integral related to Beta function and binomial












3












$begingroup$


I have not frequented the board lately due to being very busy. But, I ran across what I think is an interesting problem.



$displaystyle int_{0}^{k}(tx+1-x)^{n}dx$, where $nin mathbb{Z}^{+}$ and $t$ is a parameter independent of $x$. $k=0,1,2,3,ldots, n$



Then, use this to show that $displaystyle int_0^k x^{k}(1-x)^{n-k} ; dx=frac{1}{binom{n}{k}(n+1)}$



I managed to integrate the former by making the sub $u=x(t-1), ; dx=frac{du}{t-1}$



This gives $displaystyle frac{1}{t-1}int_0^{t-1}(u+1)^n ; du=frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}$



This appears to be the partial sum of the geometric series
$displaystyle sum_{k=0}^n t^k=frac{t^{n+1}-1}{t-1}$



Now, how can we relate this to $frac{1}{binom{n}{k}}$?.



Since $displaystyle frac{1}{(n+1)binom{n}{k}}=B(n-k+1,k+1)$, the latter integral looks like some sort of incomplete beta function. It's that upper k limit that I am kind of wondering about.



After all, $displaystyle frac{1}{(n+1)binom{n}{k}}=frac{k!(n-k)!}{(n+1)!}=frac{Gamma(k+1)Gamma(n-k+1)}{Gamma(n+2)}=B(n-k+1,k+1)$.



I assume it is right in front of me, but how does solving the first integral, which evaluates to $displaystyle frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}=frac{1}{n+1}sum_{k=0}^n t^k$ relate to the latter Beta-type integral which apparently evaluates to $displaystyle frac{1}{binom{n}{k}(n+1)}$?.



Thank you all very much for any responses. I will be sure to leave credit and thanks to everyone. If I have not, it is simply because it was overlooked. Not out of disregard.






integration special-functions







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I have not frequented the board lately due to being very busy. But, I ran across what I think is an interesting problem.



    $displaystyle int_{0}^{k}(tx+1-x)^{n}dx$, where $nin mathbb{Z}^{+}$ and $t$ is a parameter independent of $x$. $k=0,1,2,3,ldots, n$



    Then, use this to show that $displaystyle int_0^k x^{k}(1-x)^{n-k} ; dx=frac{1}{binom{n}{k}(n+1)}$



    I managed to integrate the former by making the sub $u=x(t-1), ; dx=frac{du}{t-1}$



    This gives $displaystyle frac{1}{t-1}int_0^{t-1}(u+1)^n ; du=frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}$



    This appears to be the partial sum of the geometric series
    $displaystyle sum_{k=0}^n t^k=frac{t^{n+1}-1}{t-1}$



    Now, how can we relate this to $frac{1}{binom{n}{k}}$?.



    Since $displaystyle frac{1}{(n+1)binom{n}{k}}=B(n-k+1,k+1)$, the latter integral looks like some sort of incomplete beta function. It's that upper k limit that I am kind of wondering about.



    After all, $displaystyle frac{1}{(n+1)binom{n}{k}}=frac{k!(n-k)!}{(n+1)!}=frac{Gamma(k+1)Gamma(n-k+1)}{Gamma(n+2)}=B(n-k+1,k+1)$.



    I assume it is right in front of me, but how does solving the first integral, which evaluates to $displaystyle frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}=frac{1}{n+1}sum_{k=0}^n t^k$ relate to the latter Beta-type integral which apparently evaluates to $displaystyle frac{1}{binom{n}{k}(n+1)}$?.



    Thank you all very much for any responses. I will be sure to leave credit and thanks to everyone. If I have not, it is simply because it was overlooked. Not out of disregard.






    integration special-functions







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I have not frequented the board lately due to being very busy. But, I ran across what I think is an interesting problem.



      $displaystyle int_{0}^{k}(tx+1-x)^{n}dx$, where $nin mathbb{Z}^{+}$ and $t$ is a parameter independent of $x$. $k=0,1,2,3,ldots, n$



      Then, use this to show that $displaystyle int_0^k x^{k}(1-x)^{n-k} ; dx=frac{1}{binom{n}{k}(n+1)}$



      I managed to integrate the former by making the sub $u=x(t-1), ; dx=frac{du}{t-1}$



      This gives $displaystyle frac{1}{t-1}int_0^{t-1}(u+1)^n ; du=frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}$



      This appears to be the partial sum of the geometric series
      $displaystyle sum_{k=0}^n t^k=frac{t^{n+1}-1}{t-1}$



      Now, how can we relate this to $frac{1}{binom{n}{k}}$?.



      Since $displaystyle frac{1}{(n+1)binom{n}{k}}=B(n-k+1,k+1)$, the latter integral looks like some sort of incomplete beta function. It's that upper k limit that I am kind of wondering about.



      After all, $displaystyle frac{1}{(n+1)binom{n}{k}}=frac{k!(n-k)!}{(n+1)!}=frac{Gamma(k+1)Gamma(n-k+1)}{Gamma(n+2)}=B(n-k+1,k+1)$.



      I assume it is right in front of me, but how does solving the first integral, which evaluates to $displaystyle frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}=frac{1}{n+1}sum_{k=0}^n t^k$ relate to the latter Beta-type integral which apparently evaluates to $displaystyle frac{1}{binom{n}{k}(n+1)}$?.



      Thank you all very much for any responses. I will be sure to leave credit and thanks to everyone. If I have not, it is simply because it was overlooked. Not out of disregard.






      integration special-functions







      share|cite|improve this question











      $endgroup$




      I have not frequented the board lately due to being very busy. But, I ran across what I think is an interesting problem.



      $displaystyle int_{0}^{k}(tx+1-x)^{n}dx$, where $nin mathbb{Z}^{+}$ and $t$ is a parameter independent of $x$. $k=0,1,2,3,ldots, n$



      Then, use this to show that $displaystyle int_0^k x^{k}(1-x)^{n-k} ; dx=frac{1}{binom{n}{k}(n+1)}$



      I managed to integrate the former by making the sub $u=x(t-1), ; dx=frac{du}{t-1}$



      This gives $displaystyle frac{1}{t-1}int_0^{t-1}(u+1)^n ; du=frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}$



      This appears to be the partial sum of the geometric series
      $displaystyle sum_{k=0}^n t^k=frac{t^{n+1}-1}{t-1}$



      Now, how can we relate this to $frac{1}{binom{n}{k}}$?.



      Since $displaystyle frac{1}{(n+1)binom{n}{k}}=B(n-k+1,k+1)$, the latter integral looks like some sort of incomplete beta function. It's that upper k limit that I am kind of wondering about.



      After all, $displaystyle frac{1}{(n+1)binom{n}{k}}=frac{k!(n-k)!}{(n+1)!}=frac{Gamma(k+1)Gamma(n-k+1)}{Gamma(n+2)}=B(n-k+1,k+1)$.



      I assume it is right in front of me, but how does solving the first integral, which evaluates to $displaystyle frac{t^{n+1}-1}{t-1}cdot frac{1}{n+1}=frac{1}{n+1}sum_{k=0}^n t^k$ relate to the latter Beta-type integral which apparently evaluates to $displaystyle frac{1}{binom{n}{k}(n+1)}$?.



      Thank you all very much for any responses. I will be sure to leave credit and thanks to everyone. If I have not, it is simply because it was overlooked. Not out of disregard.






      integration special-functions




      integration special-functions






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      share|cite|improve this question













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      share|cite|improve this question








      edited Mar 12 '12 at 19:37









      Michael Hardy

      1




      1










      asked Feb 11 '12 at 17:24









      CodyCody

      8,0711974




      8,0711974






















          2 Answers
          2






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          $begingroup$

          Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            I will be evaluating
            $$I_n(t)=int_0^1(tx+1-x)^nmathrm{d}x=int_0^1(1+x(t-1))^nmathrm dx$$
            Binomial theorem:
            $$I_n(t)=int_0^1sum_{r=0}^n{nchoose r}x^r(t-1)^rmathrm dx$$
            $$I_n(t)=sum_{r=0}^n{nchoose r}(t-1)^rint_0^1x^rmathrm dx$$
            $$I_n(t)=sum_{r=0}^n{nchoose r}frac{(t-1)^r}{r+1}$$
            I'm not exactly sure what you're asking for, but hopefully this helps.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.






                  share|cite|improve this answer









                  $endgroup$



                  Nevermind. I think I have it. It would appear there is a typo on that upper limit. I think It should be a 1 instead of a k.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 11 '12 at 17:59









                  CodyCody

                  8,0711974




                  8,0711974























                      0












                      $begingroup$

                      I will be evaluating
                      $$I_n(t)=int_0^1(tx+1-x)^nmathrm{d}x=int_0^1(1+x(t-1))^nmathrm dx$$
                      Binomial theorem:
                      $$I_n(t)=int_0^1sum_{r=0}^n{nchoose r}x^r(t-1)^rmathrm dx$$
                      $$I_n(t)=sum_{r=0}^n{nchoose r}(t-1)^rint_0^1x^rmathrm dx$$
                      $$I_n(t)=sum_{r=0}^n{nchoose r}frac{(t-1)^r}{r+1}$$
                      I'm not exactly sure what you're asking for, but hopefully this helps.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        I will be evaluating
                        $$I_n(t)=int_0^1(tx+1-x)^nmathrm{d}x=int_0^1(1+x(t-1))^nmathrm dx$$
                        Binomial theorem:
                        $$I_n(t)=int_0^1sum_{r=0}^n{nchoose r}x^r(t-1)^rmathrm dx$$
                        $$I_n(t)=sum_{r=0}^n{nchoose r}(t-1)^rint_0^1x^rmathrm dx$$
                        $$I_n(t)=sum_{r=0}^n{nchoose r}frac{(t-1)^r}{r+1}$$
                        I'm not exactly sure what you're asking for, but hopefully this helps.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I will be evaluating
                          $$I_n(t)=int_0^1(tx+1-x)^nmathrm{d}x=int_0^1(1+x(t-1))^nmathrm dx$$
                          Binomial theorem:
                          $$I_n(t)=int_0^1sum_{r=0}^n{nchoose r}x^r(t-1)^rmathrm dx$$
                          $$I_n(t)=sum_{r=0}^n{nchoose r}(t-1)^rint_0^1x^rmathrm dx$$
                          $$I_n(t)=sum_{r=0}^n{nchoose r}frac{(t-1)^r}{r+1}$$
                          I'm not exactly sure what you're asking for, but hopefully this helps.






                          share|cite|improve this answer









                          $endgroup$



                          I will be evaluating
                          $$I_n(t)=int_0^1(tx+1-x)^nmathrm{d}x=int_0^1(1+x(t-1))^nmathrm dx$$
                          Binomial theorem:
                          $$I_n(t)=int_0^1sum_{r=0}^n{nchoose r}x^r(t-1)^rmathrm dx$$
                          $$I_n(t)=sum_{r=0}^n{nchoose r}(t-1)^rint_0^1x^rmathrm dx$$
                          $$I_n(t)=sum_{r=0}^n{nchoose r}frac{(t-1)^r}{r+1}$$
                          I'm not exactly sure what you're asking for, but hopefully this helps.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 1 '18 at 5:13









                          clathratusclathratus

                          3,541332




                          3,541332






























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