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Question Number 143454 by mnjuly1970 last updated on 14/Jun/21
     ........nice .......integral.......    T  :=∫_0 ^( ∞) ((arctan(x))/x^( ln(x) +1) )dx=^? ((π(√π))/4)
$$ \\ $$$$\:\:\:……..{nice}\:…….{integral}……. \\ $$$$\:\:\mathscr{T}\:\::=\int_{\mathrm{0}} ^{\:\infty} \frac{{arctan}\left({x}\right)}{{x}^{\:{ln}\left({x}\right)\:+\mathrm{1}} }{dx}\overset{?} {=}\frac{\pi\sqrt{\pi}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by mindispower last updated on 14/Jun/21
x→(1/x)⇒  A=∫_0 ^∞ (((π/2)−arctan(x))/(((1/x))^(−ln(x)+1) )).(dx/x^2 )=(π/2)∫_0 ^∞ (dx/x^(ln(x)+1) )−∫_0 ^∞ ((arctan(x)dx)/x^(ln(x)+1) )  2A=(π/2)∫_0 ^∞ (dx/x^(ln(x)+1) )  ln(x)=u  A=(π/4)∫_(−∞) ^∞ (du/e^u^2  )=(π/2)∫_0 ^∞ e^(−u^2 ) du=(π/4)∫_0 ^∞ e^(−t) .t^(−(1/2)) dt  =(π/4)Γ((1/2))=(π/4)(√π)  ∫_0 ^∞ ((arctan(x))/x^(ln(x)+1) )dx=((π(√π))/4)
$${x}\rightarrow\frac{\mathrm{1}}{{x}}\Rightarrow \\ $$$${A}=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\pi}{\mathrm{2}}−{arctan}\left({x}\right)}{\left(\frac{\mathrm{1}}{{x}}\right)^{−{ln}\left({x}\right)+\mathrm{1}} }.\frac{{dx}}{{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{{ln}\left({x}\right)+\mathrm{1}} }−\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right){dx}}{{x}^{{ln}\left({x}\right)+\mathrm{1}} } \\ $$$$\mathrm{2}{A}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{{ln}\left({x}\right)+\mathrm{1}} } \\ $$$${ln}\left({x}\right)={u} \\ $$$${A}=\frac{\pi}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{{du}}{{e}^{{u}^{\mathrm{2}} } }=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} .{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{4}}\sqrt{\pi} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right)}{{x}^{{ln}\left({x}\right)+\mathrm{1}} }{dx}=\frac{\pi\sqrt{\pi}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Jun/21
thankd alot..
$${thankd}\:{alot}.. \\ $$
Commented by mindispower last updated on 14/Jun/21
pleasur
$${pleasur} \\ $$

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