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Contour Integration

Let $P(x)$ and $Q(x)$ be Polynomials of Degrees $n$ and $m$ with Coefficients $b_n$, ..., $b_0$ and $c_m$, ..., $c_0$. Take the contour in the upper half-plane, replace $x$ by $z$, and write $z\equiv Re^{i\theta}$. Then

\int_{-\infty}^\infty {P(z)\,dz\over Q(z)} = \lim_{R\to\infty} \int_{-R}^R {P(z)\,dz\over Q(z)}.
\end{displaymath} (1)

Define a path $\gamma_R$ which is straight along the Real axis from $-R$ to $R$ and makes a circular arc to connect the two ends in the upper half of the Complex Plane. The Residue Theorem then gives
$\lim_{R\to\infty} \int_{\gamma_R} {P(z)\,dz\over Q(z)}$
$ = \lim_{R\to\infty}\int_{-R}^R {P(z)\,dz\over Q(z)}+\lim_{R\to\infty} \int_0^\pi {P(R e^{i\theta})\over Q(Re^{i\theta})}iRe^{i\theta}\, d\theta$
$ = 2\pi i\sum_{\Im[z]> 0} {\rm Res} \left[{P(z)\over Q(z)}\right],\quad$ (2)
where Res denotes the Residues. Solving,

\lim_{R\to\infty}\int_{-R}^R {P(z)\,dz\over Q(z)} = 2\pi i\s...
...(R e^{i\theta})\over Q(Re^{i\theta})} iRe^{i\theta}\, d\theta.


$\displaystyle I_r$ $\textstyle \equiv$ $\displaystyle \lim_{R\to \infty} \int_0^\pi {P(Re^{i\theta})\over Q(Re^{i\theta})}iRe^{i\theta}\, d\theta$  
  $\textstyle =$ $\displaystyle \lim_{R\to \infty}\int_0^\pi {b_n(Re^{i\theta})^n+b_{n-1}(Re^{i\t...
...over c_m(Re^{i\theta} )^m+c_{m-1}(Re^{i\theta} )^{m-1}+\ldots +c_0} iR\,d\theta$  
  $\textstyle =$ $\displaystyle \lim_{R\to \infty}\int_0^\pi {b_n\over c_m} (Re^{i\theta})^{n-m}iR\,d\theta$  
  $\textstyle =$ $\displaystyle \lim_{R\to \infty }\int_0^\pi {b_n\over c_m} R^{n+1-m}i(e^{i\theta})^{n-m}\,d\theta$ (3)

and set

\epsilon \equiv -(n+1-m),
\end{displaymath} (4)

then equation (3) becomes
I_r\equiv\lim_{R\to \infty} {i\over R^\epsilon} {b_n\over c_m} \int_0^\pi e^{i(n-m)\theta}\,d\theta.
\end{displaymath} (5)

\lim_{R\to\infty} R^{-\epsilon} = 0
\end{displaymath} (6)

for $\epsilon > 0$. That means that for $-n-1+m \geq 1$, or $m\geq n+2$, $I_R=0$, so
\int_{-\infty}^\infty {P(z)\,dz\over Q(z)} = 2\pi i\sum_{\Im[z]> 0} {\rm Res} \left[{P(z)\over Q(z)}\right]
\end{displaymath} (7)

for $m\geq n+2$. Apply Jordan's Lemma with $f(x)\equiv P(x)/Q(x)$. We must have
\lim_{x\to\infty} f(x)=0,
\end{displaymath} (8)

so we require $m\geq n+1$. Then
\int_{-\infty}^\infty {P(z)\over Q(z)} e^{iaz}\,dz
= 2\pi i...
...[z]> 0}\mathop{\rm Res}\left[{{P(z)\over Q(z)} e^{iaz}}\right]
\end{displaymath} (9)

for $m\geq n+1$.

Since this must hold separately for Real and Imaginary Parts, this result can be extended to

\int_{-\infty}^\infty {P(x)\over Q(x)} \cos(ax)\,dx = 2\pi\R...
...! {\rm Res}
\left[{{P(z)\over Q(z)} e^{iaz}}\right]}\right\}
\end{displaymath} (10)

\int_{-\infty}^\infty {P(x)\over Q(x)} \sin(ax)\,dx = 2\pi\I...
... {\rm Res}
\left[{{P(z)\over Q(z)} e^{iaz}}\right]}\right\}.
\end{displaymath} (11)

It is also true that
\int_{-\infty}^\infty {P(z)\over Q(z)} \ln(az)\,dz = 0.
\end{displaymath} (12)

See also Cauchy Integral Formula, Cauchy Integral Theorem, Inside-Outside Theorem, Jordan's Lemma, Residue (Complex Analysis), Sine Integral


Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 353-356, 1953.

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© 1996-9 Eric W. Weisstein