I am aware that $\operatorname{Ei}(x)$ is indeed the antiderivative of $\dfrac{e^x}{x}$. However, the exponential integral is defined as: $$\operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt$$The integrand has $e^{-t}$ instead of $e^t$, and the limits of integration are not, for example, from $0$ to $x$. So, how exactly does it follow that it's the antiderivative of $\dfrac{e^x}{x}$?
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$\begingroup$You have \begin{align} \operatorname{Ei}(x)=\int_{\infty}^{-x}\frac{e^{-t}}t\,dt. \end{align}Then, using the Chain Rule,$$ \operatorname{Ei}'(x)=\left.\frac{e^{-t}}t\right|_{t=-x}\,\times (-1)=\frac{e^x}x. $$Since you are concerned with $\infty$, fix some $K>0$ (to avoid problems at the origin) and then $$ \operatorname{Ei}(x)=-\int_{-x}^K\frac{e^{-t}}t\,dt-\int_{K}^\infty\frac{e^{-t}}t\,dt. $$The second integral is a constant, so it disappears when you differentiate.
$\endgroup$ 2 $\begingroup$Covering what happens for any $x\ne0$.
The exponential integral is actually defined in terms of the Cauchy principal value, meaning that for $x>0$, we define
$$\operatorname{Ei}(x)=-\lim_{\epsilon\to0^+}\left(\int_{-x}^{-\epsilon}\frac{e^{-t}}t~\mathrm dt+\int_\epsilon^\infty\frac{e^{-t}}t~\mathrm dt\right)$$
For $x>0$ we may also split the integral up as follows:
$$\operatorname{Ei}(x)=\operatorname{Ei}(1)-\int_{-x}^{-1}\frac{e^{-t}}t~\mathrm dt$$
And likewise for $x<0$:
$$\operatorname{Ei}(x)=\operatorname{Ei}(-1)-\int_{-x}^1\frac{e^{-t}}t~\mathrm dt$$
which are both easily differentiated using the fundamental theorem of calculus, now that we have finite bounds, and the chain rule to get
$$\operatorname{Ei}'(x)=\frac{e^x}x$$
Note that where you choose to split the integral is arbitrary. As long as you choose to split the integral on the same side of zero as $x$ the above will work.
$\endgroup$ $\begingroup$$\DeclareMathOperator{\Ei}{Ei}$Substitute $x=-t$ so that$$\Ei(x)=-\int_{-x}^\infty\frac{e^{-t}}{t}dt=-\int_{-t}^\infty \frac{e^{-t}}{-t}(-dt)=\int_{-\infty}^x\frac{e^x}{x}dx.$$Do you see it now?
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