What is the value of $\left< e^{-|x|}\left| \left(i\frac{d} {dx}\right)^2 \right|e^{-|x|}\right> $?

Question

My approach:

\begin{equation} \begin{aligned} \left< e^{-|x|}\left| \left(i\frac{d} {dx}\right)^2 \right|e^{-|x|}\right> &= \int_{-\infty}^{\infty}e^{-|x|} \left(-\frac{d^2}{dx^2}\right)e^{-|x|}dx \\ &= -\left[ \int_{-\infty}^{0}e^{-|x|} \frac{d^2e^{-|x|}}{dx^2}dx + \int_0^\infty e^{-|x|} \frac{d^2 e^{-|x|}}{dx^2} dx\right] \\ &= -\left[ \int_{-\infty}^{0}e^{x} \frac{d^2e^{x}}{dx^2}dx + \int_0^\infty e^{-x} \frac{d^2 e^{-x}}{dx^2} dx\right] \\ &= -\left[ \int_{-\infty}^0 e^{2x}dx + \int _0^\infty e^{-2x} dx\right] \\ &= -\left[ \frac{e^{2x}}{2}\right]_{-\infty}^0 - \left[\frac{e^{-2x}}{(-2)} \right]_0^\infty \\ &= -\frac{1}{2} - \frac{1}{2} \\ &= -1 \end{aligned} \end{equation}

But:

\begin{equation} \begin{aligned} \left< e^{-|x|}\left| \left(i\frac{d} {dx}\right)^2 \right|e^{-|x|}\right> &= \left< e^{-|x|}\left| \left(i\frac{d} {dx}\right)^\dagger \left(i\frac{d} {dx}\right) \right|e^{-|x|}\right> \\ &= \|\left(i\frac{d}{dx}\right) |e^{-|x|}\rangle \|^2 \\ &\gt 0 \end{aligned} \end{equation}

My Question:

$1.$ What is wrong with my approach that I am getting a negative value?

$2.$ How to calculate $\left< e^{-|x|}\left| \left(i\frac{d} {dx}\right)^2 \right|e^{-|x|}\right>$?

Answer 1

1. The problem is that $\frac d{dx} e^{-|x|}$ has a discontinuity at $0$. Then $\frac{d^2}{dx^2}e^{-|x|}$ is not finite. So you can't ignore it when you calculate the integrals.
2. I suggest you integrate by parts. $$f(x)=e^{-|x|}\\g(x)=\frac d{dx}e^{-|x|}$$ Then use $$\int fg'dx=fg-\int f'g dx$$ You can then apply your approach and split the integral into positive and negative, since the integrand is finite around $0$.