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.
Best Answer
The first approach gets $\frac1R\int_{-\pi/2}^0\cos\varphi d\varphi=\frac1R$; the second gets$$\frac1R[x(x^2+R^2)^{-1/2}]_{-\infty}^0=-\frac1R\lim_{x\to-\infty}x(x^2+R^2)^{-1/2}=-\frac1R\cdot-1=\frac1R,$$contra your sign error. You mistakenly rewrote $x(x^2+R^2)^{-1/2}$ as $(1+R^2/x^2)^{-1/2}$, rather than $-(1+R^2/x^2)^{-1/2}$, for $x<0$. Bear in mind $\frac{\sqrt{a^2+b}}{a}=\frac{\sqrt{1+b/a^2}}{\operatorname{sgn}a}$ for $a\ne0$.