unsolved question....... ∫_0 ^∞ (cos (x^2 )−cos x)(dx/x)=(γ/2)


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

$u n s o l v e d q u e s t i o n . . . . . . .$ $\int_{0}^{\infty} (\cos (x^{2}) - \cos x) \frac{d x}{x} = \frac{γ}{2}$
	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
	$I(a,p)=∫_0 ^∞ x^p (((sin (ax^2 ))/x^2 )−((sin (ax))/x))dx=∫_0 ^∞ x^(p−2) sin (ax^2 )dx−∫_0 ^∞ x^(p−1) sin (ax)dx =(1/2)∫_0 ^∞ u^((p/2)−(3/2)) sin (au)du−∫_0 ^∞ x^(p−1) sin (ax)dx =(1/2)M_(((p/2)−(1/2),x)) (sin (ax))−M_((p,x)) (sin (ax)) M_((p,x)) (sin (ax))=a^(−p) M_((p,x)) (sin x)=a^(−p) Γ(p)sin (((pπ)/2)) I(a,p)=(1/2)a^(−(p/2)+(1/2)) Γ(((p−1)/2))sin ((π/2)(((p−1)/2)))−a^(−p) Γ(p)sin (((pπ)/2)) ;(p>−1,a>0) ∫_0 ^∞ (cos (x^2 )−cos x)(dx/x)=lim_(p→−1) lim_(a→1) ((dI(a,p))/da) =lim_(p→−1) lim_(a→1) (1/2)a^(−p−1) {−a^((p+1)/2) sin (((p−1)/4)π)Γ(((p+1)/2))+2sin (((pπ)/2))Γ(p+1)} =lim_(p→−1) (1/2){−sin (((p−1)/4)π)Γ(((p+1)/2))+2sin (((pπ)/2))Γ(p+1)} =lim_(p→−1) (1/2){−sin (((p−1)/4)π)((2/(p+1))−γ+o(1))+2sin (((pπ)/2))((1/(p+1))−γ+o(1))} =(1/2)(−γ+2γ)=(γ/2) I still dont understand ∙∙∙,Expect for Mr 1970′s answer, after all,he gives this question.$
	$I (a, p) = \int_{0}^{\infty} x^{p} (\frac{\sin ({a x}^{2})}{x^{2}} - \frac{\sin (a x)}{x}) d x = \int_{0}^{\infty} x^{p - 2} \sin ({a x}^{2}) d x - \int_{0}^{\infty} x^{p - 1} \sin (a x) d x$ $= \frac{1}{2} \int_{0}^{\infty} u^{\frac{p}{2} - \frac{3}{2}} \sin (a u) d u - \int_{0}^{\infty} x^{p - 1} \sin (a x) d x$ $= \frac{1}{2} M_{(\frac{p}{2} - \frac{1}{2}, x)} (\sin (a x)) - M_{(p, x)} (\sin (a x))$ $M_{(p, x)} (\sin (a x)) = a^{- p} M_{(p, x)} (\sin x) = a^{- p} Γ (p) \sin (\frac{p π}{2})$ $I (a, p) = \frac{1}{2} a^{- \frac{p}{2} + \frac{1}{2}} Γ (\frac{p - 1}{2}) \sin (\frac{π}{2} (\frac{p - 1}{2})) - a^{- p} Γ (p) \sin (\frac{p π}{2}); (p > - 1, a > 0)$ $\int_{0}^{\infty} (\cos (x^{2}) - \cos x) \frac{d x}{x} = \lim_{p \to - 1} \lim_{a \to 1} \frac{d I (a, p)}{d a}$ $= \lim_{p \to - 1} \lim_{a \to 1} \frac{1}{2} a^{- p - 1} {- a^{\frac{p + 1}{2}} \sin (\frac{p - 1}{4} π) Γ (\frac{p + 1}{2}) + 2 \sin (\frac{p π}{2}) Γ (p + 1)}$ $= \lim_{p \to - 1} \frac{1}{2} {- \sin (\frac{p - 1}{4} π) Γ (\frac{p + 1}{2}) + 2 \sin (\frac{p π}{2}) Γ (p + 1)}$ $= \lim_{p \to - 1} \frac{1}{2} {- \sin (\frac{p - 1}{4} π) (\frac{2}{p + 1} - γ + o (1)) + 2 \sin (\frac{p π}{2}) (\frac{1}{p + 1} - γ + o (1))}$ $= \frac{1}{2} (- γ + 2 γ) = \frac{γ}{2}$ $I s t i l l d o n t u n d e r s t a n d \cdot \cdot \cdot, E x p e c t f o r M r 1970^{'} s a n s w e r,$ $a f t e r a l l, h e g i v e s t h i s q u e s t i o n .$
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