Question Number 132459 by mnjuly1970 last updated on 14/Feb/21
![.... nice calculus .... 𝛗=∫_0 ^( ∞) ((sin(x^2 )ln(x))/x^(3/2) )dx= solution: 𝛗=^(x^2 =t) (1/2)∫_0 ^( ∞) ((sin(t)ln((√t) ))/t^(3/4) ) (dt/t^(1/2) ) =(1/4)∫_0 ^( ∞) ((sin(t)ln(t))/t^(5/4) )dt 𝛙(a)=∫_0 ^( ∞) ((sin(t).t^a )/t^(5/4) )dt=∫_0 ^( ∞) ((sin(t))/t^((5/4)−a) )dt 𝛙(a)=(𝛑/(2𝚪((5/4)−a)sin(((5π)/8)−(π/2)a))) ∴ 𝛗=((𝛙′(0))/4) ...... 𝛙′(a)=(π/2)[((Γ′((5/4)−a)sin(((5π)/8)−((πa)/2))+(π/2)Γ((5/4)−a)cos(((5π)/8)−((𝛑a)/2)))/(Γ^2 ((5/4)−a)sin^2 (((5π)/8)−((𝛑a)/2))))] 𝛙′(0)=(π/2)(((Γ′((5/4))cos((π/8))−(π/2)Γ((5/4))sin((π/8)))/(Γ^2 ((5/4))cos^2 ((π/8))))) =(π/2)(((4ψ(1+(1/4))Γ((1/4))cos((π/8))−2πΓ((1/4))sin((π/8)))/(Γ^2 ((1/4))cos^2 ((π/8))))) 𝛗=(π/4)((((2ψ((1/4))+8)Γ((1/4))cos((π/8))−πΓ((1/4))sin((π/8)))/(Γ^2 ((1/4))cos^2 ((π/8)))))](Q132459.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\:{nice}\:\:\:{calculus}\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}^{\mathrm{2}} \right){ln}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}= \\ $$$$\:\:{solution}: \\ $$$$\boldsymbol{\phi}\overset{{x}^{\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right){ln}\left(\sqrt{{t}}\:\right)}{{t}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:\frac{{dt}}{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right){ln}\left({t}\right)}{{t}^{\frac{\mathrm{5}}{\mathrm{4}}} }{dt} \\ $$$$\:\:\:\boldsymbol{\psi}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right).{t}^{{a}} }{{t}^{\frac{\mathrm{5}}{\mathrm{4}}} }{dt}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}^{\frac{\mathrm{5}}{\mathrm{4}}−{a}} }{dt} \\ $$$$\:\boldsymbol{\psi}\left({a}\right)=\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}−\boldsymbol{{a}}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$$$\:\therefore\:\:\:\boldsymbol{\phi}=\frac{\boldsymbol{\psi}'\left(\mathrm{0}\right)}{\mathrm{4}}\:...... \\ $$$$\:\:\:\boldsymbol{\psi}'\left({a}\right)=\frac{\pi}{\mathrm{2}}\left[\frac{\Gamma'\left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\pi{a}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){cos}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\boldsymbol{\pi}{a}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}−{a}\right){sin}^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}−\frac{\boldsymbol{\pi}{a}}{\mathrm{2}}\right)}\right] \\ $$$$\:\:\boldsymbol{\psi}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\left(\frac{\Gamma'\left(\frac{\mathrm{5}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{2}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{4}\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{2}\pi\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right) \\ $$$$\:\:\:\: \\ $$$$\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{4}}\left(\frac{\left(\mathrm{2}\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{8}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\pi\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right)\: \\ $$