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Question Number 192841 by TUN last updated on 29/May/23

Answered by witcher3 last updated on 02/Jun/23

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)(1+{\mathrm{e}}^{\mathrm{bx}})}⩽{\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)}={2\tan }^{-1}\left(\mathrm{a}\right)$$$$\mathrm{integral}\mathrm{cv}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_{\mathrm{a}}^{-\mathrm{a}}\frac{\mathrm{d}(-\mathrm{x})}{(1+{\mathrm{x}}^2)(1+{\mathrm{e}}^{-\mathrm{bx}})}$$$$={\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{{\mathrm{e}}^{\mathrm{bx}}}{(1+{\mathrm{x}}^2)(1+{\mathrm{e}}^{\mathrm{bx}})}\mathrm{dx}$$$$2\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{{\mathrm{e}}^{\mathrm{bx}}+1}{(1+{\mathrm{x}}^2)(1+{\mathrm{e}}^{\mathrm{bx}})}\mathrm{dx}$$$$={\int }_{-\mathrm{a}}^{\mathrm{a}}\frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)}={2\tan }^{-1}\left(\mathrm{a}\right)$$$$\mathrm{f}\left(\mathrm{a}\right)={\tan }^{-1}\left(\mathrm{a}\right)\Rightarrow \underset{\mathrm{a}\to \mathrm{\infty }}{\lim f}\left(\mathrm{a}\right)=\frac{\pi }{2}$$$$$$$$$$$$$$

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