Evaluating $\int_0^1 (1-x^2)^n dx$

Question

In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

Answer 1

I assume integration is done between $0$ and $1$.

1. For $n\in\mathbb{N}$, let's have $I_n=\int_{0}^1(1-x^2)^{n}\,dx$.

Then for any $n\in\mathbb{N}$,

$$I_{n+1}=\int_{0}^1 (1-x^2)(1-x^2)^{n}\,dx=I_n-\int_{0}^1 x^2(1-x^2)^{n}\,dx$$

Integratin par parts :

$$\int_{0}^1 x^2(1-x^2)^{n}\,dx=\int_{0}^1 x\times x(1-x^2)^{n}\,dx=[x\times\frac{-(1-x^2)^{n+1}}{2n+2}]_0^1-\int_{0}^{1}\frac{-1}{2n+2}(1-x^2)^{n+1}dx$$

Therefore $$\int_{0}^1 x^2(1-x^2)^{n}\,dx=\frac{1}{2(n+1)}I_{n+1}$$

Going back to the first equality :

$$I_{n+1}=I_n-\frac{1}{2n+2}I_{n+1}$$

Reordering everything :

$$I_{n+1}=\frac{2(n+1)}{2n+3}I_n$$

2. Since $I_0=1$, the former equality implies $I_n\neq 0$ for all $n\in\mathbb{N}$. Let $n$ be a non-negative integer.

Then

$$\frac{I_n}{I_0}=\prod_{d=0}^{n-1}\frac{I_{d+1}}{I_d}=\prod_{k=0}^{n-1}\frac{2d+2}{2d+3}=\prod_{k=0}^{n-1}\frac{2d+2}{2d+3}\frac{2d+2}{2d+2}=\frac{2^{2n}(n-1+1)!^2}{(2(n-1)+3)!}=\frac{2^{2n}n!^2}{(2n+1)!}$$

Answer 2

Beta Integral $$ \begin{align} \int_0^1\left(1-t^2\right)^n\mathrm{d}t &=\frac12\int_0^1\left(1-t\right)^nt^{-1/2}\mathrm{d}t\tag{1a}\\ &=\frac{\Gamma(n+1)\Gamma(1/2)}{2\Gamma(n+3/2)}\tag{1b}\\ &=2^{2n}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\tag{1c}\\ &=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: substitute $t\mapsto t^{1/2}$
$\text{(1b)}$: Beta Integral
$\text{(1c)}$: Gauss' Multiplication Formula
$\text{(1d)}$: convert to Binomial Coefficients via Factorials

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Integration by Parts $$ \begin{align} \int_0^1\left(1-t^2\right)^n\mathrm{d}t &=\int_0^12nt^2\left(1-t^2\right)^{n-1}\,\mathrm{d}t\tag{2a}\\ &=\frac{2n}{2n+1}\int_0^1\left(1-t^2\right)^{n-1}\mathrm{d}t\tag{2b}\\ &=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{2c}\\ &=\color{#090}{\frac1{2n+1}\frac{2^nn!}{(2n)!}}\color{#C00}{2^nn!}\tag{2d}\\[3pt] &=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: integrate by parts
$\text{(2b)}$: add $\frac{2n}{2n+1}$ of the left side of $\text{(2a)}$ to $\frac1{2n+1}$ of the right side
$\text{(2c)}$: induction
$\text{(2d)}$: evaluate the products
$\text{(2e)}$: simplify

Answer 3

$$I(n)=\int_0^1(1-x^2)^ndx$$ $u=x^2\Rightarrow dx=\frac{du}{2x}=\frac{du}{2u^{1/2}}$ and so: $$I(n)=\frac12\int_0^1(1-u)^nu^{-1/2}du=\frac12 B\left(\frac12,n+1\right)=\frac{\Gamma\left(\frac12\right)\Gamma(n+1)}{2\Gamma\left(n+\frac32\right)}$$ Now we can use the identity that: $$\Gamma\left(\frac12+(n+1)\right)=\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt{\pi}$$ $$\Gamma\left(\frac12\right)=\sqrt{\pi}$$ and so: $$\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)}=\frac{4^{n+1}(n+1)!n!}{2(2n+2)!}=\frac{2^{2n+1}(n+1)(n!)^2}{(2(n+1))!}$$ now we can use the fact that: $$(2n+2)!=(2n+2)(2n+1)!=2(n+1)(2n+1)!$$ and so we arrive at: $$\frac{2^{2n}(n!)^2}{(2n+1)!}$$

Answer 4

An idea: substitution $\;t=\sin x\,,\,\,dt=\cos x\,dx\;$ , so

$$\color{red}{I_n}:=\int_0^1(1-t^2)^ndt=\int_0^{\pi/2}(\cos x)^{2n+1}dx=\int_0^1\left(1-\sin^2x\right)(\cos x)^{2n-1} dx=$$

$$=I_{n-1}-\int_0^{\pi/2}\sin^2x(\cos x)^{2n-1}dx\stackrel{\text{by parts:}\\u=\sin x,\,v'=\sin x(\cos x)^{2n-1}}=\color{red}{I_{n-1}}+\overbrace{\left.\frac{\sin x(\cos x)^{2n}}{2n}\right|_0^1}^{=0}-\color{red}{\frac1{2n}I_n}$$

$$\implies I_n=\frac{2n}{2n+1}I_{n-1}$$

For $\;n=0\;$ we get

$$I_0=\int_0^1(1-t^2)^0dt=\int_0^1dt=1=\frac{2^0(0!)^2}{1!}\;\;\checkmark$$

Now induction...and the result follows at once:

$$I_n=\frac{2n}{2n+1}I_{n-1}\stackrel{\text{Ind. Hyp.}}=\frac{(2n)^2\cdot2^{2n-2}((n-1)!)^2}{(2n-1)!\cdot2n\cdot(2n+1)}=\frac{2^{2n}(n!)^2}{(2n+1)!}$$

Answer 5

Integrate by parts

$$I_n= \int_0^1 (1-x^2)^n dx=\int_0^1 \frac{(1-x^2)^n}{x^{2n}}d(\frac{x^{2n+1}}{2n+1})=\frac{2n}{2n+1}I_{n-1},\>\>\>I_0=1 $$ which leads to $I_n= \frac{2^{2n}(n!)^2}{(2n + 1)!}$.

Answer 6

My idea here is to use the Beta function:

$$ \int_0^1 t^{m}(1 - t)^n \;dt = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}. $$

Where the Gamma function satisfies $\Gamma(n + 1) = n!$ if $n$ is an integer, and also $\Gamma$ satisfies the recurrence $\Gamma(n + 1) = n\Gamma(n)$.

So if we substitute $t = x^2$, $dt = 2x\; dx \iff \frac{1}{2\sqrt{t}}\;dt=dx$ then

\begin{align} \int_0^1 (1 - x^2)^n \; dx &= \frac{1}{2} \int_0^1 t^{-1/2} (1 - t)^n \;dt \\ &= \frac{1}{2} \frac{\Gamma(\frac12)n!}{\Gamma(n+\frac32)}. \tag{1} \end{align}

And now using the recurrence $\Gamma(n + 1) = n\Gamma(n)$, we find that

\begin{align} \Gamma(n+\tfrac32) &= (n+\tfrac12)\Gamma(n + \tfrac12) \\ &= (n+\tfrac12)(n - \tfrac12) \Gamma(n - \tfrac12) \\ &\hspace{2.5mm}\vdots \\ &= (n+\tfrac12)(n - \tfrac12) \cdots \tfrac{5}{2} \cdot \tfrac{3}{2} \cdot \tfrac12 \cdot \Gamma(\tfrac12) \\ &= \frac{1}{2^n} (2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 \cdot \Gamma(\tfrac12) \tag{2} \end{align}

The product $(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1$ is also known as the double factorial $(2n + 1)!!$. It is well-known that we can rewrite it in terms of $(2n + 1)!$ as

\begin{align} (2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 &= \frac{(2n + 1)(2n)(2n - 1)(2n - 2) \cdots 3 \cdot 2 \cdot 1}{(2n)(2n-2) \cdots 2} \\ &= \frac{(2n+1)!}{2^n \cdot n!}. \tag{3} \end{align}

So now finally, if you combine $(1), (2), (3)$, you get the result you want.

Answer 7

Evaluate $f_n(x):=\int_0^x(1-t^2)^ndt$ with integration by parts viz. $u=(1-t^2)^n,\,v=t$ so$$f_n=x(1-x^2)^n+2n\int_0^xt^2(1-t^2)^{n-1}dt=x(1-x^2)^n+2n(f_{n-1}-f_{n+1}).$$Definite integration on $[0,\,1]$ simplifies this to $f_{n+1}=f_{n-1}-f_n/(2n)$, making b) tractable by induction.

Answer 8

Part b) can be obtained just by induction. The result is immediate for $n=1$ and \begin{align*} \int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\ = & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\ = & \frac{2^{2n}(n!)^2}{(2n+1)!}-\frac{1}{2n+2}\int_0^1(1-x^2)^{n+1} dx \end{align*}

This implies that $$ \int_0^1 (1-x^2)^{n+1}dx = \frac{2n+2}{2n+3} \frac{2^{2n}(n!)^2}{(2n+1)!}=\frac{(2n+2)^2 2^{2n} (n!)^2}{(2n+1)!(2n+2)(2n+3)}=\dfrac{2^{2n+2}((n+1)!)^2}{(2n+3)!}. $$

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\begin{align*} \int_0^1 x (-2x) (1-x^2)^n dx = &\left[x \dfrac{(1-x^2)^{n+1}}{n+1} \right]_0^1 - \frac{1}{n+1}\int_0^1 (1-x^2)^{n+1} \end{align*}

Answer 9

Here is an approach based on your attempt using a series

$$F\left( n \right)=\int\limits_{0}^{1}{{{\left( 1-{{x}^{2}} \right)}^{n}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right){{\left( -1 \right)}^{k}}}\int\limits_{0}^{1}{{{x}^{2k}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}}{2k+1}}$$ Now note $$\left( \begin{matrix} n \\ k \\ \end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$ And so $$F\left( n \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{2k+1}\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}}=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\frac{{{\left( 1+z \right)}^{n}}\arctan \left( {{z}^{-1/2}} \right)}{\sqrt{z}}dz}$$

However $$\arctan \left( z \right)=\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{{{z}^{2k+1}}}{{{\left( 1+{{z}^{2}} \right)}^{k+1}}}}$$ which yields $$F\left( n \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{1}{{{\left( z+1 \right)}^{k-n+1}}}}dz}$$ and the residue therefore provides the result $$F\left( n \right)=\frac{{{2}^{2n}}{{\left( n! \right)}^{2}}}{\left( 2n+1 \right)!}$$