Find Ratio of Integrals: $\lim_{n \to\infty} I_n/I_{n-2}$

[utkarshakash]

Homework Statement

If $I_n = \displaystyle \int^1_0 x^n. \sqrt{1-x^2} dx$ then
$\lim_{n \to \infty} \dfrac{I_n}{I_{n-2}}$ is equal to

Homework Equations

The Attempt at a Solution

Integrating by parts

$x^n \displaystyle \int \sqrt{1-x^2}dx - \int nx^{n-1} \int \sqrt{1-x^2} dx$

But integrating further is useless.

 

[Millennial]

You are integrating it the wrong way. Keep in mind that $\displaystyle \lim_{n\to\infty} \frac{I_n}{I_{n-2}} = \lim_{n\to\infty} \frac{I_{n+2}}{I_{n}}$.

 

[Millennial]

Ignore the above post, I figured out it does not lead to the solution; so you have to resort to more analytic methods. Substituting $u=x^2$ in this integral gives $\displaystyle \frac{1}{2}\int^{1}_{0} u^{(n-1)/2} (1-u)^{1/2}\,\,du$, which is in the form of the general integral $\displaystyle B(a,b) = \int^{1}_{0}x^{a-1}(1-x)^{b-1}\,dx$. This function is called the Beta function and it happens to satisfy $\displaystyle B(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!}$ for integer x and y.

Can you solve it from here?

Note: You might get fractions inside factorials while solving the question. You don't have to know their values, just use the equation $(x+1)! = (x+1)x!$ to simplify the limit. The reason you will see those is that the equation for the Beta integral is true for a meromorphic Gamma function which satisfies $\Gamma(x+1) = x!$, which extends the factorial function to all complex numbers.

 

[haruspex]

Homework Statement

If $I_n = \displaystyle \int^1_0 x^n. \sqrt{1-x^2} dx$ then
$\lim_{n \to \infty} \dfrac{I_n}{I_{n-2}}$ is equal to

Homework Equations

The Attempt at a Solution

Integrating by parts

$x^n \displaystyle \int \sqrt{1-x^2}dx - \int nx^{n-1} \int \sqrt{1-x^2} dx$

But integrating further is useless.

Try leaving one of the x with the $\sqrt{1-x^2}$ term, i.e. factor it as $x^{n-1}.x\sqrt{1-x^2}dx$, then integrate by parts very much as you did.

 

[utkarshakash]

Try leaving one of the x with the $\sqrt{1-x^2}$ term, i.e. factor it as $x^{n-1}.x\sqrt{1-x^2}dx$, then integrate by parts very much as you did.

Your method was excellent. I just want to know how do you solve these complicated problems in a jiffy?

 

[Saitama]

Try leaving one of the x with the $\sqrt{1-x^2}$ term, i.e. factor it as $x^{n-1}.x\sqrt{1-x^2}dx$, then integrate by parts very much as you did.

Using integration by parts:
$$I_n=-\frac{x^{n-1}}{3}(1-x^2)^{3/2}+\int \frac{2}{3}(n-1)x^{n-2}(1-x^2)^{3/2}$$

How do you proceed from here?

Thanks!

 

[Millennial]

$$x^{n-2}(1-x^2)^{3/2} = (x^{n-2} - x^{n})\sqrt{1-x^2}$$
Also, don't leave out the limits.

 

[Saitama]

$$x^{n-2}(1-x^2)^{3/2} = (x^{n-2} - x^{n})\sqrt{1-x^2}$$
Also, don't leave out the limits.

Thanks!

@utkarshakash: Can you post the answer?

 

[Millennial]

Honestly, having wandered around in articles and pages involving the Gamma and Beta functions, my Beta function solution was the first one that crossed my mind. However, the one provided by haruspex is so much better if you have no prior knowledge on these topics.

 

[haruspex]

Your method was excellent. I just want to know how do you solve these complicated problems in a jiffy?

In this case, by looking at what was to be proved. Since we wanted step the n by 2, and differentiation would only step it by one, we needed to leave one factor of x out of the differentiation.

 

[utkarshakash]

In this case, by looking at what was to be proved. Since we wanted step the n by 2, and differentiation would only step it by one, we needed to leave one factor of x out of the differentiation.

Ah! That makes sense. Thanks.

 

[utkarshakash]

Thanks!

@utkarshakash: Can you post the answer?

The answer is 1.

 

[Saitama]

The answer is 1.

Thanks!