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Question Number 207906 by nachosam last updated on 30/May/24

help ∫_1 ^( ∞) x^(−ln(x)) dx

$${help} \\ $$$$\int_{\mathrm{1}} ^{\:\infty} {x}^{−{ln}\left({x}\right)} {dx} \\ $$$$ \\ $$

Answered by Berbere last updated on 30/May/24

∫_0 ^∞ (e^t )^((−t)) e^t dt =∫_0 ^∞ e^(−t^2 +t) dt=∫_0 ^∞ e^(−(t−(1/2))^2 +(1/4)) dt =e^(1/4) ∫_(−(1/2)) ^∞ e^(−x^2 ) dx=e^(1/4) ∫_0 ^∞ e^(−x^2 ) dx+e^(1/4) ∫_(−(1/2)) ^0 e^(−x^2 ) dx A=e^(1/4) .((√π)/2)−e^(1/4) ((√π)/2)erf(−(1/2)) erf(x)=(2/( (√π)))∫_0 ^x e^(−t^2 ) dt⇒erf(−x)=−erf(x) A=((√π)/2)e^(1/4) (1+erf((1/2)))

$$\int_{\mathrm{0}} ^{\infty} \left({e}^{{t}} \right)^{\left(−{t}\right)} {e}^{{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} +{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}} {dt} \\ $$$$={e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}={e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}+{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{0}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${A}={e}^{\frac{\mathrm{1}}{\mathrm{4}}} .\frac{\sqrt{\pi}}{\mathrm{2}}−{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${erf}\left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}\Rightarrow{erf}\left(−{x}\right)=−{erf}\left({x}\right) \\ $$$${A}=\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{erf}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$

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