I am trying to solve integrals of the following form $$\int\frac{ {\rm e}^{\imath K \sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} {\rm d} x,$$

where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?

Substituting the square root doesn't seem to help since for $p = \sqrt{x^2+a^2}$ we have $\frac{ {\rm d}p}{{\rm d}x} = \frac x p = \frac{\sqrt{p^2-a^2}}{p}$, leading to $$\int\frac{ {\rm e}^{\imath K p}}{\sqrt{p^2-a^2}} {\rm d} p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.

Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.

Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?

edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$\int_{-\infty}^\infty\frac{ {\rm e}^{\imath K \sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} {\rm d} x = -\pi(Y_0(a K) + \jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.

  • $\begingroup$ Looks like something Bessel-ish. $\endgroup$ – Frpzzd Jul 18 '18 at 11:48
  • $\begingroup$ Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form. $\endgroup$ – Florian Jul 18 '18 at 11:50
  • $\begingroup$ If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined? $\endgroup$ – Mac's Musings Jul 18 '18 at 11:50
  • $\begingroup$ I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $\int \frac{\cos[k\sqrt{x^2+a^2}]}{\sqrt{x^2+a^2}} {\rm d}x$. This would help as well. $\endgroup$ – Florian Jul 18 '18 at 11:52
  • $\begingroup$ With limits $-\infty$ and $\infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10 $\endgroup$ – Chappers Jul 18 '18 at 12:22

Changing $x=a\sinh t$ in the integral, \begin{equation} I=\int_\beta^{\gamma}e^{ika\cosh t}\,dt \end{equation} where $\beta=\sinh^{-1}b/a$ and $\gamma=\sinh^{-1}(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<\sigma<\pi$ \begin{equation} H_0^{(1)}(ka,w)=\frac{2}{i\pi}\int_w^{\infty+i\sigma}e^{ika\cosh t}\,dt \end{equation} Then \begin{equation} I=\frac{2}{i\pi}\left[ H_0^{(1)}(ka,\beta)- H_0^{(1)}(ka,\gamma) \right] \end{equation} It can also be expressed as \begin{equation} I= K_0(-ika,\beta)- K_0(-ika,\gamma) \end{equation} where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).

  • $\begingroup$ Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like ${\rm arsinh}(b/a)$. But that's a minor thing. Thank you again! $\endgroup$ – Florian Jul 19 '18 at 8:40
  • $\begingroup$ What still confuses me is that the integral needs to run on a line with ${\rm Im}(t)>0$ (that's how I interpret the $i \sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that? $\endgroup$ – Florian Jul 19 '18 at 8:44
  • $\begingroup$ Sorry for the mistake on the integration limits, corrected now. $\endgroup$ – Paul Enta Jul 19 '18 at 9:46
  • $\begingroup$ This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $\beta$ to $\gamma$ on the real axis, 2) from $\gamma$ to $\infty +i\sigma$ and 3) from $\infty +i\sigma$ to $\beta$. $\endgroup$ – Paul Enta Jul 19 '18 at 9:50
  • $\begingroup$ Okay, that helps. Thanks! $\endgroup$ – Florian Jul 19 '18 at 10:32

Consider function:

$$y=\sqrt{x^2+a^2} \ln e^{\sqrt {x^2 +a^2}}$$

$$y'=\frac{x}{\sqrt {x^2+a^2}}.(e^ {\sqrt{x^2+a^2}})(1+\sqrt{x^2+a^2})$$

Now compare this with the relation in question we get:

$$k=-[\frac {\ln x(1+\sqrt{x^2+a^2})}{\sqrt {x^2+a^2}}+1]i$$

This is condition for initial relation to be integrable.


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