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Question Number 8691 by FilupSmith last updated on 22/Oct/16

∫_0 ^( ∞) x^(−ln(x)) dx

$$\int_{\mathrm{0}} ^{\:\infty} {x}^{−\mathrm{ln}\left({x}\right)} {dx} \\ $$

Commented by FilupSmith last updated on 22/Oct/16

x^(−ln(x)) =e^(−ln(x)ln(x)) =e^(−ln^2 (x))   u=ln(x) ⇒ x=e^u   =e^(−u^2 )   (du/dx)=(1/x) ⇒ dx=xdu=e^u du      x=0 ⇒ u=−∞  x=∞ ⇒ u=∞     =∫_(−∞) ^( ∞) e^(−u^2 ) e^u du  =∫_(−∞) ^( ∞) e^(u−u^2 ) du  continue

$${x}^{−\mathrm{ln}\left({x}\right)} ={e}^{−\mathrm{ln}\left({x}\right)\mathrm{ln}\left({x}\right)} ={e}^{−\mathrm{ln}^{\mathrm{2}} \left({x}\right)} \\ $$$${u}=\mathrm{ln}\left({x}\right)\:\Rightarrow\:{x}={e}^{{u}} \\ $$$$={e}^{−{u}^{\mathrm{2}} } \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{1}}{{x}}\:\Rightarrow\:{dx}={xdu}={e}^{{u}} {du} \\ $$$$\:\: \\ $$$${x}=\mathrm{0}\:\Rightarrow\:{u}=−\infty \\ $$$${x}=\infty\:\Rightarrow\:{u}=\infty \\ $$$$\: \\ $$$$=\int_{−\infty} ^{\:\infty} {e}^{−{u}^{\mathrm{2}} } {e}^{{u}} {du} \\ $$$$=\int_{−\infty} ^{\:\infty} {e}^{{u}−{u}^{\mathrm{2}} } {du} \\ $$$${continue} \\ $$

Answered by prakash jain last updated on 23/Oct/16

=∫_(−∞) ^∞ e^(u−u^2 ) du  =∫_(−∞) ^∞ e^(u−u^2 −(1/4)+(1/4)) du  =e^(1/4) ∫_(−∞) ^∞ e^(−(u−(1/2))^2  ) du  (u−(1/2))=v⇒u=v+(1/2)  u=∞,v=∞ ,  u=−∞,v=−∞  =e^(1/4) ∫_(−∞) ^∞ e^(−v^2 dv)   =e^(1/4) (√π)

$$=\int_{−\infty} ^{\infty} {e}^{{u}−{u}^{\mathrm{2}} } {du} \\ $$$$=\int_{−\infty} ^{\infty} {e}^{{u}−{u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}} {du} \\ $$$$={e}^{\mathrm{1}/\mathrm{4}} \int_{−\infty} ^{\infty} {e}^{−\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:} {du} \\ $$$$\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)={v}\Rightarrow{u}={v}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${u}=\infty,{v}=\infty\:,\:\:{u}=−\infty,{v}=−\infty \\ $$$$={e}^{\mathrm{1}/\mathrm{4}} \int_{−\infty} ^{\infty} {e}^{−{v}^{\mathrm{2}} {dv}} \\ $$$$={e}^{\mathrm{1}/\mathrm{4}} \sqrt{\pi} \\ $$

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