To motivate the problem: I am learning QM and want to prove that every attractive potential admits a bound energy eigenstate. Using $\psi_{\alpha} (x) = \left( \dfrac{\alpha}{\pi}\right)^{1/4} e^{-\alpha x^2 /2}$ (a bound state), one finds that (skipping some steps here) $$\langle H \rangle = \left(\frac{\alpha}{\pi}\right)^{1/2} \int_{-\infty}^{\infty} \mathrm{d} x \, e^{-\alpha x^2} \left[\frac{\hbar^2}{2m} \alpha - \frac{\hbar^2}{2m} \alpha^2 x^2 + V(x) \right], $$ where $\langle H \rangle$ is the expected value of energy, which if negative would show that one of the eigenstates of $H$ has eigenvalue $E < 0 \equiv V_{\pm}$, proving our proposition. Since $V(x)$ is negative, it seems to be a good idea to require $\alpha^2 x^2 > \alpha \,$.
It seems that an immediate issue arises here, as we're integrating over zero, whereas $\alpha > 0$ by construction. My workaround was to integrate over $(-\infty, -\epsilon)$ and $(\epsilon, \infty)$, requiring that $\alpha > 1 / \epsilon^2$. This makes both integrals negative, which--taking the limit $\epsilon \to 0$--implies that my original integral is negative. My issue with this is that it doesn't give one an explicit $\alpha$ for which the integral is negative, unless one is willing to approximate the integral by cutting off the middle part for some small $\epsilon$. I am wondering if someone could provide me an argument that constructs this specific $\alpha$ if possible.
EDIT for clarity: $V(x)$ is an attractive potential, which means it's negative everywhere, with $\vert V(x) \vert$ increasing as $\vert x \vert$ decreases, and vanishes at $\pm \infty$. Other than that, it's an arbitrary (not highly pathological) function.
It will help to split up your integral. $$\sqrt{\frac\alpha\pi}\int_{-\infty}^{\infty} e^{-\alpha x^2} \; dx = 1$$ and $$\sqrt{\frac\alpha\pi}\int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} \; dx = \frac{1}{2\alpha}$$ by the zeroth and second moments of the Normal distribution (or by just plugging them into Wolfram|Alpha, as I just did).
So we can simplify your integral to: $$\frac{\hbar^2}{2m}\left(\alpha - \alpha^2 \cdot \frac1{2\alpha} \right) + \sqrt{\frac\alpha\pi}\int_{-\infty}^{\infty} e^{-\alpha x^2} V(x) \; dx = \frac{\hbar^2}{4m}\alpha + \sqrt{\frac\alpha\pi}\int_{-\infty}^{\infty} e^{-\alpha x^2} V(x) \; dx =: I(\alpha).$$
As you can see, there isn't going to be a single value of $\alpha$ that will make this negative for all $V$. But we desire to go in the other direction: fix $V$ ($V(x) < 0$, $V$ monotonic in $|x|$, and $V(x) \to 0$ as $|x|\to \infty$) and show that there is some $\alpha>0$ such that $I(\alpha) < 0$.
The trick will be to observe that $I(0) = 0$ and to show that $\frac{dI}{d\alpha}$ is negative at $\alpha = 0$, which by the mean value theorem will give the desired result.
\begin{align*} \frac{dI}{d\alpha}& = \frac{\hbar^2}{4m} + \frac{1}{2\sqrt{\alpha}} \frac1{\sqrt\pi} \int_{-\infty}^\infty e^{-\alpha x^2} V(x) \; dx + \sqrt{\frac\alpha\pi} \int_{-\infty}^\infty e^{-\alpha x^2}(-x^2) V(x)\; dx\\ & \leq \frac{\hbar^2}{4m} + \frac{1}{2\sqrt{\alpha}} \frac1{\sqrt\pi} \int_{-\infty}^\infty e^{-x^2} V(x) \; dx \quad\text{(For $\alpha \leq 1$, since $V(x) \leq 0$)}\\ &= \frac{\hbar^2}{4m} + \frac{1}{2\sqrt{\alpha}} \mathbb{E}_{X \sim \mathcal{N}(0,1/2)}[V(X)] \\ &\to -\infty \text{ as }\alpha \to 0. \end{align*}