Figure A1.

Plots of Rð Þε as a function of the point defect's position in the wire ε for various energies. The dashed line is the small ε=w < < 1 approximation (Eq. (39)).

Single-Atom Field-Effect Transistor DOI: http://dx.doi.org/10.5772/intechopen.81526

$$\ln \mathcal{R}(\varepsilon) \equiv \lim\_{\rho \to 0} \left[ \ln \rho + 2 \int\_{\pi 3 \rho}^{\infty} \frac{\sin^2(\xi e/\rho)}{\sqrt{\xi^2 - (4\pi \rho)^2/w^2}} \exp \left( -\xi^2/4 \right) d\xi \right]. \tag{A2}$$

And due to the limit,

$$\mathcal{R}(\varepsilon) \cong \exp \lim\_{\rho \to 0} \left[ \ln \rho + 2 \int\_0^\infty \frac{\sin^2(\xi \varepsilon/\rho)}{\xi} \exp \left( -\xi^2/4 \right) d\xi \right]. \tag{A3}$$

Now, since for ρ=w < < 1

$$\ln\left(\rho/w\right) \cong \ln 2 - \gamma/2 - \int\_{\rho/w}^{\infty} \frac{\exp\left(-\xi^2/4\right)}{\xi} d\xi,\tag{A4}$$

then

<sup>2</sup> <sup>3</sup> <sup>∞</sup><sup>ð</sup> � � <sup>∞</sup><sup>ð</sup> <sup>6</sup> exp �ξ<sup>2</sup> <sup>=</sup><sup>4</sup> sin <sup>2</sup>ðξε=ρ<sup>Þ</sup> � �ξ<sup>2</sup> � <sup>7</sup> <sup>R</sup>ðεÞ ffi exp lim 4 ln <sup>2</sup><sup>w</sup> � <sup>γ</sup>=<sup>2</sup> � <sup>d</sup><sup>ξ</sup> <sup>þ</sup> <sup>2</sup> exp <sup>=</sup><sup>4</sup> <sup>d</sup>ξ<sup>5</sup> <sup>¼</sup> <sup>ρ</sup>!<sup>0</sup> <sup>ξ</sup> <sup>ξ</sup> <sup>2</sup> <sup>ρ</sup>=<sup>w</sup> <sup>0</sup> <sup>3</sup> : (A5) <sup>∞</sup> <sup>ρ</sup><sup>ð</sup> =w <sup>6</sup> <sup>ð</sup> cosð2ξε=ρ<sup>Þ</sup> � � sin <sup>2</sup>ðξε=ρ<sup>Þ</sup> � � <sup>7</sup> exp lim 4 ln <sup>2</sup><sup>w</sup> � <sup>γ</sup>=<sup>2</sup> � exp �ξ<sup>2</sup> <sup>=</sup><sup>4</sup> <sup>d</sup><sup>ξ</sup> <sup>þ</sup> <sup>2</sup> exp �ξ<sup>2</sup> <sup>=</sup><sup>4</sup> <sup>d</sup>ξ<sup>5</sup> <sup>ρ</sup>!<sup>0</sup> <sup>ξ</sup> <sup>ξ</sup> <sup>ρ</sup>=<sup>w</sup> <sup>0</sup>

Moreover, since, ρ=ε ! 0 but also ε=w ! 0, then in both integrals, the exponents can be ignored, i.e.,

$$R(\varepsilon) \cong \exp \lim\_{\rho \to 0} \left[ \ln 2w - \gamma/2 - \int\_{\rho/w}^{\infty} \frac{\cos \left(2\xi \varepsilon/\rho\right)}{\xi} d\xi + 2 \int\_{0}^{\rho/w} \frac{\sin^2(\xi \varepsilon/\rho)}{\xi} d\xi \right] \tag{A6}$$

which finally yields the analytical expression

$$R(\varepsilon) \cong \exp\left[\ln 2w - \chi/2 + \text{Ci}(2\varepsilon/w) + \varepsilon^2/w^2\right] = 4\varepsilon \exp\left(\frac{\chi}{2}\right),\tag{A7}$$

where Ci(x) is the cosine integral function [22].
