oc.optimization control - univariate prior corresponding to weighted sum of L1 and L2 penalties?

So here's my attempt at the algebra to find the normalizing constant:

$Z_{lambda,alpha} =displaystyleint exp( -lambda(alpha x^2 + (1 - alpha)|x| ) ) dx$

$ = 2 displaystyleint^{0}_{-infty} exp( -lambda(alpha x^2 + (1 - alpha)x ) ) dx$

$ = 2 displaystyleint^{0}_{-infty} exp( -lambda alpha x^2 -lambda (1 - alpha)x) dx$

$ = 2 displaystyleint^{0}_{-infty} exp(-ax^2-bx) dx$

where $a = lambda alpha$ and $b = lambda (1 - alpha)$. From Mathematica I get

$displaystyleint exp(-ax^2-bx) dx = frac{sqrt{pi}}{2 sqrt{a}} exp left( frac{b^2}{4a} right) mbox{erf} left( frac{2ax+b}{2 sqrt{a}} right)$

So continuing

$Z_{lambda,alpha} = left. 2 dfrac{sqrt{pi}}{2 sqrt{a}} exp left( dfrac{b^2}{4a} right) mbox{erf} left( dfrac{2ax+b}{2 sqrt{a}} right) right|^{0}_{-infty}$

$ = left. dfrac{sqrt{pi}}{sqrt{a}} exp left( dfrac{b^2}{4a} right) left( 2Phi left( sqrt{2} dfrac{2ax+b}{2 sqrt{a}} right) - 1 right) right|^{0}_{-infty}$

$ = left. dfrac{sqrt{pi}}{sqrt{a}} exp left( dfrac{b^2}{4a} right) left( 2Phi left( dfrac{2ax+b}{ sqrt{2a}} right) - 1 right) right|^{0}_{-infty}$

$ = left. dfrac{sqrt{pi}}{sqrt{a}} exp left( dfrac{b^2}{4a} right) left( 2Phi left( dfrac{2ax+b}{ sqrt{2a}} right) - 1 right) right|^{0}_{-infty}$

$ = 2 dfrac{sqrt{pi}}{sqrt{a}} exp left( dfrac{b^2}{4a} right) Phi left( dfrac{b}{ sqrt{2a}} right) $

$ = 2 dfrac{sqrt{pi}}{sqrt{alpha lambda}} exp left( dfrac{lambda^2(1-alpha)^2}{4lambda alpha} right) Phi left( dfrac{lambda(1-alpha)}{ sqrt{2lambda alpha}} right) $

$ = 2 dfrac{sqrt{pi}}{sqrt{alpha lambda}} exp left( dfrac{lambda(1-alpha)^2}{4 alpha} right) Phi left( dfrac{sqrt{lambda}(1-alpha)}{ sqrt{2 alpha}} right) $

Am I going wrong somewhere?