Find Ratio of Integrals $I:J$

Question

Given: $$I=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{12}$$

and

$$J=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{(x+3)^8}$$

Find Value of $\frac{I}{J}$

My attempt:The Integral $I$ is easy to evaluate using Beta Function.

So i was trying to Manipulate $J$ to convert it to $I$ as follows:

We can write $J$ as:

$$J=\int_{0}^{1}\frac{x^{6}\left ( \frac{1}{x} -1\right )^\frac{7}{2}}{x^{8}\left ( 1+\frac{3}{x} \right )^8}$$

Now put $\frac{1}{x}=t$ we get:

$$J=\int_{1}^{\infty}\frac{(t-1)^{\frac{7}{2}}}{(1+3t)^8}$$

Using integration by Parts taking $u=(t-1)^{3.5}$ and $v=\frac{1}{(1+3t)^8}$ we get:

$$J=\frac{1}{6}\times \int_{1}^{\infty}\frac{(t-1)^{\frac{5}{2}}}{(1+3t)^7}$$

Repeating Parts again and again:

$$J=\frac{1}{432}\times \int_{1}^{\infty}\frac{\sqrt{t-1}}{(1+3t)^5}\:dt$$

Any way to proceed from here?

Answer 1

As you noted, $$I=\frac 1 {12}B\left(\frac 7 2, \frac 9 2\right)$$ Also, if $|z|<1$ and $\alpha \in \mathbb R$ $$\frac 1 {(1+z)^\alpha} = \sum_{n\geq 0}\frac{(-\alpha)(-\alpha-1)...(-\alpha-n+1)}{n!}z^n\tag{1}$$ So for $\alpha =8$ $$\begin{split} J &= \frac 1 {3^8}\int_0^1\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{\left(1+\frac x 3\right)^8}\\ &=\frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}\int_0^1x^{\frac{5+n}{2}}(1-x)^{\frac{7}{2}}dx\\ &= \frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}B \left(\frac 7 2 +n,\frac 9 2\right) \end{split}$$ Now, because $$B(a+1, b) = B(a,b)\frac a {a+b}$$ we also have $$B(a+n, b)=B(a,b)\cdot\frac{a}{a+b}\cdot\frac{a+1}{a+b+1}...\frac{a+n-1}{a+b+n-1}$$ Also note that $\frac 7 2 +\frac 9 2 = 8$. Therefore $$B\left(\frac 7 2+n, \frac 9 2 \right)=B\left(\frac 7 2,\frac 9 2\right)\cdot\frac{\frac 7 2}{8}\cdot\frac{\frac 7 2+1}{9}...\frac{\frac 7 2+n-1}{8+n-1}$$ We now have $$\begin{split} J &=B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^8}\sum_{n\geq 0}\frac{(-\frac 7 2)(-\frac 7 2 - 1)...(-\frac 7 2-n+1)}{n!}\frac {1} {3^n}\\ &= B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^8}\frac 1 {(1+\frac 1 3)^{\frac 7 2}} \,\,\,\,\text{ (using (1))}\\ &=B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^{\frac 9 2}4^{\frac 7 2}} \end{split}$$

Conclusion: $$\boxed{\frac I J = \frac {3^{\frac 9 2}4^{\frac 7 2}}{12}=3^{\frac 7 2}4^{\frac 5 2}}$$

Answer 2

Well, substitute $\sqrt{t-1}=z$

then,

$\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{t-1}}dt=dz$

Now multiply the J by $\dfrac{\sqrt{t-1}}{\sqrt{t-1}}$

$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{\sqrt{t-1}}{(1+3t)^{5}}\cdot\dfrac{\sqrt{t-1}}{2\sqrt{t-1}}dt$

$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{t-1}{(1+3t)^{5}}\cdot\dfrac{1}{2\sqrt{t-1}}dt$

Now, substitute $t=z^{2}+1$ and $\dfrac{1}{2\sqrt{t-1}}dt=dz$

$J=\dfrac{1}{216}\cdot\int^{\infty}_{0}\dfrac{z^{2}}{(3z^{2}+4)}dz$

Now use Integration by parts.

Hint:

$\dfrac{d}{dx}[x-tan^{-1}(x)]=\dfrac{x^{2}}{x^{2}+1}$