I am using http://www.math.mcgill.ca/gantumur/math581w12/downloads/pseudodiff.pdf as a self study on Pseudo-Differential Operators and Distribution Theory.
In example 3 on page 6 the following statement is made: $$ \left(\delta'*\varphi\right)\left(0\right)={\int_R}-e^{-\left(1+x\right)^2}\frac{2x}{\varepsilon^3\sqrt{\pi}} e^{-\left(x/\varepsilon\right)^2}dx=\frac{2}{\left(1+\varepsilon^2\right)^{^{\frac{3}{2}}}e^{^{\frac{1}{1+\varepsilon^2}}}} \label{Eq1}\tag{Equation 1} $$ where $$ {\varphi_\varepsilon}'\left(x\right)=-\frac{2x}{\varepsilon^3\sqrt{\pi}}e^{-\left(x/\varepsilon\right)^2}=\frac{1}{\varepsilon\sqrt{\pi}}\frac{d}{dx}\left(e^{-\left(x/\varepsilon\right)^2}\right) $$ is being used to approximate the derivative of the Dirac delta function $\delta'$ and $(\delta'*\varphi)(0)$ is the convolution of $e^{-(1-x)^2}$ with ${\varphi_\varepsilon}'(x)$ at $x=0$.
My question is: How is the right most equality in \eqref{Eq1} derived?
The starting expression is: \[ {\int_R}e^{-(1+x)^2}\frac{-2x}{\varepsilon^3\sqrt{\pi}}e^{-\left(\frac{x}{\varepsilon}\right)^2}dx= \] which is equivalent to \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}{\int_R}xe^{-\left[\left(1+x\right)^2+x^2/\varepsilon\right]}dx\;\;\;. \] Since, \[ \left(1+x\right)^2+x^2/\varepsilon^2=\left(1+\frac{1}{\varepsilon^2}\right)x^2+2x+1=\left(\frac{\varepsilon^2+1}{\varepsilon^2}\right)x^2+2x+1 \] equality follows as \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}{\int_R}xe^{-{\frac{\varepsilon^2+1}{\varepsilon^2}}\left[x^2+\frac{2\varepsilon^2}{\varepsilon^2+1}x+\frac{\varepsilon^2}{\varepsilon^2+1}\right]}dx\;\;\;. \] Let $\alpha=\frac{\varepsilon}{\sqrt{\varepsilon^2+1}}$; then $\alpha^2=\frac{\varepsilon^2}{\varepsilon^2+1}$ and $\frac{1}{\alpha^2}=\frac{\varepsilon^2+1}{\varepsilon^2}$ so that \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}{\int_R}xe^{{\frac{-1}{\alpha^2}}\left(x^2+2\alpha^2x+\alpha^2\right)}dx\;\;\;. \] Completing the square on $x^2+2\alpha^2x+\alpha^2$ produces \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}{\int_R}xe^{\frac{-1}{\alpha^2}\left[\left(x+\alpha^2\right)^2+\frac{\alpha^2}{\varepsilon^2+1}\right]}dx \] \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}{\int_R}xe^{\frac{-1}{\alpha^2}\left(x+\alpha^2\right)^2-\frac{1}{\varepsilon^2+1}}dx \] \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}{\int_R}xe^{-\left(x+\alpha^2\right)^2/\alpha^2}dx \] Let $y=x+\alpha^2$, then $dy=dx$ and $x=y-\alpha^2$. Then the equality follows as \[ =\frac{-2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}{\int_R}\left(y-\alpha^2\right)e^{-y^2/\alpha^2}dy \] \[ =\frac{2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}\left[{\int_R}\alpha^2e^{-y^2/\alpha^2}dy-{\int_R}ye^{-y^2/\alpha^2}dy\right] \] and since the right most integral evaluates to zero, it follows that \[ =\frac{2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}{\int_R}\alpha^2e^{-y^2/\alpha^2}dy\;\;.\;\;\;\;\left(\text{Equation 2}\right) \] Lemma 1: \[ \text{Let}\;\;\;I_R={\int_R}\alpha^2e^{-y^2/\alpha^2}dy\;,\;\;\;\text{then} \] \[ I_R^2=\left({\int_R}\alpha^2e^{-y^2/\alpha^2}dy\right)\left({\int_R}\alpha^2e^{-x^2/\alpha^2}dx\right)={\int}{\int_R}\alpha^4e^{-\left(y^2+x^2\right)/\alpha^2}dy:dx \] where a switch to polar form produces \[ I_R^2=\alpha^4{\int\limits_{r=0}^{\infty}}\;\;{\int\limits_{\theta=0}^{2\pi}}e^{-r^2/\alpha^2}\left(r:dr\right)d\theta=2\pi\alpha^4{\int\limits_{r=0}^{\infty}}e^{-r^2/\alpha^2}r:dr=2\pi\alpha^4{\int\limits_{r=0}^{\infty}}e^{-r^2/\alpha^2}\left(\frac{-2r}{\alpha^2}dr\right)\left(\frac{-\alpha^2}{2}\right) \] and \[ I_R^2=2\pi\alpha^4\left(\frac{-\alpha^2}{2}\right){\int\limits_{r=0}^{\infty}}e^{-r^2/\alpha^2}\left(\frac{-2r}{\alpha^2}dr\right) \] which reduces to \[ I_R^2=-\pi\alpha^6{\int\limits_{r=0}^{\infty}}e^{-r^2/\alpha^2}\left(\frac{-2r}{\alpha^2}dr\right)\;\;\;. \] The above integral evaluates as \[ I_R^2=\lim_{A\to\infty}\left(\pi\alpha^6-\pi\alpha^6e^{-A^2/\alpha^2}\right)=\pi\alpha^6\;\;\;. \] Since $\alpha=\frac{\varepsilon}{\sqrt{\varepsilon^2+1}}$, it follows that \[ I_R={\int_R}\alpha^2e^{-y^2/\alpha^2}dy=\alpha^3\sqrt{\pi}=\frac{\varepsilon^3\sqrt{\pi}}{\left(\varepsilon^2+1\right)^{3/2}} \;\;\;. \] Q.E.D. for lemma 1.
I now substitute the result of lemma 1 into Equation 2 where the equality follows as \[ =\frac{2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}{\int_R}\alpha^2e^{-y^2/\alpha^2}dy=\frac{2}{\varepsilon^3\sqrt{\pi}}e^{\frac{-1}{\varepsilon^2+1}}\left(\frac{\varepsilon^3\sqrt{\pi}}{\left(\varepsilon^2+1\right)^{3/2}}\right)=\frac{2}{{\left(\varepsilon^2+1\right)^{3/2}}e^{\frac{1}{\varepsilon^2+1}}}\;\;\;. \] This is the desired result. Q.E.D.