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Question Number 205451 by MrGHK last updated on 21/Mar/24

$$\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^x\frac{dt}{e^{2t}t}$$

Answered by Berbere last updated on 21/Mar/24

$$x\stackrel{f}{\to }{e}^{-2x};over[0,t]t⩽\frac{1}{2};\mathrm{\exists }c\in ]0,t[Such$$$$\Rightarrow f\left(t\right)-f\left(0\right)={tf}^{\prime }\left(c\right)$$$$\Rightarrow e^{-2t}-1=-2{te}^{-2c};e^{-2c}⩽0\Rightarrow e^{-2t}-1⩾-2t$$$$\Rightarrow e^{-2t}⩾1-2t$$$${\int }_0^x\frac{dt}{e^{2t}.t}=f\left(x\right);f^{\prime }\left(x\right)>0fincrese\Rightarrow (\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)⩾f\left(a\right);\mathrm{\forall }a\in {\mathbb{R}}_{+})$$$$\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)⩾f\left(\frac{1}{2}\right)={\int }_0^{\frac{1}{2}}\frac{e^{-2t}}{t}dt⩾{\int }_0^{\frac{1}{2}}\frac{1}{t}(1-2t)dt$$$$={[ln\left(t\right)-2]}_0^{\frac{1}{2}}=+\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=+\mathrm{\infty }$$$$orjustsaying;\frac{1}{e^{-2x}x}\stackrel{0}{\sim }\frac{1}{x}diverge$$

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