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Question Number 156455 by SANOGO last updated on 11/Oct/21

Commented by prakash jain last updated on 11/Oct/21

$$0$$

Commented by SANOGO last updated on 11/Oct/21

$$demonstrationsipossible$$

Commented by prakash jain last updated on 11/Oct/21

$$\mathrm{put}t=\frac{1}{u}$$$$\mathrm{calculate}2I$$

Commented by prakash jain last updated on 11/Oct/21

$$onewithtandanotheronewith$$$$t=\frac{1}{u}$$

Answered by Mathspace last updated on 11/Oct/21

$$I={\int }_{\frac{1}{x}}^{x}f\left(t\right)dt\Rightarrow I{=}_{t=\frac{1}{u}}{\int }_x^{\frac{1}{x}}f\left(\frac{1}{u}\right)\frac{-du}{u^2}$$$$={\int }_{\frac{1}{x}}^x\frac{-u^{2}f\left(u\right)}{u^2}du=-{\int }_{\frac{1}{x}}^{x}f\left(u\right)du$$$$=-(F\left(x\right)-F\left(\frac{1}{x}\right))=F\left(\frac{1}{x}\right)-F\left(x\right)$$$$Festlaprimitivedef$$$$(jepensequilyaerreurdans$$$$laQ!)$$

Commented by SANOGO last updated on 11/Oct/21

$$mercibien$$

Commented by prakash jain last updated on 12/Oct/21

$$I={\int }_{1/x}^x\frac{1}{u^2}f\left(\frac{1}{u}\right)du={\int }_{1/x}^{x}f\left(u\right)du$$$$x^{2}f\left(x\right)+f\left(\frac{1}{x}\right)=0(x>0)$$$$f\left(x\right)+\frac{1}{x^2}f\left(\frac{1}{x}\right)=0$$$$2I={\int }_{1/x}^x\frac{1}{u^2}f\left(\frac{1}{u}\right)du+{\int }_{1/x}^{x}f\left(u\right)du$$

Commented by SANOGO last updated on 12/Oct/21

$$mercibien$$

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