# Forcing Convergence

1. Sep 22, 2009

### keniwas

I am trying to understand a proof about the inner product of continuous eigenstates. Since there is no guarantee that the functions are square integrable they multiply the inner product by $$e^{-\gamma |x|}$$ then take the limit as $$\gamma\rightarrow 0$$ to force the surface terms to behave at infinity.

So my question is the following: is it true that
$$\lim_{\gamma\rightarrow 0}\int_{-\infty}^{\infty}e^{-\gamma |x|}f(x)dx=\int_{-\infty}^{\infty}f(x)dx$$

for all f(x)?

I have shown it is true for multiple cases where f(x) = xn or f(x)=sin(x) ... etc... and the limits are finite (i.e. on some interval a<x<b) but I cannot figure out how to do it for a general case, nor for the case when the limits are from $$-\infty<x<\infty$$

The problem I run into in the general case is I am not sure what to do with the integration by parts after 1 round of integrating since I am left with integrals.

I suppose I could also ask, do I have to evaluate the integral before taking the limit? I am not sure what the rules on this are... is
$$\lim_{\gamma\rightarrow 0}\int_{-\infty}^{\infty}e^{-\gamma |x|}f(x)dx=\int_{-\infty}^{\infty}\lim_{\gamma\rightarrow 0}e^{-\gamma |x|}f(x)dx$$
allowed??

2. Sep 22, 2009

### g_edgar

Perhaps look up the "Dominated Convergence Theorem"...