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Let $\displaystyle f(n) = \int^{1}x^{n-1}\sin \bigg(\frac{\pi x}{2}\bigg)dx$ and $\displaystyle g(n)=\int^{1}_{0}x^{n-1}\cos\bigg(\frac{\pi x}{2}\bigg)dx$ where $n$ is a natural number and $\displaystyle \lim_{n\rightarrow \infty}\frac{(3n+1)f(n)}{(2n+1)^2g(n)}=\frac{a}{b\pi}.$ Then what is the least value of $a+b$?

Plan

$$f(n)=-\frac{2\pi }x^{n-1}\cos(\pi x/2)\bigg)\bigg|^{1}_{0}+\frac{2(n-1)}{\pi}\int^{1}_{0}x^{n-1}\cos(\pi x/2)dx$$

$$f(n)=\frac{2(n-1)}{\pi}g(n-1)=\frac{2^2}{\pi^2}(n-1)(n-2)g(n-2)$$

How do I solve it?

The Count
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jacky
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  • Dp you know the answer(the value)?? I tried finding an iterative relation for both f(n) and g(n) but seems like I'm lost. – Archis Welankar May 31 '19 at 10:00
  • For large $n$, $x^{n-1}\ll1$ everywhere but close to $1$, so taking the first few terms of the Taylor series at $x=1$ will give you the behavior for large $n$. – MathIsFun7225 May 31 '19 at 19:12

2 Answers2

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Lemma. If $f:[0,1]\to\mathbb R$ is a continuous function then $$\lim_{n\to\infty}n\int_0^1 f(x)x^{n-1}\,dx=f(1).$$

Proof. See this question.

Hence it follows that $$\lim_{n\to\infty}nf(n)=\sin(\pi/2)=1.$$ Now, you have shown that $$g(n)=\frac\pi2\frac{f(n+1)}n$$ so $$\lim_{n\to\infty}n^2g(n)=\frac\pi2$$ and the problem reduces to a simple computation.

user246336
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}x^{n - 1}\expo{\ic\pi x/2}\dd x} = \int_{0}^{1}\pars{1 - x}^{n - 1}\expo{\ic\pi\pars{1 - x}/2}\dd x \\[5mm] = &\ \ic\int_{0}^{1} \exp\pars{\vphantom{\Large A}\bracks{n - 1}\ln\pars{1 - x}} \expo{-\ic\pi x/2}\dd x \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& \ic\int_{0}^{\infty} \exp\pars{-\bracks{n - 1}x}\pars{1 - {\ic\pi \over 2}\,x - {\pi^{2} \over 8}\,x^{2}}\dd x \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& {\pi/2 \over \pars{n - 1}^{2}} + {\ic \over n - 1} \\[5mm] \implies & \bbx{\left\{\begin{array}{rcrcl} \ds{\mrm{g}\pars{n}} & \ds{\equiv} & \ds{\int_{0}^{1}x^{n - 1}\cos\pars{\pi x \over 2}\dd x} & \ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}} & \ds{\pi/2 \over \pars{n - 1}^{2}} \\[2mm] \ds{\mrm{f}\pars{n}} & \ds{\equiv} & \ds{\int_{0}^{1}x^{n - 1}\sin\pars{\pi x \over 2}\dd x} & \ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}} & \ds{1 \over n - 1} \end{array}\right.} \\[8mm] \implies & \lim_{n \to \infty}{\pars{3n + 1}\,\mrm{f}\pars{n} \over \pars{2n + 1}^{2}\,\mrm{g}\pars{n}} = \lim_{n \to \infty}{\pars{3n + 1}\bracks{1/\pars{n - 1}} \over \pars{2n + 1}^{2}\bracks{\pars{\pi/2}/\pars{n - 1}^{2}}} \\[5mm] = &\ \bbx{3 \over 2\pi} \end{align}

Felix Marin
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