Integral with symmetric infinitesimal bounds

[shinobi20]

Homework Statement

I'm reading something in my quantum physics book that says given a wavefunction ψ that is even, if we evaluate its integral from -ε to ε, the integral is 0. How can this be? I thought this is the property of odd functions.

Homework Equations

ψ=Ae^kx if x<0 and ψ=Be^-kx if x>0, ε is infinitesimal change in x

The Attempt at a Solution

By boundary conditions, say at the origin, this will give A=B then the book says we can represent the wave function as ψ=Ae^-k|x|. The wave function is even so the integral is 0 between -ε to ε.

 

[Ray Vickson]

Homework Statement

I'm reading something in my quantum physics book that says given a wavefunction ψ that is even, if we evaluate its integral from -ε to ε, the integral is 0. How can this be? I thought this is the property of odd functions.

Homework Equations

ψ=Ae^kx if x<0 and ψ=Be^-kx if x>0, ε is infinitesimal change in x

The Attempt at a Solution

By boundary conditions, say at the origin, this will give A=B then the book says we can represent the wave function as ψ=Ae^-k|x|. The wave function is even so the integral is 0 between -ε to ε.

That is false. For very small, but positive $\epsilon$ the function $\psi(x)$ is very nearly constant ($=A$) over the interval $-\epsilon \leq x \leq \epsilon$, so $\int_{-\epsilon}^{\epsilon} \psi(x) \, dx \approx 2 A \epsilon$. Of course, if $\epsilon$ is infinitesimal, so is the integral, but infinitesimal does not mean zero.