How to find integration of function, in form of hypergeometric function, given below?












0












$begingroup$


I would like to prove the left side to right hand side which is in form of hypergeometeric function. Looking for your hints, suggestions and solultions.



$$ alpha_{1} int_{0}^{1} (1-z)^{alpha_{1}+alpha_{2}-1}(1+z)^{alpha_{2}}dz= frac{1}{alpha_{1}+alpha_{2}} {}_2F_1(1-alpha_{2};2+alpha_{1}+alpha_{2}:-1) $$
where $ 2F1(a_{1},a_{2};b_{1};x)= sum_{i=0}^{infty} frac{(a_{1})_{i}(a_{2})_{i}}{(b_{1})_{i}} x_{i}$ is a hypergeometric function



Regards






integration special-functions hypergeometric-function beta-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
    $endgroup$
    – Nosrati
    Jan 5 at 11:57










  • $begingroup$
    Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
    $endgroup$
    – J.H
    Jan 5 at 17:39










  • $begingroup$
    using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
    $endgroup$
    – Nosrati
    Jan 5 at 19:32










  • $begingroup$
    @Nosrati kindly share reference for binomial expansion.
    $endgroup$
    – J.H
    Jan 13 at 11:17


















0












$begingroup$


I would like to prove the left side to right hand side which is in form of hypergeometeric function. Looking for your hints, suggestions and solultions.



$$ alpha_{1} int_{0}^{1} (1-z)^{alpha_{1}+alpha_{2}-1}(1+z)^{alpha_{2}}dz= frac{1}{alpha_{1}+alpha_{2}} {}_2F_1(1-alpha_{2};2+alpha_{1}+alpha_{2}:-1) $$
where $ 2F1(a_{1},a_{2};b_{1};x)= sum_{i=0}^{infty} frac{(a_{1})_{i}(a_{2})_{i}}{(b_{1})_{i}} x_{i}$ is a hypergeometric function



Regards






integration special-functions hypergeometric-function beta-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
    $endgroup$
    – Nosrati
    Jan 5 at 11:57










  • $begingroup$
    Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
    $endgroup$
    – J.H
    Jan 5 at 17:39










  • $begingroup$
    using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
    $endgroup$
    – Nosrati
    Jan 5 at 19:32










  • $begingroup$
    @Nosrati kindly share reference for binomial expansion.
    $endgroup$
    – J.H
    Jan 13 at 11:17
















0












0








0





$begingroup$


I would like to prove the left side to right hand side which is in form of hypergeometeric function. Looking for your hints, suggestions and solultions.



$$ alpha_{1} int_{0}^{1} (1-z)^{alpha_{1}+alpha_{2}-1}(1+z)^{alpha_{2}}dz= frac{1}{alpha_{1}+alpha_{2}} {}_2F_1(1-alpha_{2};2+alpha_{1}+alpha_{2}:-1) $$
where $ 2F1(a_{1},a_{2};b_{1};x)= sum_{i=0}^{infty} frac{(a_{1})_{i}(a_{2})_{i}}{(b_{1})_{i}} x_{i}$ is a hypergeometric function



Regards






integration special-functions hypergeometric-function beta-function







share|cite|improve this question











$endgroup$




I would like to prove the left side to right hand side which is in form of hypergeometeric function. Looking for your hints, suggestions and solultions.



$$ alpha_{1} int_{0}^{1} (1-z)^{alpha_{1}+alpha_{2}-1}(1+z)^{alpha_{2}}dz= frac{1}{alpha_{1}+alpha_{2}} {}_2F_1(1-alpha_{2};2+alpha_{1}+alpha_{2}:-1) $$
where $ 2F1(a_{1},a_{2};b_{1};x)= sum_{i=0}^{infty} frac{(a_{1})_{i}(a_{2})_{i}}{(b_{1})_{i}} x_{i}$ is a hypergeometric function



Regards






integration special-functions hypergeometric-function beta-function




integration special-functions hypergeometric-function beta-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 11:58









Nosrati

26.6k62354




26.6k62354










asked Jan 5 at 10:18









J.HJ.H

12




12








  • 1




    $begingroup$
    One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
    $endgroup$
    – Nosrati
    Jan 5 at 11:57










  • $begingroup$
    Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
    $endgroup$
    – J.H
    Jan 5 at 17:39










  • $begingroup$
    using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
    $endgroup$
    – Nosrati
    Jan 5 at 19:32










  • $begingroup$
    @Nosrati kindly share reference for binomial expansion.
    $endgroup$
    – J.H
    Jan 13 at 11:17
















  • 1




    $begingroup$
    One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
    $endgroup$
    – Nosrati
    Jan 5 at 11:57










  • $begingroup$
    Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
    $endgroup$
    – J.H
    Jan 5 at 17:39










  • $begingroup$
    using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
    $endgroup$
    – Nosrati
    Jan 5 at 19:32










  • $begingroup$
    @Nosrati kindly share reference for binomial expansion.
    $endgroup$
    – J.H
    Jan 13 at 11:17










1




1




$begingroup$
One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
$endgroup$
– Nosrati
Jan 5 at 11:57




$begingroup$
One more argument need in ${}_2F_1(1-alpha_2;2+alpha_1+alpha_2:-1)$
$endgroup$
– Nosrati
Jan 5 at 11:57












$begingroup$
Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
$endgroup$
– J.H
Jan 5 at 17:39




$begingroup$
Dear Nosrat: It seems like that b1 is missing. May you help to get Right hand side?
$endgroup$
– J.H
Jan 5 at 17:39












$begingroup$
using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
$endgroup$
– Nosrati
Jan 5 at 19:32




$begingroup$
using binomial expansion of $(1+z)^{alpha_2}=sum_{k=0}^infty frac{Gamma(.)Gamma(.)}{Gamma(.)} (-z)^k$ and with changing the order of summation and integral, find the expression as terms of beta function.
$endgroup$
– Nosrati
Jan 5 at 19:32












$begingroup$
@Nosrati kindly share reference for binomial expansion.
$endgroup$
– J.H
Jan 13 at 11:17






$begingroup$
@Nosrati kindly share reference for binomial expansion.
$endgroup$
– J.H
Jan 13 at 11:17












1 Answer
1






active

oldest

votes


















0












$begingroup$

Using binomial expansion of $(1+z)^{alpha_2}$ and with changing the order of summation and integral, we find



begin{align}
int_0^1(1-z)^{a+b-1}(1+z)^b dz
&= int_0^1(1-z)^{a+b-1}sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-z)^k}{Gamma(1+k)} dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}int_0^1(1-z)^{a+b-1}z^k dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}{bf B}(a+b,k+1)\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{Gamma(a+b)Gamma(1+k)}{Gamma(1+k)Gamma(a+b+1+k)} (-1)^k\
&= dfrac{1}{a+b}sum_{k=0}^{infty}dfrac{Gamma(-b+k)Gamma(k+1)Gamma(a+b+1)}{Gamma(-b)Gamma(1)Gamma(a+b+1+k)} dfrac{(-1)^k}{k!}\
&= dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1)
end{align}
where ${bf B}(.,.)$ is beta function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you sharing your idea and answer. It seems that original result might have some error.
    $endgroup$
    – J.H
    Jan 13 at 10:44











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Using binomial expansion of $(1+z)^{alpha_2}$ and with changing the order of summation and integral, we find



begin{align}
int_0^1(1-z)^{a+b-1}(1+z)^b dz
&= int_0^1(1-z)^{a+b-1}sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-z)^k}{Gamma(1+k)} dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}int_0^1(1-z)^{a+b-1}z^k dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}{bf B}(a+b,k+1)\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{Gamma(a+b)Gamma(1+k)}{Gamma(1+k)Gamma(a+b+1+k)} (-1)^k\
&= dfrac{1}{a+b}sum_{k=0}^{infty}dfrac{Gamma(-b+k)Gamma(k+1)Gamma(a+b+1)}{Gamma(-b)Gamma(1)Gamma(a+b+1+k)} dfrac{(-1)^k}{k!}\
&= dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1)
end{align}
where ${bf B}(.,.)$ is beta function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you sharing your idea and answer. It seems that original result might have some error.
    $endgroup$
    – J.H
    Jan 13 at 10:44
















0












$begingroup$

Using binomial expansion of $(1+z)^{alpha_2}$ and with changing the order of summation and integral, we find



begin{align}
int_0^1(1-z)^{a+b-1}(1+z)^b dz
&= int_0^1(1-z)^{a+b-1}sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-z)^k}{Gamma(1+k)} dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}int_0^1(1-z)^{a+b-1}z^k dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}{bf B}(a+b,k+1)\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{Gamma(a+b)Gamma(1+k)}{Gamma(1+k)Gamma(a+b+1+k)} (-1)^k\
&= dfrac{1}{a+b}sum_{k=0}^{infty}dfrac{Gamma(-b+k)Gamma(k+1)Gamma(a+b+1)}{Gamma(-b)Gamma(1)Gamma(a+b+1+k)} dfrac{(-1)^k}{k!}\
&= dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1)
end{align}
where ${bf B}(.,.)$ is beta function.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you sharing your idea and answer. It seems that original result might have some error.
    $endgroup$
    – J.H
    Jan 13 at 10:44














0












0








0





$begingroup$

Using binomial expansion of $(1+z)^{alpha_2}$ and with changing the order of summation and integral, we find



begin{align}
int_0^1(1-z)^{a+b-1}(1+z)^b dz
&= int_0^1(1-z)^{a+b-1}sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-z)^k}{Gamma(1+k)} dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}int_0^1(1-z)^{a+b-1}z^k dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}{bf B}(a+b,k+1)\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{Gamma(a+b)Gamma(1+k)}{Gamma(1+k)Gamma(a+b+1+k)} (-1)^k\
&= dfrac{1}{a+b}sum_{k=0}^{infty}dfrac{Gamma(-b+k)Gamma(k+1)Gamma(a+b+1)}{Gamma(-b)Gamma(1)Gamma(a+b+1+k)} dfrac{(-1)^k}{k!}\
&= dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1)
end{align}
where ${bf B}(.,.)$ is beta function.






share|cite|improve this answer









$endgroup$



Using binomial expansion of $(1+z)^{alpha_2}$ and with changing the order of summation and integral, we find



begin{align}
int_0^1(1-z)^{a+b-1}(1+z)^b dz
&= int_0^1(1-z)^{a+b-1}sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-z)^k}{Gamma(1+k)} dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}int_0^1(1-z)^{a+b-1}z^k dz\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{(-1)^k}{Gamma(1+k)}{bf B}(a+b,k+1)\
&= sum_{k=0}^{infty}dfrac{Gamma(-b+k)}{Gamma(-b)}dfrac{Gamma(a+b)Gamma(1+k)}{Gamma(1+k)Gamma(a+b+1+k)} (-1)^k\
&= dfrac{1}{a+b}sum_{k=0}^{infty}dfrac{Gamma(-b+k)Gamma(k+1)Gamma(a+b+1)}{Gamma(-b)Gamma(1)Gamma(a+b+1+k)} dfrac{(-1)^k}{k!}\
&= dfrac{1}{a+b} {}_2F_1(-b,1,a+b+1;-1)
end{align}
where ${bf B}(.,.)$ is beta function.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 20:25









NosratiNosrati

26.6k62354




26.6k62354












  • $begingroup$
    Thank you sharing your idea and answer. It seems that original result might have some error.
    $endgroup$
    – J.H
    Jan 13 at 10:44


















  • $begingroup$
    Thank you sharing your idea and answer. It seems that original result might have some error.
    $endgroup$
    – J.H
    Jan 13 at 10:44
















$begingroup$
Thank you sharing your idea and answer. It seems that original result might have some error.
$endgroup$
– J.H
Jan 13 at 10:44




$begingroup$
Thank you sharing your idea and answer. It seems that original result might have some error.
$endgroup$
– J.H
Jan 13 at 10:44


















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