How can we prove $$\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)<\int_{x}^{\infty}\frac{e^{-t}}{t}dx<e^{-x}\ln\left(1+\frac{1}{x}\right)\;, x>0$$
$\bf{My\; Try::}$ Let $\displaystyle f(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dx\;,$ Then $\displaystyle f'(x) = -\frac{e^{x}}{x}<0\;\forall x>0$
So $f(x)$ is Strictly Decreasing function.
Now How can I estimate above function, Help required, Thanks
Upper Bound Inequality
Note that we have the elementary inequalities
$$\frac{1}{x+1}\le \log\left(1+\frac1x\right)\le \frac1x \tag 1$$
Using the left-hand side inequality in $(1)$, it is easy to see that
$$\frac{1}{x}\le \log\left(1+\frac1x\right) -\left(\frac{1}{x+1}-\frac{1}{x}\right) \tag 2$$
Multiplying $(2)$ by $e^{-x}$ and integrating, we find that
$$\begin{align} \int_x^{\infty}\frac{e^{-t}}{t}\,dt& \le \int_x^\infty \left(e^{-t}\log\left(1+\frac1t\right) -e^{-t}\left(\frac{1}{t+1}-\frac{1}{t}\right)\right)\,dt\\\\ &=-\int_x^\infty \frac{d}{dt}\left(e^{-t}\log\left(1+\frac1t\right)\right)\,dt\\\\ &=e^{-x}\log\left(1+\frac1x\right) \end{align}$$
Lower Bound Inequality
To establish the lower bound, we first show that the logarithm function satisfies the inequality
$$\log\left(1+\frac2x\right)\le \frac1x+\frac1{x+2} \tag 2$$
for $x>0$.
To show that $(2)$ is valid, we proceed as follows.
Define a function $f(x)$ to be
$$f(x)=\log\left(1+\frac2x\right)- \frac1x-\frac1{x+2}$$
Then, the derivative of $f(x)$ is given by
$$\begin{align} f'(x)&=\frac{1}{x+2}-\frac1x+\frac{1}{x^2}+\frac{1}{(x+2)^2}\\\\ &=\frac{4}{x^2(x+2)^2}\\\\ &>0 \end{align}$$
Inasmuch as $f(x)$ is monotonically increasing with $\lim_{x\to \infty}f(x)=0$, then $f(x)< 0$.
Now, using $(2)$, it is easy to see that
$$\frac12\left(\frac1x-\frac{1}{x+2}\right)+\frac12\log\left(1+\frac2x\right)<\frac1x \tag 3$$
Multiplying both sides of $(3)$ by $e^{-x}$ and integrating yields
$$\begin{align} \int_x^\infty \frac{e^{-t}}{t}\,dt &> \frac12\int_x^\infty \left(e^{-t}\left(\frac1t-\frac{1}{t+2}\right)+e^{-t}\log\left(1+\frac2x\right)\right)\,dt\\\\ &=-\frac12 \int_x^\infty \frac{d}{dt}\left(e^{-t}\log\left(1+\frac2t\right)\right)\,dt\\\\ &=\frac12 e^{-x}\log\left(1+\frac2x\right) \end{align}$$
as was to be shown!
