$\displaystyle \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx} =$ $ ?$

I tried putting $x=sin(a)$ but I could do nothing about the 100 and 101 powers, they made the integration not solvable for me.

How should I do or approach this question?

marked as duplicate by StubbornAtom, Lord Shark the Unknown, amWhy integration Oct 12 at 15:39

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  • Could you put some form of the integral in your question title for potential question disambiguation, please? – Cœur Oct 12 at 6:33

$ \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx}$

Let $\displaystyle I_n = \int_{0}^{1} {(1-x^m)^n} dx$

$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)^{n+1}}dx$

$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)(1-x^m)^n} dx$

$\implies \displaystyle I_{n+1} = \int_{0}^{1} {(1-x^m)^n} dx- \int_{0}^{1}{x^m(1-x^m)^n} dx$ ...[*]

$\implies \displaystyle I_{n+1} = I_n- \int_{0}^{1}{x^m(1-x^m)^n} dx$

We will integrate $\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx$ via integration by parts. Watch closely, this is a little tricky.

$\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx=\int_{0}^{1}{x\cdot x^{m-1}(1-x^m)^n}dx$

Let $u = x \implies du = dx$

And $dv = x^{m-1}(1-x^m)^n dx$

Let $y = (1-x^m) \implies dy = -mx^{m-1} dx \implies x^{m-1} dx = -\dfrac{dy}{m}$

$\displaystyle v = \int{x^{m-1}(1-x^m)^n} dx = \int{-\dfrac{y^n}{m}}dy = \dfrac{-y^{n+1}}{m(n+1)}=\dfrac{-(1-x^m)^{n+1}}{m(n+1)}$

$\displaystyle \int_{0}^{1}{x^m(1-x^m)^n} dx$

$=\displaystyle \left[-\dfrac{x(1-x^m)^{n+1}}{m(n+1)}\right]_{0}^{1}+\dfrac{1}{m(n+1)}\int_{0}^{1}{(1-x^m)^{n+1}}dx$

$= \dfrac{I_{n+1}}{m(n+1)}$

Substituting this result into [*]

$I_{n+1} = I_n - \dfrac{I_{n+1}}{m(n+1)}$

$\implies \left[1+\dfrac{1}{m(n+1)}\right]=\dfrac{I_n}{I_{n+1}}$

$\implies \dfrac{I_n}{I_{n+1}} = \dfrac{m(n+1)+1}{m(n+1)}$

Putting $m = 50$ and $n = 100$, we have

$\displaystyle \dfrac{\int_{0}^{1}{(1-x^{50})^{100}}dx}{\int_{0}^{1}{(1-x^{50})^{101}}dx}=\dfrac{50\times101+1}{50\times 101}=\dfrac{5051}{5050}$

  • Excuse me, user601391 is NOT my account at all. – Pradyuman Dixit yesterday

Let $a$, $b>0$. Then, substituting $t=x^a$, $$I_{a,b}= \int_0^1(1-x^a)^b\,dx=\frac1a\int_0^1(1-t)^bt^{1/a-1}\,dt=\frac{B(b+1,1/a)}{a}$$ where $B$ denotes the Beta function. But the beta function is expressible in terms of the Gamma function so that $$\frac{B(b+1,1/a)}{a}=\frac{\Gamma(1/a)\Gamma(b+1)}{a\Gamma(1/a+b+1)}.$$

Therefore $$\frac{I_{a,b}}{I_{a,b+1}}=\frac{\Gamma(b+1)\Gamma(1/a+b+2)}{\Gamma(b+2) \Gamma(1/a+b+1)}=\frac{1/a+b+1}{b+1}.$$ Now let $a=50$ and $b=100$.