A limit wrong using Wolfram Alpha

Some more observations on Mathematica's behavior: it yields

$$ \frac{d}{dx} \int_1^{x} \frac{e^{-t}}{t} \, dt = \frac{e^{-x} - 1}{x} $$

restricted to $\Im(x) \neq 0$ or $\Re(x) \geq 0$. But changing things slightly gives

$$ \frac{d}{dt} \int \frac{e^{-t}}{t} \, dt = \frac{e^{-t}}{t}. $$

I had originally suspected there was something fishy with the branch cut: Mathematica computes

$$ \int_1^x \frac{e^{-t}}{t} \, dt = -\mathrm{Ei}(-1) - \log(x) - \Gamma(0,x) $$

again restricted to $\Im(x) \neq 0 \vee \Re(x) \geq 0$. However:

• The point we are interested in is away from the branch discontinuity
• I would have expected it to get the derivative right even if there were weird branch issues

(using $x+1$ in the above instead of $x$ does not make any qualitative difference)

Without limits, it computes

$$ \int \frac{e^{-t}}{t} \, dt = \mathrm{Ei}(-t) \color{gray}{+ \mathrm{constant}}$$

If you shift the integrand, you get

$$ \frac{d}{dx} \int_0^x \frac{e^{-(u+1)}}{u+1} \, du = \frac{e^{-(u+1)}}{u+1} $$

and correspondingly

$$ \lim_{y \to 0} \frac{ \int_0^y \frac{e^{-(u+1)}}{u+1} \, du }{3 y} = \frac{1}{3e} $$

(I substituted $x^2 = y$ so that wolfram would finish the calculation for me. This substitution does not make a qualitative difference in the original)

I think the key difference is that in the first version, the branch point is $t=0$, and Mathematica focuses on the behavior there -- which is inherently weird and strange because it's a branch point (and given that, I'm not sure if using the result leads to computing something correct but strange, or something ill-defined). But in the second version, the branch point is at $u=-1$, but Mathematica still focuses on $u=0$ so it gets sane results.

Mathematica is wrong because everything in sight is positive.

Unfortunately I cannot say what Mathematica is doing wrong.