<s>[INST] 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.
 [/INST] $$I=\int_{0}^{\infty}\frac{e^{-xt}}{1+t}\,dt=e^x \,\Gamma (0,x)$$
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 $0^-$.
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