...advanced calculus... evaluate ::: Ω=^(???) ∫_0 ^( ∞) cos(x^2 )ln(x)dx :::::::::


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Question Number 125922 by mnjuly1970 last updated on 15/Dec/20

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e ::: Ω \overset{? ? ?}{=} \int_{0}^{\infty} c o s (x^{2}) l n (x) d x$ $:::::::::$
	Answered by mathmax by abdo last updated on 15/Dec/20
	$I=∫_0 ^∞ cos(x^2 )lnx dx and J=∫_0 ^∞ sin(x^2 )lnx dx ⇒ I−iJ =∫_0 ^∞ e^(−ix^2 ) ln(x)dx =_((√i)x=t) ∫_0 ^∞ e^(−t^2 ) ln((t/( (√i))))(dt/( (√i))) =e^(−((iπ)/4)) ∫_0 ^∞ ( e^(−t^2 ) ln(t)−ln((√i))e^(−t^2 ) )dt =e^(−((iπ)/4)) ∫_0 ^∞ e^(−t^2 ) ln(t)dt−e^(−((iπ)/4)) ×((iπ)/4) ∫_0 ^∞ e^(−t^2 ) dt we have ∫_0 ^∞ e^(−t^2 ) dt =((√π)/2) ∫_0 ^∞ e^(−t^2 ) ln(t)dt =_(t^2 =u) ∫_0 ^∞ e^(−u) ln((√u))(du/( (√u))) =(1/2)∫_0 ^∞ u^(−(1/2)) e^(−u) ln(u)du we have Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt =∫_0 ^∞ e^((x−1)ln(t)) e^(−t) dt ⇒Γ^′ (x)=∫_0 ^∞ lnt t^(x−1) e^(−t) dt ⇒Γ^′ ((1/2))=∫_0 ^∞ e^(−t) t^(−(1/2)) ln(t)dt ⇒∫_0 ^∞ e^(−t^2 ) ln(t)dt=(1/2)Γ^′ ((1/2)) ⇒I−iJ =e^(−((iπ)/4)) {(1/2)Γ^′ ((1/2))−((iπ)/4)×((√π)/2)} =e^(−((iπ)/4)) {(1/2)Γ^′ ((1/2))−((π(√π))/8)} =((1/2)Γ^′ ((1/2))−((π(√π))/8))(cos((π/4))−isin((π/4)))⇒ ∫_0 ^∞ cos(x^2 )ln(x)dx =((√2)/2)((1/2)Γ^′ ((1/2))−((π(√π))/8))=∫_0 ^∞ sin(x^2 )ln(x)dx rest to calculate Γ^′ ((1/2))...$
	$I = \int_{0}^{\infty} \cos (x^{2}) lnx dx and J = \int_{0}^{\infty} \sin (x^{2}) lnx dx \Rightarrow$ $I - iJ = \int_{0}^{\infty} e^{- {ix}^{2}} \ln (x) dx =_{\sqrt{i} x = t} \int_{0}^{\infty} e^{- t^{2}} \ln (\frac{t}{\sqrt{i}}) \frac{dt}{\sqrt{i}}$ $= e^{- \frac{i π}{4}} \int_{0}^{\infty} (e^{- t^{2}} \ln (t) - \ln (\sqrt{i}) e^{- t^{2}}) dt$ $= e^{- \frac{i π}{4}} \int_{0}^{\infty} e^{- t^{2}} \ln (t) dt - e^{- \frac{i π}{4}} \times \frac{i π}{4} \int_{0}^{\infty} e^{- t^{2}} dt we have$ $\int_{0}^{\infty} e^{- t^{2}} dt = \frac{\sqrt{π}}{2}$ $\int_{0}^{\infty} e^{- t^{2}} \ln (t) dt =_{t^{2} = u} \int_{0}^{\infty} e^{- u} \ln (\sqrt{u}) \frac{du}{\sqrt{u}}$ $= \frac{1}{2} \int_{0}^{\infty} u^{- \frac{1}{2}} e^{- u} \ln (u) du we have Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} dt$ $= \int_{0}^{\infty} e^{(x - 1) \ln (t)} e^{- t} dt \Rightarrow Γ^{^{'}} (x) = \int_{0}^{\infty} lnt t^{x - 1} e^{- t} dt$ $\Rightarrow Γ^{^{'}} (\frac{1}{2}) = \int_{0}^{\infty} e^{- t} t^{- \frac{1}{2}} \ln (t) dt \Rightarrow \int_{0}^{\infty} e^{- t^{2}} \ln (t) dt = \frac{1}{2} Γ^{^{'}} (\frac{1}{2})$ $\Rightarrow I - iJ = e^{- \frac{i π}{4}} {\frac{1}{2} Γ^{^{'}} (\frac{1}{2}) - \frac{i π}{4} \times \frac{\sqrt{π}}{2}}$ $= e^{- \frac{i π}{4}} {\frac{1}{2} Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{8}} = (\frac{1}{2} Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{8}) (\cos (\frac{π}{4}) - isin (\frac{π}{4})) \Rightarrow$ $\int_{0}^{\infty} \cos (x^{2}) \ln (x) dx = \frac{\sqrt{2}}{2} (\frac{1}{2} Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{8}) = \int_{0}^{\infty} \sin (x^{2}) \ln (x) dx$ $rest to calculate Γ^{^{'}} (\frac{1}{2}) . . .$
	Commented by mnjuly1970 last updated on 15/Dec/20

	$g r e a t s i r . . .$
	Commented by mathmax by abdo last updated on 16/Dec/20
	$sorry at final lines I−iJ =e^(−((iπ)/4)) {(1/2)Γ^′ ((1/2))−i((π(√π))/8)} =(1/2)((1/( (√2)))−(i/( (√2))))(Γ^′ ((1/2))−((iπ(√π))/4)) =(1/2)((1/( (√2)))Γ^′ ((1/2))−((iπ(√π))/(4(√2)))−(i/( (√2)))Γ^′ ((1/2))−((π(√π))/(4(√2)))) =(1/2)((1/( (√2)))Γ^′ ((1/2))−((π(√π))/(4(√2)))−i((1/( (√2)))Γ^′ ((1/2))+((π(√π))/(4(√2)))) ⇒ ∫_0 ^∞ cos(x^2 )ln(x)dx=(1/(2(√2)))(Γ^′ ((1/2))−((π(√π))/4)) and ∫_0 ^∞ sin(x^2 )ln(x)dx =(1/(2(√2)))(Γ^′ ((1/2))+((π(√π))/4))$
	$sorry at final lines I - iJ$ $= e^{- \frac{i π}{4}} {\frac{1}{2} Γ^{^{'}} (\frac{1}{2}) - i \frac{π \sqrt{π}}{8}}$ $= \frac{1}{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) (Γ^{^{'}} (\frac{1}{2}) - \frac{i π \sqrt{π}}{4})$ $= \frac{1}{2} (\frac{1}{\sqrt{2}} Γ^{^{'}} (\frac{1}{2}) - \frac{i π \sqrt{π}}{4 \sqrt{2}} - \frac{i}{\sqrt{2}} Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{4 \sqrt{2}})$ $= \frac{1}{2} (\frac{1}{\sqrt{2}} Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{4 \sqrt{2}} - i (\frac{1}{\sqrt{2}} Γ^{^{'}} (\frac{1}{2}) + \frac{π \sqrt{π}}{4 \sqrt{2}}) \Rightarrow$ $\int_{0}^{\infty} \cos (x^{2}) \ln (x) dx = \frac{1}{2 \sqrt{2}} (Γ^{^{'}} (\frac{1}{2}) - \frac{π \sqrt{π}}{4}) and$ $\int_{0}^{\infty} \sin (x^{2}) \ln (x) dx = \frac{1}{2 \sqrt{2}} (Γ^{^{'}} (\frac{1}{2}) + \frac{π \sqrt{π}}{4})$
	Commented by mnjuly1970 last updated on 16/Dec/20

	$g r a t e f u l s i r m a x . .$
	Commented by mathmax by abdo last updated on 16/Dec/20

	$thankx sir$
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