Recall the definition of improper integrals in calculus:

\begin{align*}

\int_0^\infty f(x)dx&=\lim_{R\to\infty}\int_0^R f(x)dx,\\

\int_{-\infty}^\infty f(x)dx&=\lim_{R_1\to\infty}\int_{-R_1}^0 f(x)dx+\lim_{R_2\to\infty}\int_0^{R_2}f(x)dx.

\end{align*}

The *Cauchy Principal Value* (P.V.) is given by

$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=\lim_{R\to\infty}\int_{-R}^R f(x)dx.$$

The Cauchy principal value of an improper integral is not necessarily the same as the improper integral. For example,

$$\mathrm{P.V}\int_{-\infty}^\infty xdx=\lim_{R\to\infty}\int_{-R}^R xdx=0,$$

while

$$\int_{-\infty}^\infty xdx=\lim_{R_1\to\infty}\int_{-R_1}^0xdx+\lim_{R_2\to\infty}\int_0^{R_2}xdx=-\infty+\infty$$

is undefined. In general, if $\int_{-\infty}^\infty f(x)dx<\infty$ then $\mathrm{R.V.}\int_{-\infty}^\infty f(x)dx<\infty$, but the converse need not be true. Suppose that $f(x)$ is an even function. Then

\begin{align*}

\int_0^R f(x)dx&=\frac{1}{2}\int_{-R}^R f(x)dx,\\

\int_{-R_1}^0 f(x)dx&=\int_0^{R_1}f(x) dx.

\end{align*}

So,

$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty f(x)dx=2\int_0^\infty f(x)dx.$$

Let us consider an even function $f(x)$ of the form $f(x)=\frac{p(x)}{q(x)}$, where $p(x)$, $q(x)$ are polynomials with real coefficients no factors in common. Furthermore, we assume that $q(z)$ has no real zeros but has at least one zero above the real axis. Let us consider a positively oriented upper semicircle $C_R$ whose radius $R$ is large enough to contain all the zeros above the real axis as shown in the figure below.

$C_R$ together with the interval $[-R,R]$ form a positively oriented simple closed contour. Then by Cauchy’s Residue Theorem we have

$$\int_{-R}^R f(x)dx+\int_{C_R} f(z)dz=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z),$$

i.e.

$$\int_{-R}^R f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z)-\int_{C_R} f(z)dz.$$

If $\lim_{R\to\infty}\int_{C_R} f(z)dz=0$ then

$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).$$

If in addition $f(x)$ is even,

$$\int_{-\infty}^\infty f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z)$$

or

$$\int_0^\infty f(x)dx=\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).$$

*Example*. Let us consider the improper integral

$$\int_0^\infty\frac{x^2}{x^6+1}dx.$$

Let $f(z)=\frac{z^2}{z^6+1}$ has isolated singularities at the zeros of $z^6+1$, and is analytic everywhere else. $z^6=-1$ has solutions (the sixth roots of $-1$)

$$c_k=\exp\left[i\left(\frac{\pi}{6}+\frac{2k\pi}{6}\right)\right],\ k=0,1,\cdots,5.$$

The first three roots

$$c_0=e^{i\frac{\pi}{6}},\ c_1=i,\ c_2=e^{i\frac{5\pi}{6}}$$

lie in the upper half plane. Let us consider a positively oriented upper semicircle $C_R$ whose radius $R$ is greater than $1$.

Then

$$\int_{-R}^Rf(x)dx=2\pi(B_0+B_1+B_2)-\int_{C_R}f(z)dz,$$

where $B_k$ is the residue of $f(z)$ at $c_k$, $k=0,1,2$. $B_k$ can be found as we studied here

$$B_k=\mathrm{Res}_{z=c_k}\frac{z^2}{z^6+1}=\frac{c_k^2}{6c_k^5}=\frac{1}{6c_k^3},\ k=0,1,2.$$

Thus, we obtain

$$2\pi(B_0+B_1+B_2)=2\pi\left(\frac{1}{6i}-\frac{1}{6i}+\frac{1}{6i}\right)=\frac{\pi}{3}$$

and hence,

$$\int_{-R}^R f(x)dx=\frac{\pi}{3}-\int_{C_R}f(z)dz.$$

On $C_R$, $|z|=R$ so

$$|z^6+1|\geq ||z|^6-1|=|R^6-1|=R^6-1$$

and thereby we obtain

$$|f(z)|\leq\frac{R^2}{R^6-1}.$$

Since the length of $C_R$ is $\pi R$,

$$\left|\int_{C_R} f(z)dz\right|\leq\frac{R^2}{R^6-1}\cdot\pi R\to 0$$

as $R\to\infty$. Hence,

$$\mathrm{P.V.}\int_{-\infty}^\infty\frac{x^2}{x^6+1}dx=\lim_{R\to\infty}\frac{x^2}{x^6+1}dx=\frac{\pi}{3}.$$

Since the integrand is even,

$$\int_{-\infty}^\infty\frac{x^2}{x^6+1}dx=\frac{\pi}{3}$$

and

$$\int_0^\infty\frac{x^2}{x^6+1}dx=\frac{\pi}{6}.$$