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I am studying the Cauchy principal value. In my complex analysis textbook, I solved the following problem: Evaluate P.V. $\displaystyle \int_0^\infty \frac{x^\alpha}{x(x+1)},$ where $\alpha \in (0,1).$ The Cauchy principal value is $\displaystyle \frac{\pi}{\sin \alpha \pi}.$

I think $\displaystyle \frac{x^\alpha}{x(x+1)}$ is (usual) improper integrable on $(0,\infty)$.

My question is whether we can write $\displaystyle \int_0^\infty \frac{x^\alpha}{x(x+1)}=\frac{\pi}{\sin \alpha \pi}$ or not.

I would be grateful for any comments. Thanks in advance.

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  • $\begingroup$ We usually denote the principal value as $PV\int_{0}^\infty \frac{x^{\alpha}}{x(x+1)}dx=\frac{\pi}{sin(\alpha\pi)}$ But the integral itself is not equal to that expression. $\endgroup$ – aleden Nov 19 '17 at 16:07
  • $\begingroup$ @aleden Thanks for your answer. I am sorry. I didn't understand what you are saying exactly. Why do we use principal value, not just improper integral for above problem? $\endgroup$ – 04170706 Nov 19 '17 at 16:16
  • $\begingroup$ Cauchy principal value is a way of assigning a result to divergent integrals that is meaningful in some situations. Just like $\sum_{n=1}^\infty n=+\infty$, but $\zeta(-1)=-\frac{1}{12}$ using analytic continuation. $\endgroup$ – aleden Nov 19 '17 at 16:20
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    $\begingroup$ In this specific case, the integral exists as a Lebesgue integral, as well as as an improper Riemann integral, so speaking of the principal value is pointless. $\endgroup$ – Daniel Fischer Nov 19 '17 at 16:30
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    $\begingroup$ And to explain a little further: When the improper integral converges, the Cauchy Principle Value is the value of the improper integral. However, improper integrals can diverge, but still have a finite well-defined C.P.V. Because of how the technique you used works, it will give the C.P.V. for the integral, whether the integral converges or not (In fact, this is likely why the C.P.V. concept was introduced). $\endgroup$ – Paul Sinclair Nov 19 '17 at 21:10

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