Would appreciate any help in finding a good closed form upper bound on this integral not in terms of the exponential integral $Ei(x)$:
$$ I = \int_{t=0}^x \frac{e^t-1}{t}dt$$
So far I have tried using the Cauchy-Schwartz inequality to get something of the form $\frac{e^x-1}{\sqrt{x}}$.
Best Answer
If $x$ is close to zero the Maclaurin series says pretty much everything: $$ g(x)=\int_{0}^{x}\frac{e^t-1}{t}\,dt = \sum_{n\geq 1}\frac{x^n}{n\cdot n!} \tag{1} $$ and also proves that the LHS is an entire function. We may notice that de l'Hopital rule implies $$ \lim_{x\to +\infty}\frac{g(x)}{\frac{e^x-1}{x}} = 1 \tag{2}$$ and $\frac{e^x-1}{x}=\sum_{n\geq 1}\frac{x^n}{(n+1)\cdot n!}$ is actually a decent lower bound for $g(x)$. In order to turn it into an upper bound, we may notice that for any $n\geq 2$ we have $\frac{1}{n}\leq \frac{1}{n+1}+\frac{2}{(n+1)(n+2)}$. This leads to $$ g(x)\leq \frac{6(x+2)e^x-\left(12+18 x+12 x^2-x^3\right)}{6x^2}\tag{3} $$ which is horrible but accurate.