7.7 Isolated singularities, and the residue theorem  (Page 4/4)

Consider the integral

Let us use the Residue Theorem to compute this integral.

Of course what we need to compute is

The first thing we do is to replace the real variable $x$ by a complex variable $Z,$ and observe that the function $f\left(z\right)=1/(1+z^4)$ is analytic everywhere except at the four points $\pm e^{i\pi /4}$ and $\pm e^{3i\pi /4}.$ See part (f) of the preceding exercise. These are the four points whose fourth power is $-1,$ and hence are the poles of the function $f.$

Next, given a positive number $B,$ we consider the geometric set (rectangle) $S_B$ that is determined by the interval $[-B,B]$ and the two bounding functions $l\left(x\right)=0$ and $u\left(x\right)=B.$ Then, as long as $B>1,$ we know that $f$ is analytic everywhere in $S^0$ except at the two points $c_1=e^{i\pi /4}$ and $c_2=e^{3i\pi /4},$ so that the contour integral of $f$ around the boundary of $S_B$ is given by

Now, this contour integral consists of four parts, the line integrals along the bottom, the two sides, and the top. The magic here is that the integrals along the sides, and the integral along the top,all tend to 0 as $B$ tends to infinity, so that the integral along the bottom, which after all is what we originally were interested in, is in the limit just the sum of the residues inside the geometric set.

An historically famous integral in analysis is ${\int }_{-\infty }^{\infty }sinx/xdx.$ The techniques described above don't immediately apply to this function, for, even replacing the $x$ by a $z,$ this function has no poles, so that the Residue Theorem wouldn't seem to be much help. Though the point 0 is a singularity, it is a removable one,so that this function $sinz/z$ is essentially analytic everywhere in the complex plane. However, even in a case like this we can obtain information about integrals ofreal-valued functions from theorems about integrals of complex-valued functions.

Notice first that ${\int }_{-\infty }^{\infty }sinx/xdx$ is the imaginary part of ${\int }_{-\infty }^{\infty }{e}^{ix}/xdx,$ so that we may as well evaluate the integral of this function. Let $f$ be the function defined by $f\left(z\right)=e^{iz}/z,$ and note that 0 is a pole of order 1 of $f,$ and that the residue $R_f\left(0\right)=2\pi i.$ Now, for each $B>0$ and $\delta >0$ define a geometric set $S_{B,\delta },$ determined by the interval $[-B,B],$ as follows: The upper bounding function $u_{B,\delta }$ is given by $u_{B,\delta }\left(x\right)=B,$ and the lower bounding function $l_{B,\delta }$ is given by $l_{B,\delta }\left(x\right)=0$ for $-B\le x\le -\delta$ and $\delta \le x\le B,$ and $l_{B,\delta }\left(x\right)=\delta e^{i\pi x/\delta }$ for $-\delta <x<\delta .$ That is, $S_{B,\delta }$ is just like the rectangle $S_B$ in Example 1 above, except that the lower boundary is not a straight line.Rather, the lower boundary is a straight line from $-B$ to $-\delta ,$ a semicircle below the $x$ -axis of radius $\delta$ from $-\delta$ to $\delta ,$ and a straight line again from $\delta$ to $B.$

By the Residue Theorem, the contour integral

As in the previous example, the contour integrals along the two sides and across the top of $S_{B,\delta }$ tend to 0 as $B$ tends to infinity. Finally, according to part (e) of [link] , the contour integral of $f$ along the semicircle in the lower boundary is $\pi i$ independent of the value of $\delta .$ So,

implying then that