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If $\displaystyle A = \int^{1}_{0}x^{\frac{7}{2}}(1-x)^{\frac{5}{2}}dx$ and $\displaystyle B = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(x+3)^8}dx\;,$ then value of $AB^{-1} = $

Attempt: i have tried using gamma function $\displaystyle \int^{1}_{0}x^m(1-x)^ndx = \frac{(m-1)!\cdot (n-1)!}{(m+n-1)!}$

so $\displaystyle A=\int^{1}_{0}x^{\frac{7}{2}}(1-x)^{\frac{5}{2}}dx = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx = \frac{(\frac{3}{2})!\cdot (\frac{5}{2})!}{5!}$

wan,t be able to go further, could some help me, thanks

  • 0
    What is the origin of this problem? Both integrals can be obtained in closed form. The former is straightforward; the latter is a challenge. – 2017-02-16
  • 0
    Well, $A $ is easy, what have you tried for $B $? – 2017-02-16

1 Answers 1

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We first prove a result : $F(a,b,p)=\displaystyle\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right) \\$

Proof : We start with the integral $\displaystyle \beta(a,b) = \int\limits_0^1 x^{a-1}(1-x)^{b-1}\; dx \\$

The substitution $y = \dfrac{(p+1)x}{p+x}$ is pretty straight forward and gives the result,

$\displaystyle\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right) \\$

Now the second integral is $\displaystyle \int\limits_0^1 \dfrac{x^\frac{3}{2}(1-x)^\frac{7}{2}}{(x+3)^8}\; dx \\ = \dfrac{1}{3}\left(\int\limits_0^1 \dfrac{x^\frac{3}{2}(1-x)^\frac{7}{2}}{(x+3)^8}\;(x+3-x)\; dx\right) \\= \dfrac{1}{3}\left(F(5/2,9/2,3)-F(7/2,9/2,3)\right)\\ = \dfrac{17\pi}{21233664\sqrt{3}}\\$

This is what we needed and you can carry on from here. In fact the ratio is $\dfrac{51840\sqrt{3}}{17}$