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 [tex]e^{-\gamma |x|}[/tex] then take the limit as [tex]\gamma\rightarrow 0[/tex] to force the surface terms to behave at infinity.(adsbygoogle = window.adsbygoogle || []).push({});

So my question is the following: is it true that

[tex]\lim_{\gamma\rightarrow 0}\int_{-\infty}^{\infty}e^{-\gamma |x|}f(x)dx=\int_{-\infty}^{\infty}f(x)dx[/tex]

for all f(x)?

I have shown it is true for multiple cases where f(x) = x^{n}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 [tex]-\infty<x<\infty[/tex]

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

[tex]\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[/tex]

allowed??

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# Forcing Convergence

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