- #1
hideelo
- 91
- 15
I am looking at the integral
$$\int_0^\infty dx \: e^{-iax} - e^{iax}$$
I know that this does not converge for many reasons, but most obviously because I can rewrite it as
$$2i \int_0^\infty dx \: sin(ax) = -2i a [\cos(ax)]_0^\infty$$
which does not converge to anything.
However the book I'm following (Solutions to Kardar's Statistical Mechanics of Particles; Chapter 5, Problem 1) has a "clever" solution for this
$$\int_0^\infty dx \: e^{-iax} - e^{iax} = \lim_{\epsilon \to 0} \int_0^\infty dx \: e^{-iax-\epsilon x} - e^{iax-\epsilon x}$$
$$= \lim_{\epsilon \to 0} \left[\frac{e^{-iax-\epsilon x}}{-ia-\epsilon} - \frac{e^{iax-\epsilon x}}{ia-\epsilon}\right]_0^\infty = \frac{2i}{a}$$How the hell is this allowed? You can't just swap limits and integrals like that, or am I missing something?
$$\int_0^\infty dx \: e^{-iax} - e^{iax}$$
I know that this does not converge for many reasons, but most obviously because I can rewrite it as
$$2i \int_0^\infty dx \: sin(ax) = -2i a [\cos(ax)]_0^\infty$$
which does not converge to anything.
However the book I'm following (Solutions to Kardar's Statistical Mechanics of Particles; Chapter 5, Problem 1) has a "clever" solution for this
$$\int_0^\infty dx \: e^{-iax} - e^{iax} = \lim_{\epsilon \to 0} \int_0^\infty dx \: e^{-iax-\epsilon x} - e^{iax-\epsilon x}$$
$$= \lim_{\epsilon \to 0} \left[\frac{e^{-iax-\epsilon x}}{-ia-\epsilon} - \frac{e^{iax-\epsilon x}}{ia-\epsilon}\right]_0^\infty = \frac{2i}{a}$$How the hell is this allowed? You can't just swap limits and integrals like that, or am I missing something?