φ =∫_0 ^( ∞) log^2 (x)sin(x^2 )dx=? i had solved that already and: answ : = −((π^2 (√(2π)) )/(32))


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Question Number 130979 by mnjuly1970 last updated on 31/Jan/21

$ϕ = \int_{0}^{\infty} {l o g}^{2} (x) s i n (x^{2}) d x = ?$ $i h a d s o l v e d t h a t a l r e a d y a n d :$ $a n s w : = - \frac{π^{2} \sqrt{2 π}}{32}$
Commented by Dwaipayan Shikari last updated on 31/Jan/21

$\int_{0}^{\infty} \frac{t a n x}{x} d x \Rightarrow D i v e r g e s$
Commented by Dwaipayan Shikari last updated on 31/Jan/21

$\int_{0}^{\infty} \frac{s i n x}{x} d x = \int_{0}^{\infty} \int_{0}^{\infty} e^{- x t} s i n x d t d x$ $= \frac{1}{2 i} \int_{0}^{\infty} \frac{1}{t - i} - \frac{1}{t + i} d t$ $= \int_{0}^{\infty} \frac{1}{t^{2} + 1} d t = \frac{π}{2}$
Commented by mnjuly1970 last updated on 31/Jan/21

$y e s i t i s c o n v e r g e n t$ $i w i l l p r e p a r e$ $o t h e r q u e s t i o n . . .$
	Answered by mnjuly1970 last updated on 31/Jan/21

	Answered by mnjuly1970 last updated on 31/Jan/21

	Answered by Dwaipayan Shikari last updated on 31/Jan/21

	$I (a) = \int_{0}^{\infty} x^{a} s i n (x^{2}) d x$ $I (a) = \frac{1}{2 i} \int_{0}^{\infty} x^{a} e^{{i x}^{2}} - x^{a} e^{- {i x}^{2}} d x x^{2} = i u x^{2} = - i j$ $I (a) = \frac{1}{4} \int_{0}^{\infty} x^{a - 1} e^{- u} d u - x^{a - 1} e^{- j} d j$ $I (a) = \frac{{(i)}^{\frac{a - 1}{2}}}{4} \int_{0}^{\infty} u^{\frac{a - 1}{2}} e^{- u} + {(- i)}^{a - 1} \frac{1}{4} \int_{0}^{\infty} j^{\frac{a - 1}{2}} e^{- j} d j$ $= \frac{Γ (\frac{a + 1}{2})}{2} c o s (\frac{π}{4} (a + 1))$ $I^{'} (a) = \frac{Γ^{'} (\frac{a + 1}{2})}{4} c o s (\frac{π}{4} (a + 1)) - \frac{π}{4} \frac{Γ (\frac{a + 1}{2})}{2} s i n (\frac{π}{4} (a + 1))$ $I^{″} (a) = \frac{Γ^{″} (\frac{a + 1}{2})}{8} c o s (\frac{π}{4} (a + 1)) - \frac{π}{4} . \frac{Γ^{'} (\frac{a + 1}{2})}{4} s i n (\frac{π}{4} (a + 1)) - \frac{π}{16} Γ^{'} (\frac{a + 1}{2}) s i n (\frac{π}{4} (a + 1)) - \frac{π^{2}}{32} . Γ (\frac{a + 1}{2}) c o s (\frac{π}{4} (a + 1))$ $I^{″} (0) = \frac{Γ^{″} (\frac{1}{2})}{2 \sqrt{2}} - \frac{π Γ^{'} (\frac{1}{2})}{8 \sqrt{2}} - \frac{π^{\frac{5}{2}}}{32}$ $= \frac{π ψ (\frac{1}{2}) + ψ^{'} (\frac{1}{2}) \sqrt{π}}{2 \sqrt{2}} - \frac{π \sqrt{π}}{8 \sqrt{2}} ψ (\frac{1}{2}) - \frac{π^{\frac{5}{2}}}{32}$ $= \frac{π \sqrt{π}}{8 \sqrt{2}} (γ + l o g (4)) - \frac{π^{\frac{5}{2}}}{32} - \frac{π (γ + l o g (4)}{2 \sqrt{2}} + \frac{ψ^{'} (\frac{1}{2}) \sqrt{π}}{2 \sqrt{2}}$
	Answered by mathmax by abdo last updated on 31/Jan/21
	$Φ=_(x=(√t)) ∫_0 ^∞ ln^2 ((√t))sin(t)(2t)dt=(1/2)∫_0 ^∞ ln^2 (t) tsint dt ϕ(a) =∫_0 ^∞ t^(a+1) sint dt =∫_0 ^∞ e^((a+1)lnt) sint dt ⇒ϕ^′ (a)=∫_0 ^∞ lnt .e^((a+1)lnt) sint dt ⇒ϕ^((2)) (a) =∫_0 ^∞ ln^2 t .t^(a+1) sint dt ⇒ϕ^((2)) (0)=∫_0 ^∞ ln^2 t .tsint dt =2Φ ϕ(a) =−Im(∫_0 ^∞ t^(a+1) e^(−it) dt) and ∫_0 ^∞ t^(a+1) e^(−it) dt =_(it=u) ∫_0 ^∞ ((u/i))^(a+1) e^(−u) (du/i) =(1/i^(a+2) )∫_0 ^∞ u^(a+2−1) e^(−u) du =(1/((e^((iπ)/2) )^(a+2) ))×Γ(a+2) =e^(−((iπ)/2)(a+2)) .Γ(a+2) =Γ(a+2){cos((π/2)(a+2))−isin((π/2)(a+2))} ⇒ ϕ(a)=sin(((πa)/2)+π).Γ(a+2) =−sin(((πa)/2)).Γ(a+2) (witha>−2) Γ(a+2) =Γ(a+1 +1) =(a+1)Γ(a+1) =a(a+1)Γ(a) ⇒ ϕ(a) =−a(a+1)sin(((πa)/2)).Γ(a)=w(a).Γ(a) ⇒ ϕ^′ (a) =W^, (a).Γ(a)+w(a).Γ^′ (a) w^′ (a)=−{(2a+1)sin(((πa)/2))+(a^2 +a)(π/2)cos(((πa)/2))} ⇒ ϕ^, (a)=−{(2a+1)sin(((πa)/2))+(π/2)(a^2 +a)cos(((πa)/2))}.Γ(a) −(a^2 +a)sin(((πa)/2)).Γ^′ (a).....be continued...$
	$Φ =_{x = \sqrt{t}} \int_{0}^{\infty} \ln^{2} (\sqrt{t}) \sin (t) (2 t) dt = \frac{1}{2} \int_{0}^{\infty} \ln^{2} (t) tsint dt$ $φ (a) = \int_{0}^{\infty} t^{a + 1} sint dt = \int_{0}^{\infty} e^{(a + 1) lnt} sint dt \Rightarrow φ^{^{'}} (a) = \int_{0}^{\infty} lnt . e^{(a + 1) lnt} sint dt$ $\Rightarrow φ^{(2)} (a) = \int_{0}^{\infty} \ln^{2} t . t^{a + 1} sint dt \Rightarrow φ^{(2)} (0) = \int_{0}^{\infty} \ln^{2} t . tsint dt = 2 Φ$ $φ (a) = - Im (\int_{0}^{\infty} t^{a + 1} e^{- it} dt) and$ $\int_{0}^{\infty} t^{a + 1} e^{- it} dt =_{it = u} \int_{0}^{\infty} {(\frac{u}{i})}^{a + 1} e^{- u} \frac{du}{i}$ $= \frac{1}{i^{a + 2}} \int_{0}^{\infty} u^{a + 2 - 1} e^{- u} du = \frac{1}{{(e^{\frac{i π}{2}})}^{a + 2}} \times Γ (a + 2)$ $= e^{- \frac{i π}{2} (a + 2)} . Γ (a + 2) = Γ (a + 2) {\cos (\frac{π}{2} (a + 2)) - isin (\frac{π}{2} (a + 2))} \Rightarrow$ $φ (a) = \sin (\frac{π a}{2} + π) . Γ (a + 2) = - \sin (\frac{π a}{2}) . Γ (a + 2) (witha > - 2)$ $Γ (a + 2) = Γ (a + 1 + 1) = (a + 1) Γ (a + 1) = a (a + 1) Γ (a) \Rightarrow$ $φ (a) = - a (a + 1) \sin (\frac{π a}{2}) . Γ (a) = w (a) . Γ (a) \Rightarrow$ $φ^{^{'}} (a) = W^{,} (a) . Γ (a) + w (a) . Γ^{^{'}} (a)$ $w^{^{'}} (a) = - {(2 a + 1) \sin (\frac{π a}{2}) + (a^{2} + a) \frac{π}{2} \cos (\frac{π a}{2})} \Rightarrow$ $φ^{,} (a) = - {(2 a + 1) \sin (\frac{π a}{2}) + \frac{π}{2} (a^{2} + a) \cos (\frac{π a}{2})} . Γ (a)$ $- (a^{2} + a) \sin (\frac{π a}{2}) . Γ^{^{'}} (a) . . . . . be continued . . .$
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