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Question Number 32339 by abdo imad last updated on 23/Mar/18

$$calculate{\int }_0^{+\mathrm{\infty }}\frac{th\left(3x\right)-th\left(2x\right)}{x}dx.$$

Commented by abdo imad last updated on 24/Mar/18

$$I=lim{}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)withI\left(\xi \right)={\int }_0^{\xi }\frac{th\left(3x\right)-th\left(2x\right)}{x}dx$$$$I\left(\xi \right)={\int }_0^{\xi }\frac{th\left(3x\right)}{x}dx-{\int }_0^{\xi }\frac{th\left(2x\right)}{x}dxbutch.3x=tgive$$$${\int }_0^{\xi }\frac{th\left(3x\right)}{x}dx={\int }_0^{3\xi }\frac{th\left(t\right)}{\frac{t}{3}}\frac{dt}{3}={\int }_0^{3\xi }\frac{th\left(t\right)}{t}dtalsowehave$$$${\int }_0^{\xi }\frac{th\left(2x\right)}{x}dx={\int }_0^{2\xi }\frac{th\left(t\right)}{t}dt\Rightarrow$$$$I\left(\xi \right)={\int }_0^{3\xi }\frac{th\left(t\right)}{t}dt-{\int }_0^{2\xi }\frac{th\left(t\right)}{t}dt={\int }_{2\xi }^{3\xi }\frac{th\left(t\right)}{t}dtbut$$$$\mathrm{\exists }c\in ]2\xi ,3\xi [/I\left(\xi \right)=th\left(c\right){\int }_{2\xi }^{3\xi }\frac{dt}{t}=ln\left(\frac{3}{2}\right)th\left(c\right)\Rightarrow$$$${lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)=ln\left(3\right)-ln\left(2\right).(lookthat{lim}_{c\to +\mathrm{\infty }}thc=1)$$$$$$

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