the cosecant as the answer.
sequences-and-series
asked Jan 15 at 14:23
ErdbergErdberg
166
$endgroup$
add a comment |
$begingroup$
Has anyone encountered the summation
$$pi csc(pi alpha) stackrel{?}{=} sum_{k = 0}^infty binom{alpha}{k} dfrac{(-1)^k}{alpha-k}, quad alpha neq mathbb{Z}$$
somewhere, and if yes, could you give me an idea how to derive such expansion? I got the series expansion after integrating some stuff and tried Mathematica to give me hints as to how this series behaves, and it quickly gave the cosecant as the answer.
sequences-and-series
asked Jan 15 at 14:23
ErdbergErdberg
166
$endgroup$
$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45
add a comment |
$begingroup$
Has anyone encountered the summation
$$pi csc(pi alpha) stackrel{?}{=} sum_{k = 0}^infty binom{alpha}{k} dfrac{(-1)^k}{alpha-k}, quad alpha neq mathbb{Z}$$
somewhere, and if yes, could you give me an idea how to derive such expansion? I got the series expansion after integrating some stuff and tried Mathematica to give me hints as to how this series behaves, and it quickly gave the cosecant as the answer.
sequences-and-series
asked Jan 15 at 14:23
ErdbergErdberg
166
$endgroup$
Has anyone encountered the summation
$$pi csc(pi alpha) stackrel{?}{=} sum_{k = 0}^infty binom{alpha}{k} dfrac{(-1)^k}{alpha-k}, quad alpha neq mathbb{Z}$$
somewhere, and if yes, could you give me an idea how to derive such expansion? I got the series expansion after integrating some stuff and tried Mathematica to give me hints as to how this series behaves, and it quickly gave the cosecant as the answer.
sequences-and-series
sequences-and-series
asked Jan 15 at 14:23
ErdbergErdberg
166
asked Jan 15 at 14:23
ErdbergErdberg
166
asked Jan 15 at 14:23
ErdbergErdberg
166
asked Jan 15 at 14:23
ErdbergErdberg
166
asked Jan 15 at 14:23
ErdbergErdberg
166
166
$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45
add a comment |
$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45
$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45
add a comment |
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$begingroup$
What did you use when you said "after integrating some stuff?"
$endgroup$
– Tom Himler
Jan 15 at 14:39
$begingroup$
Originally I was trying to compute $$int_0^1 dfrac{u^alpha}{u+c},du $$ with $alpha >1$ non-integer and $c>0$. With the substitution $v = u+c$ and using the generalized binomial expansion for $(v-c)^alpha$, the result gives me a multiple of a hypergeometric function minus a multiple of the above series.
$endgroup$
– Erdberg
Jan 15 at 14:44
$begingroup$
I would not be shocked if the beta function could be used to prove this result. Let me see what I can do. It looks like it comes from the reflection formula for the Gamma function.
$endgroup$
– Tom Himler
Jan 15 at 14:45