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Question Number 163574 by Zaynal last updated on 08/Jan/22

$$\mathit{S}\mathit{h}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{a}\mathit{t};$$$${\int }_1^{\mathrm{\infty }}\frac{\mathit{I}\mathit{n}\mathit{x}}{{\mathit{x}}^4}\mathit{d}\mathit{x}=\frac{1}{9}$$

Answered by Ar Brandon last updated on 08/Jan/22

$${\int }_1^{\mathrm{\infty }}\frac{\ln x}{x^4}dx=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}{\int }_1^{\mathrm{\infty }}\frac{x^{\alpha }}{x^4}dx=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}{\left[\frac{x^{\alpha -3}}{\alpha -3}\right]}_1^{\mathrm{\infty }}$$$$=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}\frac{1}{3-\alpha }={\mid }_{\alpha =0}\frac{1}{{(3-\alpha )}^2}=\frac{1}{3^2}=\frac{1}{9}$$

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