# Integral

#### sbhatnagar

##### Active member
Evaluate: $$\dfrac{\displaystyle (5050)\int_{0}^{1}(1-x^{50})^{100} dx}{\displaystyle \int_0^1 (1-x^{50})^{101} dx}$$

Hint:
Integration by Parts

#### MarkFL

Staff member
Interesting problem!

Here is a related conjecture I wish to prove when I have more time later:

$\displaystyle \frac{2n(2n+1)}{2}\cdot\frac{\int_0^1(1-x^n)^{2n}\,dx}{\int_0^1(1-x^n)^{2n+1}\,dx}=\frac{2n(2n+1)}{2}+1$ where $\displaystyle n\in\mathbb{N}$

#### sbhatnagar

##### Active member
Let $I_1 = \int_{0}^1(1-x^{50})^{100}dx$ and $I_2 = \int_{0}^1(1-x^{50})^{101}dx$.

\begin{align*} I_2 &= \int_{0}^1 (1-x^{50})^{101}dx \\ I_2 &= \left[x(1-x^{50})^{101}\right]_0^1-\int (x)(101)(-50x^{49})(1-x^{50})^{100}dx \quad \text{(Integration by Parts)}\\ I_2 &=0+(5050)\int x^{50}(1-x^{50})^{100}dx \\ I_2 &=-(5050)\int_0^1 (1-x^{50}-1)(1-x^{50})^{100}dx \\ I_2 &= -5050\int_{0}^1 (1-x^{50})^{101}dx+5050\int_0^1 (1-x^{50})^{100}dx \\ I_2 &= -5050I_2+5051I_1 \\ \frac{5050I_1}{I_2} &=5051\end{align*}

#### MarkFL

Staff member

Let:

$\displaystyle I_1=\int_0^1(1-x^n)^{2n}\,dx$

$\displaystyle I_2=\int_0^1(1-x^n)^{2n+1}\,dx$

where $\displaystyle n\in\mathbb{N}$

Using integration by parts, we find:

$\displaystyle I_2=\left[x(1-x^n)^{2n+1} \right]_0^1+n(2n+1)\int_0^1x^n(1-x^n)^{2n}\,dx$

$\displaystyle I_2=0-n(2n+1)\int_0^1(1-x^n-1)(1-x^n)^{2n}\,dx$

$\displaystyle I_2=n(2n+1)(I_1-I_2)$

$\displaystyle n(2n+1)I_1=\left(n(2n+1)+1 \right)I_2$

Hence:

$\displaystyle n(2n+1)\cdot\frac{\int_0^1(1-x^n)^{2n}\,dx}{\int_0^1(1-x^n)^{2n+1}\,dx}=n(2n+1)+1$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\dfrac{\displaystyle \int_{0}^{1}(1-x^{n})^{2n} dx}{\displaystyle \int_0^1 (1-x^{n})^{2n+1} dx}$$

Here we can use the beta function but first we need to do a t-sub :

$$\text{Let : } x^n = t \,\,\, x = \sqrt [n]{t}\,\,\, \Rightarrow \,\, dx = \frac{1}{n}t^{\frac{1}{n}-1}$$

$$\text{So the integrand becomes : } \frac{\int^1_0 \,t^{\frac{1}{n}-1}\, (1-t)^{2n} \, dt}{\int^1_0 \,t^{\frac{1}{n}-1}\, (1-t)^{2n+1}\, dt}$$

Now using beta function :
$$I =\frac{\beta{(\frac{1}{n}, 2n+1)}}{\beta{(\frac{1}{n},2n+2)}}=\, \frac{\frac{\Gamma{(\frac{1}{n})\Gamma{(2n+1)}} }{\Gamma{(\frac{1}{n}+2n+1)}}}{\frac{\Gamma{(\frac{1}{n})\Gamma{(2n+2)}} }{\Gamma{(\frac{1}{n}+2n+2)}}} = \frac{\Gamma{(\frac{1}{n}+2n+2)}\Gamma{(2n+1)}}{ \Gamma {(\frac{1}{n}+2n+1)} \Gamma {(2n+2)}}= \frac {1}{n(2n+1)}+1$$