I will write
$$ J_{\pm} = \int_{0}^{\infty} xe^{-x}\cos\left(x\pm\frac{x^2}{\alpha}\right)\,\mathrm{d}x $$
so that your integral takes the form $\frac{1}{2}(J_{+} + J_{-})$. Then
\begin{align*}
J_{\pm}
&= \operatorname{Re}\left[ \int_{0}^{\infty} x\exp\left( -x + ix \pm \frac{ix^2}{\alpha}\right) \,\mathrm{d}x. \right]
\end{align*}
Now write $\mathbb{H}_{\text{right}} = \{ z \in \mathbb{C} : \operatorname{Re}(z) > 0 \}$. Then for each $a \in \mathbb{H}_{\text{right}}$, the map $z \mapsto \int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x$ is analytic on $\mathbb{H}_{\text{right}}$ and continuous on $\overline{\mathbb{H}_{\text{right}}}$. Moreover, if $a, z \in (0, \infty)$, then with $b = a^2/4z$,
\begin{align*}
\int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x
&= \int_{0}^{\infty} x \exp\bigg( -b \left(\frac{2z x}{a}+1\right)^2 + b \bigg) \, \mathrm{d}x \\
&= \frac{be^{b}}{z} \int_{1}^{\infty} (u-1) e^{-bu^2} \, \mathrm{d}u \\
&= \frac{be^{b}}{z} \left( \int_{1}^{\infty} u e^{-bu^2} \, \mathrm{d}u - \int_{0}^{\infty} e^{-bu^2} \, \mathrm{d}u + \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \right) \\
&= \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u.
\end{align*}
Then by the principle of analytic continuation, this holds for all $a \in \mathbb{H}_{\text{right}}$ and $z \in \overline{\mathbb{H}_{\text{right}}}$. Then plugging $a = 1-i$ and $z = z_{\pm} = \pm i/\alpha$, we get $b = b_{\pm} = \mp \alpha /2 \in \mathbb{R}$.
\begin{align*}
J_{\pm}
&= \operatorname{Re}\bigg[ \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \bigg] \\
&= -\frac{\sqrt{\pi} \, e^{b}}{4} \operatorname{Re}\bigg[ \frac{a}{z^{3/2}} \bigg],
\end{align*}
By noting that $a_{+}/z_{+}^{3/2} = -\sqrt{2}\,\alpha^{3/2}$ and $a_{-}/z_{-}^{3/2} = i\sqrt{2}\,\alpha^{3/2}$, we get
$$ J_{+} = \frac{\sqrt{2\pi}}{4} \alpha^{3/2} e^{-\alpha/2}, \qquad J_{-} = 0. $$
This complete the proof.
Best Answer
If $a=0$ then $\int_0^\infty e^{ax} x^b \, dx$ diverges because (at least) one of the integrals $$ I_1 = \int_0^1 x^b \, dx \, , \quad I_2 = \int_1^\infty x^b \, dx $$ diverges. ($I_1$ diverges if $b \le -1$, and $I_2$ diverges if $b \ge -1$.)
If $a > 0$ then choose an integer $k$ with $k+b > 0$, so that $$ e^{ax} x^b \ge \frac{(ax)^k}{k!} x^b = \frac{a^k}{k!} x^{k+b} \, . $$ This implies the divergence of $\int_1^\infty e^{ax} x^b \, dx$. (This works for all values of $b$.) Roughly speaking, we are using that the exponential function “grows faster than any polynomial” for $x \to \infty$.