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In a physics problem I am confronted with a divergent integral

$$ \int_{-\infty}^\infty x \sin x \, dx = \sin x - x \cos x \bigg|_{-\infty}^\infty \approx 0$$

How to regularize it?

in order to regularize this sum I would argue this is zero. another possibility is

$$ \int_{-L}^L x \sin x \, dx = \sin x - x \cos x \bigg|_{-L}^L = 2( \sin L - L\cos L)$$

which is oscillating. if $L \in 2 \pi \mathbb{ Z }$ the integral is $\int = \pm L$ if $L \in \pi/2+ 2 \pi \mathbb{ Z }$ then $\int = \pm 2$.

so even if this integral is oscillatory maybe theory of distributions can save us.

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closed as unclear what you're asking by Did, Adam Hughes, user91500, Rohan, Daniel W. Farlow Jan 1 '17 at 10:58

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  • $\begingroup$ Since $\int_{-\infty}^{\infty} x^s \sin(x) \, \mathrm{d}x = (1 - (-1)^s) \cos(\pi s / 2) \Gamma(s+1)$ when $\mathrm{Re}[s] \in (-2,0)$, extrapolating to $s=1$ gives you the value $0$. I have no idea what $\sin(x) - x \cos(x) \Big|_{-\infty}^{\infty} \approx 0$ means but maybe it is related. $\endgroup$ – user399601 Dec 31 '16 at 18:18
  • $\begingroup$ One has to be careful with the meaning of $x^s$ when $x<0, \, s \in (-2,0)$. $\endgroup$ – Olivier Oloa Dec 31 '16 at 18:23
  • $\begingroup$ I wrote a symbol "$\approx$" open to interpretation - since I am trying to assign meaning to divergent integral. $\endgroup$ – cactus314 Dec 31 '16 at 18:25
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    $\begingroup$ What is the question? $\endgroup$ – Did Dec 31 '16 at 18:25
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    $\begingroup$ @OlivierOloa I think it's OK as long as $x^s$ and $(-1)^s$ are from the same branch of logarithm. But taking $2 \int_0^{\infty} x^s \sin(x) \, \mathrm{d}x = \cos(\pi s / 2) \Gamma(s+1)$ would probably be better. $\endgroup$ – user399601 Dec 31 '16 at 18:28
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Since the integrand is even, consider $\int_0^\infty x \sin x \, dx $. For this integral there's a convenient regularization: multiply the integrand by $e^{-cx}$ where $c>0$. This integrates without much work (write $\sin x = (e^{ix}-e^{-ix})/(2i)$, then integrate $xe^{ax}$ by parts, and simplify). The antiderivative is $$ - \frac{e^{-cx}}{\left(c^{2} + 1\right)^{2}} \left((c^{3} x+cx+c^2-1) \sin x + (c^{2} x +x + 2 c)\cos x \right) $$
so the integral from $0$ to $\infty $ is $$\frac{2c}{(c^2+1)^2}$$ and this tends to $0$ as $c\to 0$.

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  • $\begingroup$ Good catch (+1). In the same vein $$\int_{-\infty}^{\infty } \frac{x\sin x}{1+c^2x^2} \, dx =\pi\cdot \frac{e^{-1/c}}{c^2} \to 0$$ as $c \to 0^+$. $\endgroup$ – Olivier Oloa Dec 31 '16 at 19:06
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Suppose we replace the integrand by an even function $f(x,p)$ such that $f(x,0) = x\sin(x)$, and suppose that the series expansion coefficients of the expansion of $f(x,p)$ around $x = 0$ tends to those of $x\sin(x)$ for $p$ to zero uniformly. Then if $\int_0^\infty f(x,p)dx$ exists for $p>0$, then Glaisher's theorem says that:

$$\int_0^{\infty}f(x,p)dx = \frac{\pi}{2}c_{-\frac{1}{2}}(p)$$

where for integer $n$, $c_n(p)$ is the coefficient of $(-1)^nx^{2n}$ in the series expansion of $f(x,p)$ around zero. For fractional $n$ the $c_n$ are defined using analytic continuation. We can then define the regularized integral by taking the limit of $p$ to zero in the above expression, this then doesn't depend on the regularization method is the series coefficients tend to those of the original integrand uniformly. In this case, one then finds that the integral is given as $\frac{\pi}{2 (-2)!} = 0$.

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  • $\begingroup$ Well, the factorial function is defined only for positive integers. We can extend the definition using Gamma function or any other extension. For the case that we use the gamma function then the result is zero, but if we use the Hadamard gamma function then is not. $\endgroup$ – Masacroso Dec 31 '16 at 19:16
  • $\begingroup$ @Masacroso Yes, the analytic continuation is not unique here, it has to be done in the "standard way" where one replaces factorials by gamma functions. $\endgroup$ – Count Iblis Dec 31 '16 at 19:21
  • $\begingroup$ Other thing to say: if we use the gamma function then $\Gamma(-3)$ doesnt not exists, so the equality you posted is not an equality, it is a limit. $\endgroup$ – Masacroso Dec 31 '16 at 19:24
  • $\begingroup$ @Masacroso Yes, but you can get around that complication by invoking that $1/\Gamma(z)$ is an entire function, and we get straight to the reciprocal of the gamma function here. $\endgroup$ – Count Iblis Dec 31 '16 at 19:28
  • $\begingroup$ @Masacroso: Just because something doesn't exist, doesn't mean that it can't be created. :-$)$ For instance, we can regularize the $\Gamma$ function at negative integers by evaluating $\Gamma^\star(x)=\displaystyle\lim_{h\to0}\frac{\Gamma(x+h)+\Gamma(x-h)}2.$ $\endgroup$ – Lucian Dec 31 '16 at 23:49

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