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 $\frac{e^x-1}{\sqrt{x}}$.
On
$$\int_0^x\dfrac{e^t-1}{t}dt=\int_0^x\int_0^1e^{at}dadt=\int_0^1\int_0^xe^{at}dtda\leq\int_0^1\int_0^xe^{ax}dtda=e^x-1$$
On
The nice upper bound given by Jack’s answer fits in a whole sequence of ever tighter upper bounds. For any $n \in \mathbb{N}$ define the polynomial $$e_n(x)=\sum_{k=0}^n\frac{x^k}{k!}.$$ Repeating partial integration $n$ times shows that $$\int_0^x\frac{e^t-1}t\mathrm{d}t = \sum_{k=1}^n (k-1)! \frac{e^x-e_k(x)}{x^k}+n! \int_0^x \frac{e^t-e_n(t)}{t^{n+1}}\mathrm{d}t.$$ Now any upper bound for the integral in the right hand side results in an upper bound for the left hand side. Simply using the trapezium rule results in $$\int_0^x\frac{e^t-1}t\mathrm{d}t \leq \sum_{k=1}^n(k-1)! \frac{e^x-e_k(x)}{x^k} + n!\frac{e^x-e_n(x)}{2x^n} + \frac{x}{2n+2}.$$ The first couple of bounds are then: $$\tag{$n$=0} \frac{e^x-1}2 + \frac{x}2$$ $$\tag{$n$=1} \frac{3(e^x-(x+1))}{2x} + \frac{x}4$$ $$\tag{$n$=2} \frac{(x+2)e^x - (2x^2+3x+2)}{x^2} + \frac{x}6$$ $$\tag{$n$=3} \frac{6(x^2+x+5)e^x - (14 x^3 + 27 x^2 + 36 x + 30)}{6x^3} + \frac{x}8$$
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.