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I am trying to evaluate $$\int_{-\infty}^{\infty} \frac{\sin(b x)\sin(\sqrt{a+x^2})}{x \sqrt{a+x^2}} dx$$ for $a>0$ and real $b$. When $|b|>1$, we can split $\sin(bx)$ into complex exponentials, one term decays towards $+i\infty$, the other towards $-i\infty$, and we can use the residue theorem to get $$\pi \frac{\text{sign}(b)\sin(\sqrt{a})}{\sqrt{a}}$$ I have no idea how to handle the $|b|<1$ case.

Details $$\int_{-\infty}^{\infty} \frac{\sin(b x)\sin(\sqrt{a+x^2})}{x \sqrt{a+x^2}} dx=\frac{i}{2}\int_{-\infty}^{\infty} \frac{\exp(- i b x)\sin(\sqrt{a+x^2})}{x \sqrt{a+x^2}} dx - \frac{i}{2}\int_{-\infty}^{\infty} \frac{\exp( i b x)\sin(\sqrt{a+x^2})}{x \sqrt{a+x^2}} dx$$ Since $\sin(\sqrt{a+x^2})$ grows at most as $\exp(|\Im x|)$, depending on sign $b$, one of these terms decays exponentially towards $+i\infty$, the other towards $-i\infty$, and we can use the residue theorem for each term separately, closing the curve along the upper half plane for one term and the lower half plane for the other. The residue at $x=0$ is the only one we need (there is no pole at $\pm\sqrt{-a}$), it is $\frac{\sin(\sqrt{a})}{\sqrt{a}}$.

Wouter
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  • Can you add a proof of your answer for $|b|>1$? – user Mar 16 '24 at 20:00
  • Yes. I have checked that result numerically, too. – Wouter Mar 16 '24 at 20:17
  • The immediate problem with extending the approach to $|b|<1$ is that the contribution due to the semicircle at infinity does not vanish. Currently, I can't even show that it is finite, though it must be... – Wouter Mar 17 '24 at 12:21
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    The $|b|<1$ case is equivalent to finding a closed-form expression for $\frac{\pi}{2} \int_{0}^{b} J_{0} \left(\sqrt{a}\sqrt{1-t^{2}} \right) \mathrm dt. $ See this question. – Random Variable Mar 17 '24 at 16:36

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