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I'm currently studying the Dirac delta function using a textbook which unfortunately provides only partial solutions to its explanations. Why does $\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$ if $ f_\epsilon=\frac{1}{2\epsilon}e^{-\frac{|x|}{\epsilon}}$? Any help is very much appreciated.

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  • $\begingroup$ @ViktorGlombik What makes you think so? $\endgroup$ – lisyarus Apr 18 at 11:05
  • $\begingroup$ Nevermind, I thought of something else. $\endgroup$ – Viktor Glombik Apr 18 at 11:07
  • $\begingroup$ What is $g$ here? The answer depends on the type of $g$ you want to consider. $\endgroup$ – Kabo Murphy Apr 18 at 11:28
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Use a change of variable: $$ \int_{-\infty}^{\infty} \frac{1}{2\epsilon} \, e^{-|x|/\epsilon} g(x) \, dx = \{ y = x/\epsilon \} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-|y|} \, g(\epsilon y) \, dy \\ \to \frac{1}{2} \int_{-\infty}^{\infty} e^{-|y|} \, g(0) \, dy = \frac{1}{2} \left( \int_{-\infty}^{\infty} e^{-|y|} \, dy \right) \, g(0) = g(0) $$ The dominated convergence theorem has been used when taking the limit.

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Set $f(x) = \frac{1}{2}e^{-|x|}$, so $\frac{1}{\epsilon} f(x/\epsilon) =f_\epsilon$ and $||f||_{L^1} = 1$ for $\epsilon\in (0,1]$, so for any test functions $g\in C_c^\infty$ you have $$\int\frac{1}{\epsilon} f(x/\epsilon)g(x)dx = \int f(x)g(\epsilon x)dx$$ $$=\int f(x)[g(0) + g'(0)\epsilon x + ...]dx$$ $$ =g(0)\int f(x)dx + \int_{}f(x)[\epsilon xg'(0) + \frac{g''(0)(x\epsilon)^2}{2}+ ...]dx$$ The expansion is valid by the support and regularity assumption and $f\in \mathcal{S}$. So you have the limit $$\int\frac{1}{\epsilon} f(x/\epsilon)g(x)dx \rightarrow g(0)\int f(x)dx = g(0)$$ hence $f_\epsilon \rightarrow \delta$ in the distributional sense

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