I am trying to compute $$-\hbar^2 \kappa^2 \int_{-\infty}^{\infty} \exp(-\kappa^2 |x|) \frac{d^2}{dx^2} \exp(-\kappa^2|x|)dx$$
where $\kappa = \frac{\sqrt{m \alpha}}{\hbar}$. Thus far, I have defined $f(x) = \exp(-\kappa^2 |x|)$ and have asserted
$$ f(x) = \begin{cases} \exp(-\kappa^2 x) & : x > 0 \\ \exp(\kappa^2 x) & : x < 0 \end{cases} $$
to show
$$ \dot{f}(x)\begin{cases} -\kappa^2 \exp(-\kappa^2 x) & : x > 0 \\ \kappa^2 \exp(\kappa^2 x) & : x < 0 \end{cases}= \kappa^2 f(x) (-\text{sgn}(x)) $$
where I have been intentionally sloppy about the behavior of the system at $0$. (I am currently reading a physics textbook and the theory of distributions was never introduced in a rigorous context.) We have previously defined the step function as
$$ \theta(x) = \begin{cases} 1 & : x > 0 \\ 0 & : x < 0 \end{cases} $$
where (as the author stated) $\theta(0)$ is irrelevant. Furthermore, I have already proven that $\frac{d\theta}{dx} = \delta(x)$. In terms of this function,
$$ \dot{f}(x) = \kappa^2 f(x) (\theta(-x) - \theta(x)) $$
Thus, we integrate by parts (once again sloppily assuming that convergence is guaranteed and that the boundary conditions align).
$$ \begin{align} -\hbar^2\kappa^2 \int_{-\infty}^{\infty} f(x) \frac{d^2 f}{dx^2} dx &= -\hbar^2\kappa^2 \left[f(x) \frac{df}{dx}\right]_{-\infty}^{\infty} \\ & - \hbar^2 \kappa^4 \left[\int_{-\infty}^{\infty} f(x) \frac{df}{dx} (\theta(-x) - \theta(x)) - f(x)^2 (\dot{\theta}(-x) + \dot{\theta}(x))dx\right] \\ & = - \hbar^2 \kappa^4 \left[\int_{-\infty}^{\infty} \kappa^2 f(x)^2 - f(x)^2 (\dot{\theta}(-x) + \dot{\theta}(x))dx\right] \\ & = - \hbar^2 \kappa^4 \left[\int_{-\infty}^{\infty} \kappa^2 f(x)^2 dx - 2 f(0)^2 \right] \\ & = - \hbar^2 \kappa^4 \left[\int_{-\infty}^{\infty} \kappa^2 f(2x) dx - 2 \right] \\ & = 2 \left(\frac{m \alpha}{\hbar}\right)^2 - \hbar^2 \kappa^6 \int_{-\infty}^{\infty} f(2x) dx \\ & = 2 \left(\frac{m \alpha}{\hbar}\right)^2 - 2 \hbar^2 \kappa^6 \int_{0}^{\infty} \exp(-2\kappa^2 x) dx \\ & = 2 \left(\frac{m \alpha}{\hbar}\right)^2 + \hbar^2 \kappa^4 \int_{0}^{\infty} \exp(-2\kappa^2 x) dx \\ & = 2 \left(\frac{m \alpha}{\hbar}\right)^2 - \hbar^2 \kappa^4\\ & = \left(\frac{m \alpha}{\hbar}\right)^2 \end{align} $$
The book tells me that the correct answer is $\left(\frac{m \alpha}{\hbar}\right)^2$, so I assume my general approach is correct. How can I make the use of distributions more rigorous? How can I try to quantify asymptotic growth rates for $f(x)$? I even tried setting up a recursion relationship by introducing
$$I_{\alpha} = \int_{-\infty}^{\infty} f_{\alpha}(x) \frac{d^2 f_{\alpha}}{dx^2} dx$$
where $f_{\alpha} (x) = \exp(-\alpha\kappa^2 |x|)$ and then trying to use a generating function to find series from
$$I_{2^n} = 2^n k^4 I_{2^{n+1}} + 2 \kappa^2$$
However, it did not appear to be particularly helpful. I feel that the book heavily suggested differentiating the Integrand, so I decided that I would either use Feynman's trick or integration by parts. The latter felt more natural given the second derivative in the middle of the integrand.
Edit: I should clarify that I am asking my question on Math SE (as opposed to Physics SE) because I deeply dissatisfied with the lack of mathematical explanation in the derivation. Does anyone have advice for making the use of distributions rigorous. Every time I say sloppily, is there a theorem or result I should quote? I have not yet had a formal course in the theory of distributions although I had to learn a little bit when I studied Green's functions in PDEs.
Are there any combinatorial approaches for describing distributions? I may be overzealous to apply generating functions, but I had hoped that they would provide bounds on more general wavefunctions that do not arise from simple potentials such as the delta function.
You are making the situation far too complicated. We don't neet the theory of distributions. The integrand is an even function and $\kappa^2 >0.$ Let$c=\kappa^2.$ The integral is $$2\int_0^\infty e^{-cx}c^2e^{-cx}dx=2c^2\int_0^\infty e^{-2cx}dx$$ $$=\frac{2c^2}{2c}=c$$