unsolved-question-0-cos-x-2-cos-x-dx-x-2-

Question Number 136966 by Ñï= last updated on 28/Mar/21

u n s o l v e d q u e s t i o n \dots \dots .

\int_{0}^{\infty} (\cos (x^{2}) - \cos x) \frac{d x}{x} = \frac{γ}{2}

Answered by mathmax by abdo last updated on 29/Mar/21

i think this integral is divergent!

Commented by Ñï= last updated on 29/Mar/21

I (a, p) = \int_{0}^{\infty} x^{p} (\frac{\sin ({a x}^{2})}{x^{2}} - \frac{\sin (a x)}{x}) d x = \int_{0}^{\infty} x^{p - 2} \sin ({a x}^{2}) d x - \int_{0}^{\infty} x^{p - 1} \sin (a x) d x

= \frac{1}{2} \int_{0}^{\infty} u^{\frac{p}{2} - \frac{3}{2}} \sin (a u) d u - \int_{0}^{\infty} x^{p - 1} \sin (a x) d x

= \frac{1}{2} M_{(\frac{p}{2} - \frac{1}{2}, x)} (\sin (a x)) - M_{(p, x)} (\sin (a x))

M_{(p, x)} (\sin (a x)) = a^{- p} M_{(p, x)} (\sin x) = a^{- p} Γ (p) \sin (\frac{p π}{2})

I (a, p) = \frac{1}{2} a^{- \frac{p}{2} + \frac{1}{2}} Γ (\frac{p - 1}{2}) \sin (\frac{π}{2} (\frac{p - 1}{2})) - a^{- p} Γ (p) \sin (\frac{p π}{2}); (p > - 1, a > 0)

\int_{0}^{\infty} (\cos (x^{2}) - \cos x) \frac{d x}{x} = \lim_{p \to - 1} \lim_{a \to 1} \frac{d I (a, p)}{d a}

= \lim_{p \to - 1} \lim_{a \to 1} \frac{1}{2} a^{- p - 1} {- a^{\frac{p + 1}{2}} \sin (\frac{p - 1}{4} π) Γ (\frac{p + 1}{2}) + 2 \sin (\frac{p π}{2}) Γ (p + 1)}

= \lim_{p \to - 1} \frac{1}{2} {- \sin (\frac{p - 1}{4} π) Γ (\frac{p + 1}{2}) + 2 \sin (\frac{p π}{2}) Γ (p + 1)}

= \lim_{p \to - 1} \frac{1}{2} {- \sin (\frac{p - 1}{4} π) (\frac{2}{p + 1} - γ + o (1)) + 2 \sin (\frac{p π}{2}) (\frac{1}{p + 1} - γ + o (1))}

= \frac{1}{2} (- γ + 2 γ) = \frac{γ}{2}

I s t i l l d o n t u n d e r s t a n d \cdot \cdot \cdot, E x p e c t f o r M r 1970^{'} s a n s w e r,

a f t e r a l l, h e g i v e s t h i s q u e s t i o n .

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