Gamma reflection formula proof












3












$begingroup$


For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$



Then we have
$$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
Is this a correct way to prove the famous Euler's reflection formula ?






proof-verification special-functions gamma-function







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$endgroup$

















    3












    $begingroup$


    For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
    and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$



    Then we have
    $$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
    Is this a correct way to prove the famous Euler's reflection formula ?






    proof-verification special-functions gamma-function







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
      and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$



      Then we have
      $$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
      Is this a correct way to prove the famous Euler's reflection formula ?






      proof-verification special-functions gamma-function







      share|cite|improve this question











      $endgroup$




      For $xinmathbb{C}setminus0,-1,...$ If we know that $$Gamma(x)=lim_{nto +infty} frac{n^x.n!}{prod_{k=0}^n(k+x)}$$
      and $$sin(pi x)=pi xprod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)$$



      Then we have
      $$ Gamma(x)Gamma(1-x)=lim_{nto +infty} frac{n^x.n!.n^{1-x}.n!}{prod_{k=0}^n(k+x)(k+1-x)} =frac1xlim_{nto +infty} frac{n}{n+1-x}prod_{k=1}^nfrac{k^2}{k^2-x^2}=$$$$=frac1xleft(prod_{k=1}^{infty}left(1- frac{x^2}{k^2}right)right)^{-1}=frac{pi}{sin(pi x)} $$
      Is this a correct way to prove the famous Euler's reflection formula ?






      proof-verification special-functions gamma-function




      proof-verification special-functions gamma-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 1:46









      Yadati Kiran

      2,1261622




      2,1261622










      asked Dec 16 '17 at 4:38









      KamouloxKamoulox

      185




      185






















          1 Answer
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          2












          $begingroup$

          Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My bad.. Thanks!
            $endgroup$
            – Kamoulox
            Dec 16 '17 at 4:49










          • $begingroup$
            would the downvoter care to comment on what is wrong with this answer?
            $endgroup$
            – robjohn♦
            Dec 16 '17 at 4:58












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          1 Answer
          1






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          active

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          2












          $begingroup$

          Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My bad.. Thanks!
            $endgroup$
            – Kamoulox
            Dec 16 '17 at 4:49










          • $begingroup$
            would the downvoter care to comment on what is wrong with this answer?
            $endgroup$
            – robjohn♦
            Dec 16 '17 at 4:58
















          2












          $begingroup$

          Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My bad.. Thanks!
            $endgroup$
            – Kamoulox
            Dec 16 '17 at 4:49










          • $begingroup$
            would the downvoter care to comment on what is wrong with this answer?
            $endgroup$
            – robjohn♦
            Dec 16 '17 at 4:58














          2












          2








          2





          $begingroup$

          Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!






          share|cite|improve this answer









          $endgroup$



          Other than the fact that the formula for $$sin pi x = pi x , , prod_{k=1}^{infty} left(1-frac{x^2}{k^2}right)$$ should be corrected, I think your proof is fine!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '17 at 4:43









          RohanRohan

          27.8k42444




          27.8k42444












          • $begingroup$
            My bad.. Thanks!
            $endgroup$
            – Kamoulox
            Dec 16 '17 at 4:49










          • $begingroup$
            would the downvoter care to comment on what is wrong with this answer?
            $endgroup$
            – robjohn♦
            Dec 16 '17 at 4:58


















          • $begingroup$
            My bad.. Thanks!
            $endgroup$
            – Kamoulox
            Dec 16 '17 at 4:49










          • $begingroup$
            would the downvoter care to comment on what is wrong with this answer?
            $endgroup$
            – robjohn♦
            Dec 16 '17 at 4:58
















          $begingroup$
          My bad.. Thanks!
          $endgroup$
          – Kamoulox
          Dec 16 '17 at 4:49




          $begingroup$
          My bad.. Thanks!
          $endgroup$
          – Kamoulox
          Dec 16 '17 at 4:49












          $begingroup$
          would the downvoter care to comment on what is wrong with this answer?
          $endgroup$
          – robjohn♦
          Dec 16 '17 at 4:58




          $begingroup$
          would the downvoter care to comment on what is wrong with this answer?
          $endgroup$
          – robjohn♦
          Dec 16 '17 at 4:58


















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