$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx$

Question

Evaluate $$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx$$

My try:

$$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx=\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {e^{i2nx}(1+e^{-i2x})^{2n}}{2^{2n}e^{x\ln 2}} dx$$

I write this using that $\cos x=\frac {e^{ix}+e^{-ix}}{2}$ and $2^x=e^{x\ln 2}$

We also know that $$\prod_{k=1}^n \frac {2k}{2k-1}=\frac {2^{2n}(n!)^2}{(2n-1)!}$$ Using this along with binomial theorem we get $$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {e^{i2nx}(1+e^{-i2x})^{2n}}{2^{2n}e^{x\ln 2}} dx=\lim_{n\to\infty} \frac {(n!) ^2}{(2n-1)!}\left(\sum_{r=0}^{2n} \binom {2n}{r}\left(\int_{-1}^{\infty} e^{x(2i(n-r)-\ln 2)} dx\right)\right) $$

$$=\lim_{n\to\infty} \frac {(n!) ^2}{(2n-1)!}\left(\sum_{r=0}^{2n} \binom {2n}{r} \left[\frac {e^{x(2i(n-r)-\ln 2}}{ 2i(n-r)-\ln 2)} \right]_{-1}^{\infty}\right)$$

And now I am stuck here. Any suggestions or a different method are openly welcomed.

Answer 1

We introduce the following result.

Proposition. Let $f$ be a bounded measurable function on $[-\pi/2, \pi/2]$ which is continuous at $0$. Then

$$ \lim_{n\to\infty} \left(\prod_{k=1}^n\frac{2k}{2k-1}\right) \int_{-\pi/2}^{\pi/2}f(x)\cos^{2n}(x)\,dx = \pi f(0). $$

We defer the proof to the end and rejoice its consequence now. We have

\begin{align*} \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} \, dx &= \int_{-1}^{\pi/2} \frac {(\cos x)^{2n}}{2^x} \, dx + \sum_{k=1}^{\infty} \int_{-\pi/2}^{\pi/2} \frac {(\cos x)^{2n}}{2^{x+k\pi}} \, dx \\ &= \int_{-\pi/2}^{\pi/2} \left( \frac{1}{2^x} \mathbf{1}_{[-1,\pi/2]}(x) + \sum_{k=1}^{\infty} \frac {1}{2^{x+k\pi}} \right) (\cos x)^{2n} \, dx. \end{align*}

So by the above proposition,

\begin{align*} \left(\prod_{k=1}^n\frac{2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} \, dx &\xrightarrow[n\to\infty]{} \pi \sum_{k=0}^{\infty} \frac {1}{2^{k\pi}} = \frac{\pi 2^{\pi}}{2^{\pi} - 1}. \end{align*}

---

Proof. By the substitution $x = u/\sqrt{n}$, we may write

$$ \int_{-\pi/2}^{\pi/2} f(x)\cos^{2n}(x) \, dx = \frac{1}{\sqrt{n}} \underbrace{ \int_{-\pi\sqrt{n}/2}^{\pi\sqrt{n}/2} f\left(\frac{u}{\sqrt{n}}\right) \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \, du}_{\text{(*)}}. $$

Now we make several observations.

• It is easy to see from Stirling's formula that $\prod_{k=1}^{n} \frac{2k}{2k-1} \sim \sqrt{\pi n}$, see this for instance.

• There exists $c > 0$ for which $\cos x \leq 1 - cx^2$ for all $|x| \leq \pi/2$. For instance, one may utilize the inequality $\sin x \geq \frac{2}{\pi}x$, valid for $0 \leq x \leq \pi/2$, to show that $c = \frac{1}{\pi}$ works. Together with the inequality $1+x \leq e^x$ which holds for all $x \in \mathbb{R}$,

  $$ \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \leq \left(1 - \frac{cu^2}{n} \right)^n \leq e^{-cu^2} $$

  for each $u$ satisfying $|u| \leq \pi\sqrt{n}/2$. This shows that the integrand of $\text{(*)}$, extended to all of $\mathbb{R}$ by setting its value to $0$ outside $[-\pi\sqrt{n}/2, \pi\sqrt{n}/2]$, is bounded by an integrable dominating function.

• For each fixed $u$, Taylor's theorem tells that

  $$ f\left(\frac{u}{\sqrt{n}}\right) \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) = (f(0) + o(1)) \left( 1 - \frac{u^2}{2n} + \mathcal{O}\left(\frac{1}{n^2}\right) \right)^{2n} \xrightarrow[n\to\infty]{} f(0) e^{-u^2}. $$

Combining altogether, by the dominated convergence theorem,

$$ \int_{-\pi\sqrt{n}/2}^{\pi\sqrt{n}/2} f\left(\frac{u}{\sqrt{n}}\right) \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \, du \xrightarrow[n\to\infty]{} \int_{-\infty}^{\infty} f(0)e^{-u^2} \, du = \sqrt{\pi}f(0). $$

Together with the observation $\frac{1}{\sqrt{n}} \prod_{k=1}^{n} \frac{2k}{2k-1} \to \sqrt{\pi} $, the desired conclusion follows.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The integral integrand yields its greatest contribution 'around' $\ds{x \approx 0}$ such that the integral asymptotic evaluation, as $\ds{n \to \infty}$, involves the Laplace Method: \begin{align} \int_{-1}^{\infty}\mrm{f}\pars{x}{\cos^{2n}\pars{x} \over 2^{x}}\,\dd x & = \int_{-1}^{\infty}\mrm{f}\pars{x} \expo{2n\ln\pars{\cos\pars{x}} - x\ln\pars{2}}\,\dd x \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\mrm{f}\pars{0} \int_{-\infty}^{\infty}\expo{-nx^{2} - \ln\pars{2}x}\dd x \\[5mm] & \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {\root{\pi}\,\mrm{f}\pars{0} \over n^{1/2}} \end{align}

Then,

\begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}\prod_{k = 1}^{n} {2k \over 2k - 1}\int_{-1}^{\infty}{\cos^{2n}\pars{x} \over 2^x}\,\dd x} = \lim_{n \to \infty}{n! \over \prod_{k = 1}^{n}\pars{k - 1/2}} {\root{\pi}\mrm{f}\pars{0} \over n^{1/2}} \\[5mm] = &\ \root{\pi}\mrm{f}\pars{0}\lim_{n \to \infty} {n! \over \pars{1/2}^{\large\overline{n}}}{1 \over n^{1/2}} = \root{\pi}\mrm{f}\pars{0}\lim_{n \to \infty} {n! \over \pars{n - 1/2}!/\pars{-1/2}!}\,{1 \over n^{1/2}} \\[5mm] = &\ \pi\,\mrm{f}\pars{0}\lim_{n \to \infty} {\root{2\pi}n^{n + 1/2}\expo{-n} \over \root{2\pi}\pars{n - 1/2}^{n}\expo{-n + 1/2}}\,{1 \over n^{1/2}} \\[5mm] = &\ \pi\,\mrm{f}\pars{0}\lim_{n \to \infty} {n^{n + 1/2} \over n^{n}\bracks{1 - \pars{1/2}/n}^{n}\expo{1/2}} \,{1 \over n^{1/2}} = \bbx{\pi\,\mrm{f}\pars{0}} \end{align}

Note that $\ds{\pars{-1/2}! = \Gamma\pars{1/2} = \root{\pi}}$ and $\ds{\lim_{n \to \infty}\bracks{1 - \pars{1/2}/n}^{\, n} = \expo{-1/2}}$.