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?

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    $\begingroup$ $d\theta$ 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

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.

  • $\begingroup$ 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

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 $I_n=\frac {\alpha_n} 2$

  • $\begingroup$ 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

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


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.

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

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