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Question Number 123454 by mnjuly1970 last updated on 25/Nov/20

\dots n i c e c a l c u l u s \dots

c a l c u l a t e :::

Ω \overset{? ? ?}{=} \int_{0}^{\infty} \sqrt{x} \underset{n = 1}{\prod^{\infty}} (c o s (\frac{x}{2^{n}})) d x

Answered by Olaf last updated on 25/Nov/20

\sin 2 θ = 2 \sin θ \cos θ

\cos θ = \frac{\sin 2 θ}{2 \sin θ}

Let θ = \frac{x}{2^{n}}

\cos (\frac{x}{2^{n}}) = \frac{\sin (\frac{x}{2^{n - 1}})}{2 \sin (\frac{x}{2^{n}})}

\underset{n = 1}{\prod^{p}} \cos (\frac{x}{2^{n}}) = \underset{n = 1}{\prod^{p}} \frac{\sin (\frac{x}{2^{n - 1}})}{2 \sin (\frac{x}{2^{n}})} = \frac{\sin x}{2^{p} \sin (\frac{x}{2^{p}})}

2^{p} \sin (\frac{x}{2^{p}}) \underset{\infty}{\sim} 2^{p} \times \frac{x}{2^{p}} = x

\Rightarrow \underset{n = 1}{\prod^{\infty}} \cos (\frac{x}{2^{n}}) = \frac{\sin x}{x}

Ω = \int_{0}^{\infty} \sqrt{x} \underset{n = 1}{\prod^{\infty}} \cos (\frac{x}{2^{n}}) d x

Ω = \int_{0}^{\infty} \sqrt{x} . \frac{\sin x}{x} . d x

Ω = \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} d x

Let t = \sqrt{x}, d t = \frac{d x}{2 \sqrt{x}}

Ω = 2 \int_{0}^{\infty} \sin (t^{2}) d t

With \int_{0}^{\infty} \sin (t^{2}) d t = \frac{1}{2} \sqrt{\frac{π}{2}}

(Fresnel integral)

\Rightarrow Ω = \sqrt{\frac{π}{2}}

Commented by mnjuly1970 last updated on 26/Nov/20

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