Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The author of my old complex analysis textbook frequently asks the reader to calculate the Cauchy principal values of absolutely convergent real-valued integrals, e.g., EDIT: $\displaystyle\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx $. For a long time I thought that meant that EDIT: $\displaystyle\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx \ne \int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx $. But that's not correct, right?

share|improve this question
1  
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
1  
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
I meant $\text{PV} \int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} \ dx $. Sorry. –  Random Variable Mar 28 '12 at 21:31
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.