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Question Number 142475 by cherokeesay last updated on 01/Jun/21

Answered by qaz last updated on 01/Jun/21

$$\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)+{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)=0$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)+\frac{1}{{\mathrm{x}}^2}\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=0$$$$\mathrm{I}={\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}={\int }_{\mathrm{x}}^{\frac{1}{\mathrm{x}}}\mathrm{f}\left(\frac{1}{\mathrm{t}}\right)(-\frac{1}{{\mathrm{t}}^2})\mathrm{dt}={\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}\mathrm{f}\left(\frac{1}{\mathrm{t}}\right)\frac{\mathrm{dt}}{{\mathrm{t}}^2}$$$$=\frac{1}{2}{\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}[\mathrm{f}\left(\mathrm{t}\right)+\mathrm{f}\left(\frac{1}{\mathrm{t}}\right)\frac{1}{{\mathrm{t}}^2}]\mathrm{dt}=0$$

Commented by cherokeesay last updated on 01/Jun/21

$$thankyousir.$$

Answered by mathmax by abdo last updated on 01/Jun/21

$$\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)+{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=-{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{I}={\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}{=}_{\mathrm{t}=\frac{1}{\mathrm{z}}}{\int }_{\mathrm{x}}^{\frac{1}{\mathrm{x}}}\mathrm{f}\left(\frac{1}{\mathrm{z}}\right)(-\frac{\mathrm{dz}}{{\mathrm{z}}^2})$$$$={\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}-{\mathrm{z}}^2\mathrm{f}\left(\mathrm{z}\right)\frac{\mathrm{dz}}{{\mathrm{z}}^2}=-{\int }_{\frac{1}{\mathrm{x}}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=-\mathrm{I}\Rightarrow 2\mathrm{I}=0\Rightarrow \mathrm{I}=0$$

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