# The Cauchy principal value of an integral

The author of my complex analysis textbook asks the reader to find the Cauchy principal value of absolutely convergent real-valued integrals such as $\displaystyle\int_{{\color{red}{-\infty}}}^\infty \frac{\cos x}{1+x^{2}} \ dx$.

For a long time I thought that meant that $$\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx = \lim_{R \to \infty} \int_{-R}^{R} \frac{\cos x}{1+x^{2}} \ dx \ne \int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx.$$

But doesn't an absolutely convergent real-valued integral equal its Cauchy principal value?

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What definition of the principal value is your "old complex analysis textbook" using? Did you try comparing what happens if you try the "principal value" route as opposed to the route you are more accustomed to? – J. M. Mar 28 '12 at 20:48
By PV I meant Cauchy principal value. I think the author just didn't want to discuss when you can drop the PV sign. – Random Variable Mar 28 '12 at 20:56
I know what the $PV$ notation stands for; I am asking what definition your author is using, so that there's something to discuss (and I am familiar with at least two definitions)... – J. M. Mar 28 '12 at 21:03
Good, so I assume you want to use the first definition given in MathWorld. Do you think anything changes if you evaluate at different values of $c$? – J. M. Mar 28 '12 at 21:16

Generally PV is used in a "two-sided" situation: $PV \int_{-\infty}^\infty$ (where it means the limit of $\int_{-R}^R$ as $R \to +\infty$), or $PV \int_a^b$ if there is a singularity at $c \in (a,b)$ (where you take the limit of $\int_a^{c-\epsilon} + \int_{c+\epsilon}^b$ as $\epsilon \to 0+$).
In elementary complex analysis you most often encounter $\int_{-\infty}^\infty$. $PV \int_0^\infty \frac{\cos x}{1+x^2}\ dx$ doesn't really make sense as a principal value integral, but it actually comes from $PV \int_{-\infty}^\infty \frac{\cos x}{1+x^2} \ dx$, where the PV does make sense, by using symmetry. Of course the PV is not necessary here because the integral converges absolutely, and that will occur in very many of the examples.
I meant $\text{PV} \int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} \ dx$. Sorry. – Random Variable Mar 28 '12 at 21:31
What about for conditionally convergent integrals like $\int_{-\infty}^{\infty} \frac{\sin x}{x} \ dx$? What's the justification for dropping the PV sign if you use a principal value approach to evaluate it? – Random Variable Mar 28 '12 at 22:10
That one actually converges as an improper Riemann integral, i.e. $\int_0^N \frac{\sin x}{x}\ dx$ and $\int_{-N}^0 \frac{\sin x}{x}\ dx$ have limits as $N \to \infty$ (which, by symmetry, are the same limit), although it doesn't converge absolutely. – Robert Israel Mar 28 '12 at 23:09