Value of ratio of $2$ definite Integration












-1












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Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$




Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$



put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$



So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$



Now i am struck here , I did not understand how to solve it



Could some help me to solve it, Thanks






definite-integrals







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  • 2




    $begingroup$
    Bhai beta function ki property check kar le wikipedia se. Best of luck.
    $endgroup$
    – NewBornMATH
    Dec 11 '18 at 15:24










  • $begingroup$
    bhai tu sahi bola ye beta baba hee hain
    $endgroup$
    – deleteprofile
    Dec 11 '18 at 15:41
















-1












$begingroup$



Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$




Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$



put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$



So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$



Now i am struck here , I did not understand how to solve it



Could some help me to solve it, Thanks






definite-integrals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Bhai beta function ki property check kar le wikipedia se. Best of luck.
    $endgroup$
    – NewBornMATH
    Dec 11 '18 at 15:24










  • $begingroup$
    bhai tu sahi bola ye beta baba hee hain
    $endgroup$
    – deleteprofile
    Dec 11 '18 at 15:41














-1












-1








-1





$begingroup$



Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$




Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$



put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$



So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$



Now i am struck here , I did not understand how to solve it



Could some help me to solve it, Thanks






definite-integrals







share|cite|improve this question









$endgroup$





Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$




Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$



put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$



So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$



Now i am struck here , I did not understand how to solve it



Could some help me to solve it, Thanks






definite-integrals




definite-integrals






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asked Dec 11 '18 at 15:02









DXTDXT

5,6742630




5,6742630








  • 2




    $begingroup$
    Bhai beta function ki property check kar le wikipedia se. Best of luck.
    $endgroup$
    – NewBornMATH
    Dec 11 '18 at 15:24










  • $begingroup$
    bhai tu sahi bola ye beta baba hee hain
    $endgroup$
    – deleteprofile
    Dec 11 '18 at 15:41














  • 2




    $begingroup$
    Bhai beta function ki property check kar le wikipedia se. Best of luck.
    $endgroup$
    – NewBornMATH
    Dec 11 '18 at 15:24










  • $begingroup$
    bhai tu sahi bola ye beta baba hee hain
    $endgroup$
    – deleteprofile
    Dec 11 '18 at 15:41








2




2




$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24




$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24












$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41




$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41










1 Answer
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$begingroup$

With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    With $a = 1004$ and $b = 2010$ and $B$ the beta function,
    $$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
    J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
    }
    $$
    and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
    In your case I get
    $$ 2^{2010} I/J = 4020$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      With $a = 1004$ and $b = 2010$ and $B$ the beta function,
      $$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
      J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
      }
      $$
      and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
      In your case I get
      $$ 2^{2010} I/J = 4020$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        With $a = 1004$ and $b = 2010$ and $B$ the beta function,
        $$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
        J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
        }
        $$
        and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
        In your case I get
        $$ 2^{2010} I/J = 4020$$






        share|cite|improve this answer









        $endgroup$



        With $a = 1004$ and $b = 2010$ and $B$ the beta function,
        $$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
        J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
        }
        $$
        and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
        In your case I get
        $$ 2^{2010} I/J = 4020$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 15:15









        Robert IsraelRobert Israel

        321k23211465




        321k23211465






























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