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This integral does not converge, how is it solved that it converges in the sense of distribution? if there is, thank you.

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  • $\begingroup$ The integral of $(k/\sqrt {k^2 + a} - \operatorname {sgn} k) e^{i x k}$ converges. The integral of $\operatorname {sgn}(k) e^{i x k}$ makes sense if interpreted as the Fourier transform of the distribution $\operatorname {sgn} k$ or as the distributional limit $$\lim_{A \to \infty} \int_{-A}^A \operatorname {sgn}(k) e^{i x k} dk = \lim_{A \to \infty} \frac {2 i (1 - \cos A x)} x = 2 i \mathcal P {\left( \frac 1 x \right)}.$$ A closed form can be found as the distributional derivative of the Fourier transform of $1/\sqrt {k^2 + a}$. $\endgroup$ – Maxim Sep 16 '20 at 11:47
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This integral doesn't converge absolutely, but it does converge for most values of $a$. First, suppose that $a=b^2>0$ let us note that it is well known that

$$\int_{-\infty}^{\infty}dk\frac{e^{ikx}}{\sqrt{k^2+b^2}}=\int_{-\infty}^{\infty}dt\frac{e^{it|b|x}}{\sqrt{t^2+1}}=2K_0(|b|x)$$

where $K_0(x)$ is a modified Bessel function. It is obvious that the integral diverges when $b\to 0$. We can produce the integral in question from this one by taking a derivative with respect to $x$ which gives

$$\int_{-\infty}^{\infty}dk\frac{ik e^{ikx}}{\sqrt{k^2+b^2}}=2\frac{dK_0(|b|x)}{dx}=-2|b|K_1(|b|x)$$

which again converges just fine for any non-zero values of $x,b$.

Also the integral converges for $a<0$ since the branch points on the real axis are integrable.

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  • $\begingroup$ The integral of $e^{i x k}/\sqrt {k^2 + b^2}$ is $2 K_0(|b x|)$ (for $x, b \in \mathbb R \setminus \{ 0 \}$). The integral of $k e^{i x k}/\sqrt {k^2 + b^2}$ diverges because the integral of $(k/\sqrt {k^2 + b^2} - \operatorname {sgn} k) e^{i x k}$ converges. $\endgroup$ – Maxim Sep 16 '20 at 14:39

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