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

$$help$$$${\int }_1^{\mathrm{\infty }}{x}^{-ln\left(x\right)}dx$$$$$$

Answered by Berbere last updated on 30/May/24

$${\int }_0^{\mathrm{\infty }}{\left(e^t\right)}^{(-t)}{e}^{t}dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-t^2+t}dt={\int }_0^{\mathrm{\infty }}{e}^{-{(t-\frac{1}{2})}^2+\frac{1}{4}}dt$$$$=e^{\frac{1}{4}}{\int }_{-\frac{1}{2}}^{\mathrm{\infty }}{e}^{-x^2}dx=e^{\frac{1}{4}}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx+e^{\frac{1}{4}}{\int }_{-\frac{1}{2}}^0{e}^{-x^2}dx$$$$A=e^{\frac{1}{4}}.\frac{\sqrt{\pi }}{2}-e^{\frac{1}{4}}\frac{\sqrt{\pi }}{2}erf(-\frac{1}{2})$$$$erf\left(x\right)=\frac{2}{\sqrt{\pi }}{\int }_0^x{e}^{-t^2}dt\Rightarrow erf(-x)=-erf\left(x\right)$$$$A=\frac{\sqrt{\pi }}{2}{e}^{\frac{1}{4}}(1+erf\left(\frac{1}{2}\right))$$

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