Integrals of the form $\int_{- \infty}^{+ \infty} f(x) \; dx$ .

In calculus, we define an improper (real) integral with two infinite limits as follows:

$$\int_{- \infty}^{+ \infty} f(x) \; dx = \underset{R \rightarrow + \infty}{\lim} \int_{-R}^{R} f(x) \; dx$$

We can compute such integrals using integrals of functions of a complex variable. Consider the loop $\mathcal{C}$ consisting of the segment $[-R,R]$ on the $x-$ axis and the upper semicircle $\mathcal{C}_1$ defined by $\vert z\vert=R$ (with $\operatorname{Im}\nolimits (z)>0$ ),as in Figure 1, choosing $R$ such that this loop encloses all the singular points of the complex function $f(z)$ which are in the upper half-plane.

Figure 1: Upper semicircle.

Then we compute the integral

$$\oint_{\mathcal{C}} f(z) \; dz = \int_{-R}^{R} f(x) \; dx + \int_{\mathcal{C}_1} f(z) \; dz$$

We can show that

$$\underset{R \rightarrow + \infty}{\lim} \int_{\mathcal{C}_1} f(z) \; dz = 0$$

thus

$$\int_{-R}^{R} f(x) \; dx = \oint_{\mathcal{C}} f(z) \; dz = 2 \pi i \underset{z_0=\text{pole}}{\sum} \text{Res } (f(z), z_0)$$ (9.1)

Example 9.3.1   We compute the integral

$$I= \int_{-R}^{R} \frac {1}{x^4+1} \; dx$$

Let $f(z)=\frac {1}{z^4+1}$ . This function has 4 poles in the complex plane, namely the solutions of the equation $z^4+1=0$ . These poles are:

$$\frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{2}, \; \frac {\sqrt{2}}{2}-... ...sqrt{2}}{2}-i\frac {\sqrt{2}}{2}, \; -\frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{2}.$$ (9.2)

Thus we have:

$$f(z) =\frac {1}{z^4+1}$$

$$\quad =\frac {1}{ \left[z-\left( \frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{... ...ght] \left[ z-\left( -\frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{2}\right) \right] }$$

$$\quad =\frac {1}{ [z-e^{i \pi /4}][z-e^{3i \pi /4}][z-e^{5i \pi /4}][z-e^{7i \pi /4}]. }$$

Let $R$ be a positive real number such that all of these poles with positive imaginary part are inside the loop defined as above, namely the first and the fourth of the numbers listed in 6; these are

$$e^{i \pi /4}=\frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{2} \qquad e^{3i \pi /4}=-\frac {\sqrt{2}}{2}+i\frac {\sqrt{2}}{2}.$$ (9.3)

Now we compute the residues of $f(z)$ at each of these points; in what follows, $\mathcal{C}_2$ and $\mathcal{C}_3$ are simple loops obtained by ``cutting'' $\mathcal{C}$ into two parts, such that $e^{i \pi /4}$ is an interior point of $\mathcal{C}_2$ , but not of $\mathcal{C}_3$ , and $e^{3i \pi /4}$ is an interior point of $\mathcal{C}_3$ , but not of $\mathcal{C}_2$ (cf Figure 2).

Figure 2: Two simple loops in an upper semicircle.

$$2 \pi i (f(z),e^{i \pi /4}) = \oint_{\mathcal{C}_2} \frac {1}{(z-e^{3i \pi /4})(z-e^{5i \pi /4})(z-e^{7i \pi /4})(z-e^{i \pi /4})} \;dz$$

$$\quad = \oint_{\mathcal{C}_2} \frac {\; \; \frac {1}{(z-e^{3i \pi /4})(z-e^{5i \pi /4})(z-e^{7i \pi /4})} \; \; } {z-e^{i \pi /4}} \;dz$$

$$\quad =\left[ \frac {1}{(z-e^{3i \pi /4})(z-e^{5i \pi /4})(z-e^{7i \pi /4})} \right] _{z=e^{i \pi /4}}$$

$$\quad =-\frac {\sqrt{2}}{8}(1+i).$$

$$2 \pi i (f(z),e^{3i \pi /4}) \quad = \oint_{\mathcal{C}_3} \frac {\; \; \frac {1}{(z-e^{i \pi /4})(z-e^{5i \pi /4})(z-e^{7i \pi /4})} \; \; } {z-e^{3i \pi /4}} \;dz$$

$$\quad = \left[ \frac {1}{(z-e^{i \pi /4})(z-e^{5i \pi /4})(z-e^{7i \pi /4})} \right] _{z=e^{3i \pi /4}}$$

$$\quad = \frac {\sqrt{2}}{8}(1-i).$$

$$2 \pi i [ (f(z),e^{i \pi /4}) + (f(z),e^{3i \pi /4})] =-i \frac {\sqrt{2}}{4}.$$

Thus, we have:

$$I= \int_{-R}^{R} \frac {1}{x^4+1} \; dx = \frac {\pi \sqrt{2}}{2}. = 2 \pi i \cdot \left( -i \frac {\sqrt{2}}{4} \right) = \frac {\pi \sqrt{2}}{2}.$$