Question

$\huge \displaystyle \sf \red{\int_{0}^{ \frac{\pi}{4} } x \prod \limits_{k = 1}^{ \infty } \cos( \frac{x}{2 {}^{k} } ) \: dx}$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \: I = \sf {\int_{0}^{ \frac{\pi}{4} } x \prod \limits_{k = 1}^{ \infty } \cos \left( \frac{x}{2^{k} } \right) \: dx}$

Consider the series

$\displaystyle S= \sf {cos \left( \dfrac{x}{2 } \right) \cdot \: cos \left( \dfrac{x}{2^{2} } \right) \cdot \:cos \left( \dfrac{x}{2^{3} } \right) \cdots \: cos \left( \dfrac{x}{2^{n} } \right) }$

$\displaystyle \lim_{n \to \infty} S= \sf {\lim_{n \to \infty}cos \left( \dfrac{x}{2 } \right) \cdot \: cos \left( \dfrac{x}{2^{2} } \right) \cdot \:cos \left( \dfrac{x}{2^{3} } \right) \cdots \: cos \left( \dfrac{x}{2^{n} } \right) }$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty}cos \left( \dfrac{x}{2^{n} } \right) \cdot \: cos \left( \dfrac{x}{2^{n - 1} } \right) \cdot \:cos \left( \dfrac{x}{2^{n - 2} } \right) \cdots \: cos \left( \dfrac{x}{2 } \right) }$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty}cos \left( \dfrac{x}{2^{n} } \right) \cdot \: cos \left(2 \cdot \dfrac{x}{2^{n } } \right) \cdot \:cos \left( {2}^{2} \cdot\dfrac{x}{2^{n} } \right) \cdots \: cos \left( {2}^{n - 1} \cdot\dfrac{x}{2^{n} } \right) }$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty} \dfrac{sin \left( {2}^{n} \cdot \dfrac{x}{ {2}^{n} } \right)}{ {2}^{n} \cdot \: sin \left( \dfrac{x}{ {2}^{n} } \right)}}$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty} \dfrac{sin \left( x\right)}{ {2}^{n} \cdot \: sin \left( \dfrac{x}{ {2}^{n} } \right)}}$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty} \dfrac{x \cdot \: sin \left( x\right)}{ {2}^{n} \cdot x \cdot\: sin \left( \dfrac{x}{ {2}^{n} } \right)}}$

$\displaystyle \implies \lim_{n \to \infty} S= \sf {\lim_{n \to \infty} \dfrac{ \: sin ( x) \cdot \dfrac{x}{ {2}^{n} } }{ x \cdot\: sin \left( \dfrac{x}{ {2}^{n} } \right)}}$

$\displaystyle \implies \lim_{n \to \infty} S= \sf { \dfrac{sin(x)}{x}\lim_{n \to \infty} \dfrac{ \dfrac{x}{ {2}^{n} } }{ sin \left( \dfrac{x}{ {2}^{n} } \right)}}$

$\displaystyle \implies \lim_{n \to \infty} S= \sf { \dfrac{sin(x)}{x}}$

Now,

$\displaystyle \: I = \sf {\int_{0}^{ \frac{\pi}{4} } x \cdot \dfrac{ sin(x) }{x} \: dx}$

$\displaystyle \: \implies I = \sf {\int_{0}^{ \frac{\pi}{4} } sin(x) \: dx}$

$\implies I = - \sf { [cos(x)]_{0}^{ \frac{\pi}{4} } }$

$\implies I = - \sf { \left[cos \left( \dfrac{\pi}{4} \right) - cos(0) \right] }$

$\implies I = \sf { cos(0) - cos \left( \dfrac{\pi}{4} \right) }$

$\implies I = \sf { 1 - \dfrac{1}{ \sqrt{2} } }$

$\implies I = \sf { \dfrac{ \sqrt{2} - 1}{ \sqrt{2} } }$

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: \displaystyle \rm {\int_{0}^{ \dfrac{\pi}{4} } x \prod \limits_{k = 1}^{ \infty } \cos( \frac{x}{2 {}^{k} } ) \: dx}$

Let we first evaluate

$\rm :\longmapsto\:\prod \limits_{k = 1}^{ \infty } \cos( \frac{x}{2 {}^{k} } )$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{n \to \infty }\rm \prod \limits_{k = 1}^{n} \cos( \frac{x}{2 {}^{k} } )$

$\rm \: = \:\displaystyle\lim_{n \to \infty }\rm \cos\bigg[\dfrac{x}{2} \bigg]cos\bigg[\dfrac{x}{ {2}^{2} } \bigg]cos\bigg[\dfrac{x}{ {2}^{3} } \bigg] - - cos\bigg[\dfrac{x}{ {2}^{n} } \bigg]$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \: cosx = \dfrac{sin2x}{2 \: sinx}}}$

So, using this, we get

$\rm \:=\displaystyle\lim_{n \to \infty }\rm \dfrac{sin\bigg[x \bigg]}{2 \: sin\bigg[\dfrac{x}{2} \bigg]}\dfrac{sin\bigg[\dfrac{x}{2} \bigg]}{2 \: sin\bigg[\dfrac{x}{4} \bigg]} - - \dfrac{sin\bigg[\dfrac{x}{ {2}^{n - 1} } \bigg]}{2 \: sin\bigg[\dfrac{x}{ {2}^{n} } \bigg]}$

$\rm \:=\displaystyle\lim_{n \to \infty }\rm \dfrac{sin\bigg[x \bigg]}{ {2}^{n} \: sin\bigg[\dfrac{x}{ {2}^{n} } \bigg]}$

$\rm \:=\displaystyle\lim_{n \to \infty }\rm \dfrac{sin\bigg[x \bigg]}{ \dfrac{ {2}^{n} }{x} \: sin\bigg[\dfrac{x}{ {2}^{n}} \bigg] \times x}$

We know,

$\boxed{\tt{ \displaystyle\lim_{x \to \infty }\rm \frac{sin\bigg[\dfrac{1}{x} \bigg]}{ \dfrac{1}{x} } = 1}}$

So, using this, we get

$\rm \:= \: \dfrac{sinx}{x} \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{ \dfrac{ {2}^{n} }{x} \: sin\bigg[\dfrac{x}{ {2}^{n}} \bigg] }$

$\rm \: = \: \dfrac{sinx}{x}$

Hence,

$\rm \: \implies \: \boxed{\tt{ \prod \limits_{k = 1}^{ \infty } \cos\bigg( \dfrac{x}{2 {}^{k} } \bigg) = \dfrac{sinx}{x} }}$

Now, Consider

$\rm :\longmapsto\: \displaystyle \rm {\int_{0}^{ \dfrac{\pi}{4} } x \prod \limits_{k = 1}^{ \infty } \cos( \frac{x}{2 {}^{k} } ) \: dx}$

$\rm \: = \: \displaystyle \rm {\int_{0}^{ \dfrac{\pi}{4} } x \times \frac{sinx}{x} \: dx}$

$\rm \: = \: \displaystyle \rm {\int_{0}^{ \dfrac{\pi}{4} } sinx \: dx}$

$\rm \: = \: - \bigg |cosx\bigg| _{0}^{ \dfrac{\pi}{4} }$

$\rm \: = \: - \bigg |cos\dfrac{\pi}{4} - cos0 \bigg|$

$\rm \: = \: - \bigg |\dfrac{1}{ \sqrt{2} } - 1 \bigg|$

$\rm \: = \: 1 - \dfrac{1}{ \sqrt{2} }$

$\rm \: = \: \dfrac{ \sqrt{2} - 1}{ \sqrt{2} }$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle \rm {\int_{0}^{ \dfrac{\pi}{4} } x \prod \limits_{k = 1}^{ \infty } \cos( \frac{x}{2 {}^{k} } ) \: dx} = \frac{ \sqrt{2} - 1}{ \sqrt{2} }}}$

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EXPLORE MORE

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$