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How to solve the following integral?

$$\int_{-\infty}^{\infty} i (k^2 + a)^{1/2} e^{ikx} dk$$

and some literature on this, thanks.

Sumanta
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Ricardo
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    Any thoughts, please try to write, if any. – Sumanta Sep 17 '20 at 19:53
  • Related: https://math.stackexchange.com/q/3827524/168433 – md2perpe Sep 17 '20 at 19:58
  • This integral gives one primitive function to the one in https://math.stackexchange.com/q/3827524/168433 (modulo a factor $i$). – md2perpe Sep 17 '20 at 20:00
  • I know that the derivative of this function is the function that is in the link that you put, but I can't see how it affects the Bessel function, I am not clear about it, please help – Ricardo Sep 17 '20 at 20:31

1 Answers1

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Suppose $a > 0$ and let $a = b^2$ with the understanding that $b > 0$.

Denote the integral in question

$$ I(b,x) = \int_{-\infty}^{\infty} i (k^2 + b^2)^{1/2} e^{ikx} dk $$ We have $$\frac{\partial I(b,x)}{\partial b} ={i b} \int_{-\infty}^{\infty} \frac{1}{(k^2 + b^2)^{1/2}} e^{ikx} dk$$ Now it has been shown here that $$\int_{-\infty}^{\infty}dk\frac{e^{ikx}}{\sqrt{k^2+b^2}} = 2K_0(b x)$$

where $K_0(x)$ is a modified Bessel function. So we have $$\frac{\partial I(b,x)}{\partial b} =2 i b K_0(b x)$$

Now treat $x$ as a constant and substitute $b \to bx = c$ to get

$$\frac{\partial I(c,x)}{\partial c} =\frac{2 i}{x^2} c K_0(c)$$

We recover $$ I(c,x) = \int \frac{\partial I(c,x)}{\partial c} dc =\frac{2 i}{x^2}\int c K_0(c) dc = \frac{2 i}{x^2} (-c K_1 (c))$$ with $K_1(c)$ again a modified Bessel function.

Resubstituting $c = bx$ gives the result

$$ I(b,x) = - \frac{2 i b}{x} K_1 (b x)$$

Andreas
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  • Thank you very much. The substitution of b for bx = c did not occur to me. Now I will study more to Bessel functions. Thank you again! – Ricardo Sep 18 '20 at 18:37
  • Same as in the linked answer, it's not true that $\int_{-\infty}^\infty e^{i x k}/\sqrt {k^2 + b^2} , dk = 2 K_0(b x)$ for all $b > 0$ and $x \in \mathbb R \setminus { 0 }$, and $\int_{-\infty}^\infty \sqrt {k^2 + b^2} , e^{i x k} dk$ diverges if understood as an improper integral (or even as a p.v. integral). – Maxim Oct 13 '20 at 16:17