# Principal value integral

I am reading A. Zee, QFT in a nutshell, and in appendix 1 he has:

Meanwhile the principal value integral is defined by: $$\int dx\,{\cal P}{1\over x}f(x)~=~ \lim_{\epsilon \rightarrow 0} \int dx\, {x\over x^2+\epsilon^2}f(x)$$

Please can someone explain to me why this is the case? As I understood it the principal value integral is rather defined as $$\int_a^b dx\,{\cal P}{1\over x}f(x)~=~ \lim_{\epsilon \rightarrow 0^+} \int_a^{-\epsilon} dx\, {1\over x}f(x)+\lim_{\epsilon \rightarrow 0^+} \int_{\epsilon}^b dx\, {1\over x}f(x),$$ where $a<0<b$. But as far as I can see these two definitions are not equivalent.

-

The short answer is that the two principal value definitions agree on sufficiently well-behaved functions, but may disagree on sufficiently singular functions. For instance, on one hand $$\lim_{\epsilon\searrow 0} \int_{\mathbb{R}\backslash[-\epsilon,\epsilon]} \frac{\mathrm{d}x}{x^3}~=~0$$

is zero, while on the other hand

$$\lim_{\epsilon\searrow 0} \int_{\mathbb{R}} \frac{\mathrm{d}x}{x(x^2+\epsilon^2)}$$

is not well-defined, since the integrand is not integrable at $x=0$.

I) Here we would like to investigate further the definition of principal value $P\int\! \mathrm{d} x$.

Definition. Let $\chi=(\chi_{\epsilon})_{\epsilon>0}$ be a family of functions $\chi_{\epsilon}:\mathbb{R}\to [0,1]\subseteq \mathbb{R}$ that are:

1. even functions $\chi_{\epsilon}(x)~=~\chi_{\epsilon}(-x),$

2. Lebesgue measurable functions,

3. $\chi_{\epsilon}(x)\nearrow 1$ pointwise almost everywhere for $\epsilon \searrow 0$.

Let us refer to such a function $\chi_{\epsilon}$ as a kernel function.

Examples of kernel functions $\chi_{\epsilon}$ are for instance:

1. the characteristic function $$\chi_{\epsilon}(x) ~=~\chi^{\rm std}_{\epsilon}(x) ~:=~ 1_{\mathbb{R}\backslash[-\epsilon,\epsilon]}(x)$$ for the set $\mathbb{R}\backslash[-\epsilon,\epsilon]$. (This choice $\chi^{\rm std}$ will lead to the standard definition of principal value.)

2. the continuous function $$\chi_{\epsilon}(x) ~=~ \chi^{a,b}_{\epsilon}(x) ~:=~ \frac{|x|^a}{|x|^a+ \epsilon^b},$$ where $a,b>0$ are two positive constants. (The choice $\chi^{2,2}$ will lead to the other definition of principal value mentioned by OP.)

3. the constant unit function $\chi_{\epsilon}(x) ~=~ 1$. (Unsurprisingly, this latter choice will turn out to be not so useful.)

II)

Definition. Define the set $V(\chi)$ of $\chi$-admissible functions as $$V(\chi)~:=~\left\{ f: \mathbb{R} \to \mathbb{C} ~\left|~ \begin{array}{c} f~\text{is Lebesgue measurable},\cr \forall \epsilon>0~:~~ \chi_{\epsilon} f~\in~ {\cal L}^1(\mathbb{R}), \cr \text{and} \cr \left(\int\! \mathrm{d}x~ \chi_{\epsilon}(x) f(x)\right)_{\epsilon>0} \text{is convergent for}~ \epsilon\searrow 0 \end{array}\right.\right\}.$$

Definition. If a function $f\in V(\chi)$ is $\chi$-admissible, we define the $\chi$-based principal value as $$P(\chi)\int\! \mathrm{d} x f(x)~:=~\lim_{\epsilon\searrow 0} \int\! \mathrm{d}x~ \chi_{\epsilon}(x) f(x).$$

Here ${\cal L}^1(\mathbb{R})$ denotes the set of functions that are Lebesgue integrable, i.e. functions that are Lebesgue measurable and whose absolute value has a finite integral. ${\cal L}^1(\mathbb{R})$ is an example of an ${\cal L}^p$ space.

III) It is not hard to see that:

1. If $f\in{\cal L}^1(\mathbb{R})$ is Lebesgue integrable, then it is $\chi$-admissible $f\in V(\chi)$, and the principal value $$P(\chi)\int\! \mathrm{d} x ~f(x)~=~ \int\! \mathrm{d} x ~f(x)$$ is just the ordinary Lebesgue integral because of the Lebesgue dominated convergence theorem.

2. The set $V(\chi)$ of $\chi$-admissible functions is a $\mathbb{C}$-vector space.

3. If a function $f\in V(\chi)$ is $\chi$-admissible, so is the mirrored function $(x\mapsto f(-x))\in V(\chi)$, with same principal value.

4. If an $\chi$-admissible function $f\in V(\chi)$ is odd, then $P(\chi)\int\! \mathrm{d} x~f(x) ~=~ 0$.

Thus it is enough to investigate even and odd functions.

Finally, let us investigate power functions $x\mapsto x^p$, $p\in\mathbb{R}$, which play an important role in practice as building blocks.

IV) Even functions. Let

$$g_{p,K}(x) ~:=~ 1_{[-K,K]}(x) |x|^p~=~g_{p,K}(-x)$$

be a truncated power function, where $p\in\mathbb{R}$ is a real power, and where $K>0$ is a positive truncation constant.

It is not hard to show that in the case of Example 1, 2, or 3,

$$g_{p,K}\in V(\chi) \qquad \Leftrightarrow \qquad p>-1\qquad \Leftrightarrow \qquad g_{p,K}\in {\cal L}^1(\mathbb{R}).$$

In the affirmative case $p>-1$, the principal value definitions based on the three Examples 1, 2, and 3 agree:

$$P(\chi)\int\! \mathrm{d} x ~g_{p,K}~=~\int\! \mathrm{d} x ~g_{p,K}~=~ \frac{2K^{p+1}}{p+1}.$$

V) Odd functions. Let

$$h_{p,K}(x) ~:=~ {\rm sgn}(x) 1_{[-K,K]}(x) |x|^p~=~-h_{p,K}(-x)$$

be a truncated power function, where $p\in\mathbb{R}$ is a real power, and where $K>0$ is a positive truncation constant. In the three Examples 1, 2, and 3, we get

1. $h_{p,K}\in V(\chi^{\rm std})$ always,
2. $h_{p,K}\in V(\chi^{a,b}) \qquad \Leftrightarrow \qquad p+a>-1$,
3. $h_{p,K}\in V(1) \qquad \Leftrightarrow \qquad p>-1\qquad \Leftrightarrow \qquad h_{p,K}\in {\cal L}^1(\mathbb{R}).$
-

Note that the right spelling is "principal value".

The formulae aren't identical but the results are the same whenever both definitions yield a well-defined expression. What matters is that we remove the leading logarithmic divergence on both sides from $x=0$ and we do so in a symmetric way with respect to $x\to -x$.

If you denote the second definition-based integral $Cut(\epsilon)$, $$Cut(\epsilon) = \left(\int_{a}^{-\epsilon}+\int_{\epsilon}^b\right) \frac{dx}x f(x)$$ then I claim that there exists a weighting function $g(y)$ such that $$\int_0^{\infty} g(y) Cut(y) dy = \int dx\,\frac{x}{x^2+\epsilon^2}f(x)$$ so it reduces to the first definition-based integral. The function $g(y)$ is supported for $y$ of the same order as $\epsilon$ so the limit has the same effect on both expressions.

You may also see the equivalence of both expressions if you just Taylor-expand $f(x)$ near zero. Assuming that $f(x)$ is finite and well-behaved near $x=0$, it's easy to prove that both definitions yield the same result. The real purpose of the "principal value" terminology deals with branches of functions of complex variables. So you may also imagine that $f(x)$ is a meromorphic or holomorphic function of a complex $x$. The indefinite integral $\int f(x)/x$ has a logarithmic singularity around $x=0$ and one needs to define on which branch we are at. The principal value takes the average of the results one would get on the $+i\pi$ and $-i\pi$ branches for the logarithm of the negative numbers.

-
 Thanks for your help. I will correct the spelling. I think the $f(x) x/(x^2+\epsilon^2)$ in the question and answer should have integration limits from a to b. In that case I can see from examples that the two expressions are equivalent. I tried doing the Taylor series expansion to prove it but couldn't quite get the result. If anyone can point me to a reference where this is proved I would appreciate it. – physicsphile Jul 3 '12 at 18:02 Dear physicsphile, please, this is really trivial. You won't find textbooks written about this stuff. Write $f(x)=a_0+a_1x+a_2 x^2+a_3 x^3+\dots$. Then integrate $f(x)/x$ from $a$ to $b$. Obviously, almost all the terms behave in the same way whether you omit the $(-\epsilon,+\epsilon)$ interval or not, whether you add the extra factor $x^2/(x^2+\epsilon^2)$, or not. The only exception is the absolute term $a_0$, i.e. integral of $a_0/x$, and both rules give you $a_0\ln|b/a|$ as the result of the integral in the limit $\epsilon\to 0$. Just try it. – Luboš Motl Jul 4 '12 at 4:36 I see Weinberg in "The Quantum Theory of Fields I" describes its similarly to you on pg 113: "The function ${\cal P}/x\equiv x/(x^2+\epsilon^2)$ is just $1/x$ for $|x|\gg\epsilon$, and vanishes for $x\rightarrow 0$, so for $\epsilon\rightarrow 0$ it behaves just like the `principal value function' ${\cal P}/x$, which allows us to give meaning to integrals of $1/x$ times any smooth function of $x$, by excluding an infinitesimal interval around $x=0$." – physicsphile Jul 5 '12 at 6:31 Good (but not really necessary) to see a confirmation by a famous Gentleman. – Luboš Motl Jul 5 '12 at 6:46