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Let $\sigma>0$ , $1<\alpha\leq 2$, and $-1\leq \beta \leq 1$. I am looking for a closed-form solution (or something near) for the following integral.

$$\frac{1}{2 \pi } \text{PV}\int_{-\infty }^{\infty } \left(\pi \delta (u)-\frac{i e^{-i K u}}{u}\right) \left(\mu +\alpha \beta \sigma \tan \left(\frac{\pi \alpha }{2}\right) \left(\sigma \sqrt{u^2}\right)^{\alpha -1}-\frac{i \alpha \left(\sigma \sqrt{u^2}\right)^{\alpha }}{u}\right) \exp \left(i \left(-\mu u+\left(\sigma \sqrt{u^2}\right)^{\alpha } \left(-\frac{\beta \sqrt{u^2} \tan (\frac{\pi \alpha }{2})}{u}+i\right)\right)\right) \, du$$ (where PV indicates the Cauchy principal value and $\delta(.)$ is the Dirac delta function).

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    $\begingroup$ Isn't the integrand divergent as $u \to 0$ as $u^{-2}$ and hence non-integrable? $\endgroup$ Jan 17, 2016 at 19:21
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    $\begingroup$ I revised to add principal value. $\endgroup$
    – Nero
    Jan 17, 2016 at 22:53
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    $\begingroup$ NNT, what are K, i, $\mu$? $\endgroup$
    – BCLC
    Feb 7, 2016 at 5:45
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    $\begingroup$ why $\sqrt{u^2}$ and not simply $|u|$ ? Because of complex numbers? $\endgroup$
    – BCLC
    Feb 7, 2016 at 5:47
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    $\begingroup$ @BCLC : I'am wondering if he directly copied a maple formula without fully understanding it $\endgroup$
    – reuns
    Feb 7, 2016 at 5:52

1 Answer 1

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This is an improper integral near $u=0$. The numerator goes to 1 near $u=0$ but the denominator makes the integral divergent like $\int_0^1 u^{-2}du$ . I suggest revisiting the two original functions whose Fourier transforms yielded your integral to see that they are in fact integrable and that their convolutions therefore is defined. (For example, are the two functions $L^1$ integrable?) Another suggestion: calculate or estimate your integral over the real line but with the interval $[-\delta,\delta]$ removed. Then see if the limit as $\delta\to0$ exists.

Good luck! Sam

PS, to make matters simpler, take $\beta=K=\mu=0$ and $\alpha=2, \sigma=1$. Might be easier to see what's going on in order to determine whether it can exist.

PSS, my comment has become obsolete since the integral keeps changing.

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    $\begingroup$ I reexpressed the integral above by inverting the order in the convolution. But $\frac{\text{sign}(x)}{x^2}$ is integrable by PV, i.e. $\int_{-\infty}^{-\epsilon} \frac{\text{sign}(u)}{u^2} du$ and flipping $\int_\epsilon^{\infty}$. $\endgroup$
    – Nero
    Jan 18, 2016 at 15:01

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