Prove that $\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x})$ for $x > 0$ 
Question: Let $x> 0$, show that
$$\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x}).$$

I tried the integration by parts, but somehow couldn't get the term involving natural logarithm of $x$. I am looking forward to some hint for approaching this problem.
 A: $$I=\int_{0}^{\infty}\frac{e^{-xt}}{1+t}\,dt=e^x \,\Gamma (0,x)$$
If $x$ is large, we have
$$I=\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3}-\frac{6}{x^4}+O\left(\frac{1}{x^5}\right)$$ while
$$J=\frac{1}{2} \log \left(1+\frac{2}{x}\right)=\frac{1}{x}-\frac{1}{x^2}+\frac{4}{3 x^3}-\frac{2}{x^4}+O\left(\frac{1}{x^5}\right)$$
$$I-J=\frac{2}{3 x^3}-\frac{4}{x^4}+O\left(\frac{1}{x^5}\right)$$
The case where $x$ is small seems to be more delicate.
What we know is that
$$I-J=\frac{1}{2} \log \left(\frac{1}{2 x}\right)-\gamma+O\left(x^1\right)$$
What remains to prove is that
$$(I-J)'=e^x \Gamma (0,x)-\frac{x+1}{x(x+2)}$$ is a increasing function which never cancels and tends to $0^-$.
A: Remark: I found a simpler proof.
Using the known identity (for $u\ge 0$; easy to prove by taking derivative)
$$\ln(1 + u) = \int_0^\infty \frac{1 - \mathrm{e}^{-us}}{s}\mathrm{e}^{-s}\mathrm{d} s,$$
we have
\begin{align*}
 &\int_0^\infty \frac{\mathrm{e}^{-x t}}{1 + t}\,\mathrm{d} t - \frac12\ln(1 + 2/x)\\[6pt]
 =\, & \int_0^\infty \frac{\mathrm{e}^{-x t}}{1 + t}\,\mathrm{d} t - \int_0^\infty \frac{1 - \mathrm{e}^{-(2/x)s}}{2s}\mathrm{e}^{-s}\mathrm{d} s\\[6pt]
 \overset{s = xt} =\, & \int_0^\infty \frac{\mathrm{e}^{-x t}}{1 + t}\,\mathrm{d} t - \int_0^\infty 
 \frac{1 - \mathrm{e}^{-2t}}{2t}\mathrm{e}^{-xt}\mathrm{d} t\\[6pt]
 =\,&\int_0^\infty \mathrm{e}^{-xt}
 \left(\frac{1}{1 + t} - \frac{1 - \mathrm{e}^{-2t}}{2t} \right)\mathrm{d} t\\[6pt]
 =\, & \int_0^\infty \frac{\mathrm{e}^{-xt}}{2t}
 \left( \mathrm{e}^{-2t} + \frac{t - 1}{t + 1} \right)\mathrm{d} t\\[6pt]
 \ge\,& 0
\end{align*}
where we have used
$\mathrm{e}^{-2t} + \frac{t - 1}{t + 1} \ge 0$
for all $t > 0$ (easy to prove by taking logarithm and taking derivative).
We are done.
