3
$\begingroup$

The integral in hand is $$ I(n) = \frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}\, dx $$ I dont know whether it has closed-form or not, but currently I only want to know its asymptotic behavior. Setting $x=\cos\theta$, then $$ I(n) = \frac{1}{\pi}\int_{0}^{\pi/2} \Big[(1+2\cos\theta)^{2n}+(1-2\cos\theta)^{2n}\Big]\, d\theta $$ The second term can be neglected, therefore $$ I(n) \sim \frac{1}{\pi}\int_{0}^{\pi/2}(1+2\cos\theta)^{2n}\, d\theta $$ How can I move on?

$\endgroup$
  • 1
    $\begingroup$ $d\theta$ in the last 2 integrals? $\endgroup$ – d.k.o. Jun 22 '15 at 5:25
  • $\begingroup$ $(1+2cos(\theta))\ge1$ on $[0,\pi/2]$ $\endgroup$ – d.k.o. Jun 22 '15 at 5:26
  • $\begingroup$ Maybe binomial theorem is useful to simplify the original integrand $\endgroup$ – Dmoreno Jun 22 '15 at 5:47
  • $\begingroup$ @d.k.o.: Yes, I made a mistake. $\endgroup$ – Roger209 Jun 22 '15 at 6:00
  • $\begingroup$ @Dmoreno: It is a good point, but I donnot think it will work inpractice. We may not find the closed form of the series acutally. $\endgroup$ – Roger209 Jun 22 '15 at 6:02
7
$\begingroup$

To compute the asymptotic expansion of the integral, we split it into two pieces

$$\frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx = \frac{1}{\pi}\left(\int_{-1}^0 + \int_0^1\right)\frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx $$ Over the interval $[-1,0]$, we have $|1+2x|\le 1$, so the contribution there is bounded.

$$\mathcal{I}_1 \stackrel{def}{=} \left|\frac{1}{\pi}\int_{-1}^0 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx\right| \le \frac{1}{\pi}\int_{-1}^0 \frac{dx}{\sqrt{1-x^2}} = \frac12 $$ Over the interval $[0,1]$, introduce variable

$$1 + 2x = 3 e^{-t} \quad\iff\quad x = \frac{3e^{-t}-1}{2}$$ We have

$$\mathcal{I}_2 \stackrel{def}{=} \frac{1}{\pi}\int_{0}^1 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx = \frac{3^{2n+1}}{2\pi}\int_{0}^{\log 3} e^{-(2n+1)t} \left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2} dt$$ Notice near $t = 0$, the complicated mess of the square bracket has following Taylor series expansion: $$\left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2} = \frac{1}{\sqrt{3t}}\left( 1+ \frac{5}{8}t+ \frac{49}{384}t^2-\frac{29}{3072}t^3 + \cdots \right)\tag{*1}$$ The whole expression is in the form which we can apply Watson's Lemma and read off the asympotic expansion:

$$\begin{align} \mathcal{I}_2 \;\approx &\; \frac{3^{2n+1}}{2\sqrt{3}\pi} \left( \frac{\Gamma\left(\frac12\right)}{\sqrt{2n+1}} + \frac{5}{8}\frac{\Gamma\left(\frac32\right)}{\sqrt{2n+1}^3} + \frac{49}{384}\frac{\Gamma\left(\frac52\right)}{\sqrt{2n+1}^5} - \frac{29}{3072}\frac{\Gamma\left(\frac72\right)}{\sqrt{2n+1}^7} + \cdots \right)\\ \;\approx &\; \frac{3^{2n}\sqrt{3}}{2\sqrt{\pi(2n+1)}} \left( 1 + \frac{5}{16(2n+1)} + \frac{49}{512(2n+1)^2} - \frac{145}{8192(2n+1)^3} + \cdots\right) \end{align} $$ Since $\mathcal{I}_1$ is always of $O(1)$, above asymbolic expansion for $\mathcal{I}_2$ is also the one for $\mathcal{I}_1 + \mathcal{I}_2$. i.e. the one you are looking for.

If one want more terms for the asymptotic expansion, one just need to throw the LHS of $(*1)$ to an CAS, crank out more terms of the Taylor expansion and repeat above process.

$\endgroup$
  • $\begingroup$ Great! Thanks for your excellent answer. If $n\gg 1$, the leading term is therefore $I\sim 3^{2n}\sqrt{\frac{3}{8\pi n}}$, which is in according with Claude's comment below. $\endgroup$ – Roger209 Jun 22 '15 at 7:16
  • $\begingroup$ @Roger209 yes. if you just need the leading term. A cheap way to get the same answer is let $x = \cos\theta$ and approximate the integral as $\frac{1}{\pi}\int_0^\pi (3-\theta^2)^{2n} d\theta \approx \frac{3^{2n}}{\pi}\int_0^\infty e^{-\frac{2n}{3}\theta^2} d\theta$ . $\endgroup$ – achille hui Jun 22 '15 at 7:31
  • $\begingroup$ @achillehui. This is a very elegant solution indeed ! Cheers :-) $\endgroup$ – Claude Leibovici Jun 22 '15 at 10:10
3
$\begingroup$

This is not an answer but it is too long for a comment

For the antiderivative $$J_n = \frac{1}{\pi}\int \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}\, dx$$ there is a "closed" for which involves the Appell hypergeometric function of two variables $\frac 2{1-x}$ and $\frac 3{2(1-x)}$ which not very useful.

It is more ugly for the integral $$I_n = \frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}\, dx$$ but computing the first terms and searching $OEIS$ it appears that this is sequence $A082758$ which corresponds to the sum of the squares of the trinomial coefficients.

According to this page, Vaclav Kotesovec proposed in 2012, beside the reccurence relation $$n (2 n-1) I_n=\left(14 n^2+n-12\right) I_{n-1}+3 \left(14 n^2-71 n+78\right) I_{n-2}-27 (n-2) (2 n-5) I_{n-3}$$ an approximation formula $$I_n\approx \frac{3^{2 n+\frac{1}{2}}}{2 \sqrt{2 \pi n} }$$

Written as $$I_n = \frac{1}{\pi}\int_{0}^{\pi/2} \Big[(1+2\cos\theta)^{2n}+[(1-2\cos\theta)^{2n}\Big]\, d\theta$$ using the binomial expansion and power reduction for the cosines, the antiderivative write $$\frac 1 \pi \Big(\alpha_n \theta+\sum_{k=0}^n \beta_k \sin(2k\theta)\Big)$$ and so $I_n=\frac {\alpha_n} 2$

$\endgroup$
  • $\begingroup$ Thanks very much, Claude. I appreciate your "comment". Thanks for letting me know Vaclav Kotesovec's work. $\endgroup$ – Roger209 Jun 22 '15 at 6:48
  • $\begingroup$ You are very welcome ! This is a very interesting problem, indeed. $\endgroup$ – Claude Leibovici Jun 22 '15 at 7:00
2
$\begingroup$

Let's look at

$$(1)\,\,\,\,\int_{0}^{\pi/2}(1+2\cos t)^{n}\, dt = 3^n\int_{0}^{\pi/2}(1/3+2(\cos t)/3)^{n}\, dt.$$

For the last integral, we can look at $\int_{0}^{b}(1/3+2\cos t/3)^{n}\, dt$ for any small $b>0,$ the rest of the integral decreasing exponentially as $n\to \infty.$ Now near $0,\cos t \sim 1-t^2/2,$ so let's substitute that in and simplify. We get

$$\int_0^b(1-t^2/3)^n\,dt.$$

Let $t=(\sqrt {3}s)/\sqrt {n}.$ The above becomes

$$\frac{\sqrt {3}}{\sqrt n}\int_0^{b\sqrt n /\sqrt 3}(1-s^2/n)^n\,ds.$$

That last integral $\to \int_0^{\infty}e^{-s^2}ds = \sqrt \pi/2.$ It follows that $(1)$ is asymptotic to

$$3^n\cdot \frac{\sqrt {3}}{\sqrt n}\cdot \frac{\sqrt \pi}{2}.$$ Not a full expansion, certainly some details left out, but this I think this approach gives intuition.

$\endgroup$
  • $\begingroup$ Yes, it is rather intuitive and transparent. A good solution! $\endgroup$ – Roger209 Jun 23 '15 at 1:21

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.