5
$\begingroup$

If $$I_n = \int_0^1{x^n\sqrt{1-x^2}}\, \mathrm{d}x,$$ then find $$\lim_{n\to\infty}\frac{I_n}{I_{n-2}}.$$

$\endgroup$
2
  • $\begingroup$ Are you familiar with the Beta Function? A substitution $x = \sin u$ puts this into a classic form, at which point we find that $I_n = \frac{\sqrt{\pi} \,\Gamma(\frac{n+1}{2})}{4 \Gamma(\frac{n}{2}+2)}$. Clearly the constants are irrelevant here, and the Gamma Functions can be simplified using well known properties to evaluate the limit $\endgroup$ Apr 28, 2017 at 4:55
  • $\begingroup$ Using a trigonometric substitution, the expression reduces to a Wallis integral. $\endgroup$
    – Lucian
    Sep 1, 2017 at 14:15

1 Answer 1

10
$\begingroup$

Putting $x=\sin t$ we get $$I_{n} =\int_{0}^{\pi/2}\sin^{n}t\cos^{2}t\,dt$$ and using integration by parts we can show that $$(n+2)I_{n}=(n-1)I_{n-2}$$ and the desired limit is $1$.


More generally we can show that if $$J_{m, n} = \int_{0}^{\pi/2}\cos^{m}x\sin^{n}x\,dx$$ then $$(m + n)J_{m, n} = (m - 1)J_{m - 2, n} = (n - 1)J_{m, n - 2}$$ We have \begin{align} (m + 1)J_{m, n} &=-\int_{0}^{\pi/2}\sin^{n - 1}\frac{d}{dx}(\cos^{m + 1}x)\,dx\notag\\ &= -[\sin^{n - 1}x\cos^{m + 1}x]_{0}^{\pi/2} + \int_{0}^{\pi/2}(n - 1)\sin^{n - 2}x\cos^{m + 2}x\,dx\notag\\ &= (n - 1)\int_{0}^{\pi/2}\sin^{n - 2}x\cos^{m}x(1 - \sin^{2}x)\,dx\notag\\ &= (n - 1)J_{m, n - 2} - (n - 1)J_{m, n}\notag \end{align} And thus we get $$(m + n)J_{m, n} = (n - 1)J_{m, n - 2}$$ The other reduction formula is available by interchanging the roles of $m, n$ and noting that this does not affect $J_{m,n}$. For the current question we can see that $I_{n} = J_{2, n}$.

$\endgroup$
10
  • $\begingroup$ Oh, of course: Set up the Beta Integral, but don't complete it. Nicely done Paramanand! $\endgroup$ Apr 28, 2017 at 5:12
  • $\begingroup$ @BrevanEllefsen, I see now what you were referring to there. Thanks Paramanand. $\endgroup$ Apr 28, 2017 at 5:15
  • $\begingroup$ @AjaySubramanian: I hope you know how I arrived at the reduction formula for $I_{n} $ using integration by parts. This is pretty standard and available in any introductory calculus text. $\endgroup$
    – Paramanand Singh
    Apr 28, 2017 at 5:18
  • $\begingroup$ Thanks @BrevanEllefsenb from the question it was clear that just the reduction formula was needed. But I can understand your approach. Once one is aware of Beta function it is difficult to stop before that. $\endgroup$
    – Paramanand Singh
    Apr 28, 2017 at 5:20
  • 1
    $\begingroup$ @Stephan: Note that $\sqrt{1-\sin^{2}t}=\cos t$ and the other $\cos t$ comes from $dx=\cos t\, dt$. Remember the rule of substitution $$\int f(x) \, dx=\int f(g(t)) g'(t) \, dt$$ where $x=g(t) $. $\endgroup$
    – Paramanand Singh
    Apr 28, 2017 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.