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Consider the following integral, and it's closed forms:

$$\displaystyle \int_{\mathbb{R}} \frac{\tan^{-1}\left(\frac{\sqrt{x^2+a^2}}{b}\right) \, \text{d}x}{(x^2+b^2) \left( \frac{\sqrt{x^2+a^2}}{b} \right)} = \begin{cases} \frac{\pi}{\sqrt{a^2-b^2}}\left(2\cos^{-1} \left(\frac{b}{a}\right) - \cos^{-1} \left(\frac{b^2}{a^2}\right) \right) & \text{ if } |a|>|b| \\ \frac{\pi}{\sqrt{b^2-a^2}}\left(2\cosh^{-1} \left(\frac{b}{a}\right) - \cosh^{-1} \left(\frac{b^2}{a^2}\right) \right) & \text{ if } |a|<|b| \end{cases}$$

Does there exist a meaningful physical/mechanical/geometric interpretation of the integral?

Ideally, such an interpretation would not be too contrived and artificial, but that isn't guaranteed.

Probability would also work but that doesn't seem like it would fit such an integral.

  • Can you please clarify your integral notation for my benefit at least. Are you integrating twice over $x$ (a repeated integral) with the second integral becoming a definite integral over the interval [a,b]? – James Arathoon Apr 24 '18 at 23:11
  • Sorry I typed the square out of habit it’s meant to be a one dimensional integral – Jack Tiger Lam Apr 24 '18 at 23:48
  • Thanks. Have you seen this later related question https://math.stackexchange.com/questions/2751743/int-infty-infty-fracb-tan-1-big-frac-sqrtx2a2b-bigx2 – James Arathoon Apr 24 '18 at 23:56
  • Not relevant to my question. I already know how to arrive at the solution. – Jack Tiger Lam Apr 24 '18 at 23:57
  • In terms of physical meaning there may be a related surface, or perhaps even volume integral, which gives rise to the integral given above. Perhaps a potential theory problem with rotational symmetry (perhaps cylindrical form). – James Arathoon Apr 25 '18 at 00:20

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