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Suppose we assumed that the parameter $\alpha$ is real, and then evaluated some definite integral $I(\alpha)$, which depends on $\alpha$. Can we then claim that $I(\alpha)$ also holds for $\alpha \in S \subseteq \mathbb C$, and not just for real $\alpha$? Under what conditions could such an "interpolation" be justified? Is this a particular instance of the Identity Theorem from complex analysis? Any reference would be appreciated. Thank you in advance.

Edit: Consider the following integral $$I(\alpha) = \int_0^\infty \frac{\ln x}{x^2 + \alpha^2}dx, \quad \alpha > 0.$$ We find (without using complex analysis) that $I(\alpha) = \pi\ln\alpha/2\alpha$. However, if we now suppose that $\alpha = ib$ with $b \in \mathbb R$, say, then, at least for some values of $b$, the real part of the principle value of $\pi\ln\alpha/2\alpha$ corresponds to the value of $I(\alpha)$. I know that this example does not completely satisfy the criteria I outlined above, because we are taking the real part, but is there a reason why $\Re[\pi\ln\alpha/2\alpha] = I(\alpha)$ whenever $\alpha = ib$. We can, of course, prove this using contour integration, but I am hoping that there is a theorem that explains this. Overall, I evaluate $I(\alpha)$ for real $\alpha > 0$, and then substituted $\alpha = ib$ with $b$ real, and for some reason $\Re[\pi\ln\alpha/2\alpha] = I(\alpha)$. I am wondering if this is a particular instance of a more general principle. Also, note that, technically, $I(\alpha)$ is the Cauchy principle value of $\int_0^\infty dx\ln x/(x^2 + \alpha^2)$.

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  • $\begingroup$ What would it mean for $I(\alpha)$ to "hold"? What is $S$? $\endgroup$ – Zev Chonoles Sep 11 '13 at 4:53
  • $\begingroup$ Here, $S$ is a subset of $\mathbb C$, which is not necessarily $\mathbb R$. Also, suppose we were able to evaluate $I(\alpha)$ without assuming that $\alpha$ is real and found that $I(\alpha) = \varphi(\alpha)$, say. Now, if we could show that $I(\alpha) = \varphi(\alpha)$ is true whenever $\alpha$ is real, under what conditions could we claim that the result is also true for $\alpha \in S$ (and we know it must be true because we proved it without assuming $\alpha \in \mathbb R$)? Hopefully, it is clear what I meant by "holds." I apologize for the confusion. $\endgroup$ – glebovg Sep 11 '13 at 6:27
  • $\begingroup$ @ZevChonoles Does the example I added help? $\endgroup$ – glebovg Sep 12 '13 at 22:43

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