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The integral

$I = \int\limits_0^\infty \mathrm{d}x \, x \sin(x)$

does not converge. In physics we often use the principle of a convergence generating factor, in this example

$I = \lim\limits_{\epsilon \to 0} \int\limits_0^\infty \mathrm{d}x \, x \sin(x) \exp(-\epsilon x) = \lim\limits_{\epsilon \to 0} \frac{2\epsilon}{(1+\epsilon^2)^2} = 0$

I obviously now have 2 different solutions, one is that it does not converge, one is that it is 0. My question is of course, which one is correct?

As a physicist I always use such techniques without the proper knowledge if I am allowed to use it (here, I did not check if all requirements are fulfilled to flip the limes and the integral). I am searching for an explanation for this discrepancy in the solutions. From the given solution I know that the integral should indeed deliver a 0.

DaPhil
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  • That's not the actual value of the integral (the actual value doesn't exist), but a regularized solution (some sort of "average" around which the divergent partial result oscillates). This is similar to the discussion about Cauchy principal values, or, if you like, the Zeta regularization of divergent series, such as $1+2+3+4+5\cdots=-1/12$. It's not the value of the sum/integral. It's just a generalized result to a modified problem (modified conditions). You shouldn't even use $=$ sign without specifying that it's a regularized value. – orion May 14 '14 at 09:28
  • The both are correct. It does not converge AND its value is 0. – Anixx Sep 14 '19 at 13:59

1 Answers1

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The two integrals you address are two different stories. The first one does not exist (it does not converge on then range of integration).

For the second one, let us consider first the antiderivative $$I=\int x \sin(x) e^{-a x}~dx$$ After a series of integrations by parts, one could find $$I=\frac{e^{-a x} \left(-a \left(a^2 x+a+x\right) \sin (x)-(a (a x+2)+x) \cos (x)+\sin (x)\right)}{\left(a^2+1\right)^2}$$ If now we compute the integral $$J=\int\limits_0^\infty x \sin(x) e^{-a x}~dx$$ the value $$J=\frac{2 a}{\left(a^2+1\right)^2}$$ can be obtained provided that $\Re(a)>0$.

I hope and wish that this clarifies something for you.

Claude Leibovici
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