Both sides are zero for a sufficiently well behaved odd function.

For an even function $\displaystyle f(x)$ such that $\displaystyle f(0)=0$3. The result is a result of the fundamental theorem of calculus, and the

Laplace transform of a derivative for an even function zero at the origin(sufficiently well behaved).

$\displaystyle \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)+f(0)=$$\displaystyle 2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)$

..............$\displaystyle =2\int_{0}^{\infty}f'(x) dx=2\left[\lim_{x \to \infty}f(x) - \lim_{x \to 0+}f(x)\right]$

but as $\displaystyle f$ is even (and sufficiently well behaved) the last limit

above is $\displaystyle 0$, and $\displaystyle \lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)$

For a constant we have:

$\displaystyle \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx\ k\ e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx\ k\ e^{-\epsilon |x|}=$$\displaystyle 2 \lim_{\epsilon \to 0+} \epsilon\ \mathcal{L}k(\epsilon)=2 \lim_{\epsilon \to 0^+} \epsilon\ \frac{k}{\epsilon}=2\ k

$,

which is the required result in this case.

(Note we could have done this for an arbitrary even function by keeping track of the f(0) terms rather than setting them to zero, then we would not have had to handle the constant term separately)

RonL4. Hence the result is true for a sufficiently well behaved function.

I can amplify the details if you need (though I doubt that I will feel like

going for full rigour)

RonL