Calculating $\int_{-\infty}^{+\infty}\mathrm{d}x\frac{\exp(-x^2)}{x^2+1}$ with the Cauchy formula

Question

Chances are, I am missing something trivial here, but I ran into a problem when trying to calculate the following improper integral:

$$ I=\int_{-\infty}^{+\infty}\mathrm{d}x\frac{\exp(-x^2)}{x^2+1} . $$

My approach was to treat it as a complex integral, and use the Cauchy formula:

$$ I'=\int_{\Gamma}\mathrm{d}z\frac{\exp(-z^2)}{z^2+1}=2\pi i \left(\frac{\exp(-z^2)}{z+i}\right)_{z=i}=\pi e , $$

where $\Gamma$ is a closed curve (with the usual infinite-radius semicircle on the upper half-plane, denoted as $\gamma$). We can partition $I'$ as

\begin{align*} I'=& I+\int_{\gamma}\mathrm{d}z\frac{\exp(-z^2)}{z^2+1}= \\=& I+\underbrace{\lim_{R\rightarrow\infty}\int_0^{2\pi}\mathrm{d}\phi \frac{iR\exp(i\phi)\exp(-R^2\exp(i2\phi))}{R^2\exp(i2\phi)+1}}_{=0}\\ =&I . \end{align*}

Based on this, $I=\pi e$. However, the actual answer should be

$$ I=\pi e (1-\mathrm{erf}(1)) . $$

It seems as if I missed a $\pi e\mathrm{erf}(1)$ term on the right hand side, but I cannot see how...

I know the integral is evaluated in this SE post. But I would be grateful if someone could correct my Cauchy formula-based reasoning, if possible.

From what I see, $\exp\Big(-R^2\exp(\text{i}2\phi)\Big)$ can be exponentially big, so the claim that the limit as $R\to \infty$ of its integral being zero is most likely untrue. I think a better way to solve the problem is to note that the function $J:\mathbb{R}{\geq 0}\to \mathbb{R}$ defined by $$J(a):=\int{\mathbb{R}},\frac{\exp(-ax^2)}{x^2+1},\text{d}x$$ for every $a\geq 0$ satisfies $$J(a)-J'(a)=\int_{\mathbb{R}},\exp(-ax^2),\text{d}x,$$ with $J(0)=\pi$. Of course, $J(1)=I$. — Batominovski, Dec 22 '19 at 17:47
Related: Methods to solve $\int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}:dx$ and On $\int_0^\infty \frac{\exp(-x^2)}{1+x^2}dx=\frac{\pi e}2\text{erfc}(1)$ and Evaluating the integral $\int_{0}^{\infty} \frac{\exp(-u^2)}{1+u^2} , du$ — Maximilian Janisch, Dec 22 '19 at 17:49

Calculating $\int_{-\infty}^{+\infty}\mathrm{d}x\frac{\exp(-x^2)}{x^2+1}$ with the Cauchy formula

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