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Question Number 130979 by mnjuly1970 last updated on 31/Jan/21
φ =∫_0 ^( ∞) log^2 (x)sin(x^2 )dx=? i had solved that already and: answ : = −((π^2 (√(2π)) )/(32))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\phi\:=\int_{\mathrm{0}} ^{\:\infty} {log}^{\mathrm{2}} \left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx}=? \\ $$$$\:\:\:\:\:{i}\:{had}\:{solved}\:{that}\:{already}\:\:{and}: \\ $$$$\:\:\:{answ}\:\::\:=\:−\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{2}\pi}\:}{\mathrm{32}} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\: \\ $$
Commented by Dwaipayan Shikari last updated on 31/Jan/21
∫_0 ^∞ ((tanx)/x)dx ⇒Diverges
$$\int_{\mathrm{0}} ^{\infty} \frac{{tanx}}{{x}}{dx}\:\:\Rightarrow{Diverges} \\ $$
Commented by Dwaipayan Shikari last updated on 31/Jan/21
∫_0 ^∞ ((sinx)/x)dx=∫_0 ^∞ ∫_0 ^∞ e^(−xt) sinx dtdx =(1/(2i))∫_0 ^∞ (1/(t−i))−(1/(t+i))dt =∫_0 ^∞ (1/(t^2 +1))dt=(π/2)
$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {sinx}\:{dtdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}−{i}}−\frac{\mathrm{1}}{{t}+{i}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 31/Jan/21
yes it is convergent i will prepare other question...
$${yes}\:{it}\:{is}\:{convergent} \\ $$$$\:{i}\:{will}\:{prepare} \\ $$$$\:{other}\:{question}… \\ $$
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)=∫_0 ^∞ x^a sin(x^2 )dx I(a)=(1/(2i))∫_0 ^∞ x^a e^(ix^2 ) −x^a e^(−ix^2 ) dx x^2 =iu x^2 =−ij I(a)=(1/4)∫_0 ^∞ x^(a−1) e^(−u) du −x^(a−1) e^(−j) dj I(a)=(((i)^((a−1)/2) )/4)∫_0 ^∞ u^((a−1)/2) e^(−u) +(−i)^(a−1) (1/4)∫_0 ^∞ j^((a−1)/2) e^(−j) dj =((Γ(((a+1)/2)))/2)cos((π/4)(a+1)) I′(a)=((Γ′(((a+1)/2)))/4)cos((π/4)(a+1))−(π/4) ((Γ(((a+1)/2)))/2)sin((π/4)(a+1)) I′′(a)=((Γ′′(((a+1)/2)))/8)cos((π/4)(a+1))−(π/4).((Γ′(((a+1)/2)))/4)sin((π/4)(a+1))−(π/(16))Γ′(((a+1)/2))sin((π/4)(a+1))−(π^2 /(32)).Γ(((a+1)/2))cos((π/4)(a+1)) I′′(0)=((Γ′′((1/2)))/(2(√2)))−((πΓ′((1/2)))/( 8(√2)))−(π^(5/2) /(32)) =((πψ((1/2))+ψ′((1/2))(√π))/(2(√2)))−((π(√π))/(8(√2)))ψ((1/2))−(π^(5/2) /(32)) =((π(√π))/(8(√2)))(γ+log(4))−(π^(5/2) /(32))−((π(γ+log(4))/(2(√2)))+((ψ′((1/2))(√π))/(2(√2)))
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}^{\mathrm{2}} } −{x}^{{a}} {e}^{−{ix}^{\mathrm{2}} } {dx}\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={iu}\:\:\:\:{x}^{\mathrm{2}} =−{ij} \\ $$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {x}^{{a}−\mathrm{1}} {e}^{−{u}} {du}\:−{x}^{{a}−\mathrm{1}} {e}^{−{j}} {dj} \\ $$$${I}\left({a}\right)=\frac{\left({i}\right)^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} +\left(−{i}\right)^{{a}−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {j}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {e}^{−{j}} {dj} \\ $$$$=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\ $$$${I}'\left({a}\right)=\frac{\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}\:\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\ $$$${I}''\left({a}\right)=\frac{\Gamma''\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{8}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}.\frac{\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}}{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{16}}\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}.\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\ $$$${I}''\left(\mathrm{0}\right)=\frac{\Gamma''\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\pi\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\mathrm{8}\sqrt{\mathrm{2}}}−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\frac{\pi\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\pi\sqrt{\pi}}{\mathrm{8}\sqrt{\mathrm{2}}}\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}\:\:\:\: \\ $$$$=\frac{\pi\sqrt{\pi}}{\mathrm{8}\sqrt{\mathrm{2}}}\left(\gamma+{log}\left(\mathrm{4}\right)\right)−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}−\frac{\pi\left(\gamma+{log}\left(\mathrm{4}\right)\right.}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
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...
$$\Phi=_{\mathrm{x}=\sqrt{\mathrm{t}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{t}}\right)\mathrm{sin}\left(\mathrm{t}\right)\left(\mathrm{2t}\right)\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)\:\mathrm{tsint}\:\mathrm{dt} \\ $$$$\varphi\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{sint}\:\mathrm{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{lnt}} \mathrm{sint}\:\mathrm{dt}\:\Rightarrow\varphi^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{lnt}\:.\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{lnt}} \mathrm{sint}\:\mathrm{dt} \\ $$$$\Rightarrow\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \mathrm{ln}^{\mathrm{2}} \mathrm{t}\:.\mathrm{t}^{\mathrm{a}+\mathrm{1}} \mathrm{sint}\:\mathrm{dt}\:\Rightarrow\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{ln}^{\mathrm{2}} \mathrm{t}\:.\mathrm{tsint}\:\mathrm{dt}\:=\mathrm{2}\Phi \\ $$$$\varphi\left(\mathrm{a}\right)\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{it}} \mathrm{dt}\right)\:\mathrm{and}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{it}} \mathrm{dt}\:=_{\mathrm{it}=\mathrm{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{u}}{\mathrm{i}}\right)^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \:\frac{\mathrm{du}}{\mathrm{i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{a}+\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{u}^{\mathrm{a}+\mathrm{2}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \:\mathrm{du}\:=\frac{\mathrm{1}}{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\mathrm{a}+\mathrm{2}} }×\Gamma\left(\mathrm{a}+\mathrm{2}\right) \\ $$$$=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)} .\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=\Gamma\left(\mathrm{a}+\mathrm{2}\right)\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)\right)\right\}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)=\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}+\pi\right).\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=−\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:\:\:\left(\mathrm{witha}>−\mathrm{2}\right) \\ $$$$\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=\Gamma\left(\mathrm{a}+\mathrm{1}\:+\mathrm{1}\right)\:=\left(\mathrm{a}+\mathrm{1}\right)\Gamma\left(\mathrm{a}+\mathrm{1}\right)\:=\mathrm{a}\left(\mathrm{a}+\mathrm{1}\right)\Gamma\left(\mathrm{a}\right)\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)\:=−\mathrm{a}\left(\mathrm{a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{a}\right)=\mathrm{w}\left(\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)\:=\mathrm{W}^{,} \left(\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)+\mathrm{w}\left(\mathrm{a}\right).\Gamma^{'} \left(\mathrm{a}\right) \\ $$$$\mathrm{w}^{'} \left(\mathrm{a}\right)=−\left\{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)+\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\frac{\pi}{\mathrm{2}}\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\varphi^{,} \left(\mathrm{a}\right)=−\left\{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\right\}.\Gamma\left(\mathrm{a}\right) \\ $$$$−\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma^{'} \left(\mathrm{a}\right)…..\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$

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