5.3 Application of real integral

1.

(a)

$$\int_{-\infty}^{\infty}\frac{1}{x^2 + x + 1} dx$$

Let the closed curve $C$ be the straight line connecting $-R$ and $R$ on the real axis, and $C_{R}$ be the curveof the radius $R$ with center 0 connecting $R$ and $-R$. Then the integral can be expressed as follows.

$$\int_{C}\frac{1}{z^2 + z + 1} dz = \int_{-R}^{R}\frac{1}{x^2 + x + 1} dx + \int_{C_{R}}\frac{1}{z^2 + z + 1} dz$$

Now we find $\int_{C}\frac{1}{z^2 + z + 1} dz$. The singularities are $z = \frac{-1 \pm \sqrt{3}i}{2} = e^{-\frac{2\pi i}{3}}$. But $z = \frac{-1 + \sqrt{3}i}{2} = e^{-\frac{2\pi i}{3}}$ is inside of the curve $C$. Thus by the residue theorem, we have

$$\int_{C}\frac{1}{(z^2 +1)(z^2 +4)} dz = 2\pi i(Res[e^{-\frac{2\pi i}{3}}])$$

Thus,

$$Res[\frac{-1 + \sqrt{3}i}{2}] = \lim_{z \to \frac{-1 + \sqrt{3}i}{2}}(z - \frac{-1 + \sqrt{3}i}{2})\frac{1}{z^2 + z + 1}$$

$$= \lim_{z \to \frac{-1 + \sqrt{3}i}{2}}\frac{1}{2z + 1} = \frac{1}{\sqrt{3} i} ( )$$

and

$$\int_{C}\frac{1}{z^2 + z + 1} dz = 2\pi i(\frac{1}{\sqrt{3}i}) = \frac{2\pi}{\sqrt{3}}$$

Next, if we can show $\int_{C_{R}}\frac{1}{z^2 + z+ 1} dz \rightarrow 0 (R \to \infty)$, then

$$\int_{-\infty}^{\infty}\frac{1}{x^2 + x + 1} dx = \int_{C}\frac{1}{z^2 + z + 1} dz = \frac{2\pi}{\sqrt{3}}$$

and we have found the real integral. Note that $\int_{C_{R}}\frac{1}{z^2 + z+ 1} dz \rightarrow 0 (R \to \infty)$繧堤､ｺ縺99.

$$\vert z_{1} + z_{2}\vert \leq \vert z_{1}\vert + \vert z_{2}\vert, \vert z_{1} - z_{2}\vert \geq \vert z_{1}\vert - \vert z_{2}\vert$$

Then

$$\vert z^2 + z + 1\vert = \vert z^2 - (-z-1)\vert \geq \vert z^2\vert - \vert-(z+1)\vert = \vert Z^2\vert - \vert z\vert -1 = R^2 - R -1$$

Thus,

$$\vert\frac{1}{z^2 + z + 1}\vert \leq \frac{1}{\vert z^2\vert - \vert z\vert -1}$$

$$\leq \frac{1}{R^2 - R -1} \to 0 (R \to \infty)$$

(b)

$$\int_{0}^{\infty}\frac{1}{(x^2 + 1)(x^2 + 4)} dx = \frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(x^2 + 1)(x^2 + 4)} dx$$

Let the closed curve $C$ be the straight line connecting $-R$ and $R$ on the real axis, and $C_{R}$ be the curve of the radius $R$ with center 0 connecting $R$ and $-R$. Then the integral can be expressed as follows.

$$\int_{C}\frac{1}{(z^2 + 1)(z^2 + 4)} dz = \int_{-R}^{R}\frac{1}{(x^2 +1)(x^2 +4)} dx + \int_{C_{R}}\frac{1}{(z^2 + 1)(z^2 + 4)} dz$$

Now we find $\int_{C}\frac{1}{(z^2 +1)(z^2 +4)} dz$. The singularites are $z = \pm i, \pm 2i$. But $z = i, 2i$ are only points inside of $C$. Thus by the residue theorem, we can find the integral by

$$\int_{C}\frac{1}{(z^2 +1)(z^2 +4)} dz = 2\pi i(Res[i] + Res[2i])$$

Here,

$$Res[i] = \lim_{z \to i}(z - i)\frac{1}{(z^2 + 1)(z^2 + 4)} = \lim_{z \to i}\frac{1}{(z+i)(z^2 +4)} = \frac{1}{2i(3)} = \frac{1}{6i}$$

$$Res[2i] = \lim_{z \to 2i}(z - 2i)\frac{1}{(z^2 + 1)(z^2 + 4)} = \lim_{z \to 2i}\frac{1}{(z^2 + 1)(z + 2i)} = \frac{1}{-3(4i)} = \frac{-1}{12i}$$

implies

$$\int_{C}\frac{1}{(z^2 +1)(z^2 +4)} dz = 2\pi i(\frac{1}{6i} - \frac{1}{12i}) = 2\pi i (\frac{1}{12 i}) = \frac{\pi}{6}$$

Next if we can show $\int_{C_{R}}\frac{1}{(z^2 + 1)(z^2 + 4)} dz \rightarrow 0 (R \to \infty)$, then

$$\int_{-\infty}^{\infty}\frac{1}{(x^2 +1)(x^2 +4)} dx = \int_{C}\frac{1}{(z^2 + 1)(z^2 + 4)} dz = \frac{\pi}{6}$$

and we have found the real integral. So, we need to show $\int_{C_{R}}\frac{1}{(z^2 + 1)(z^2 + 4)} dz \rightarrow 0 (R \to \infty)$.

$$\vert z^2 + 1\vert = \vert z^2 - (-1)\vert \geq \vert z^2\vert - \vert-1\vert = \vert Z^2\vert - 1 = R^2 -1$$

$$\vert z^2 + 4\vert = \vert z^2 - (-4)\vert \geq \vert z^2\vert - \vert-4\vert = \vert z^2\vert - 4 = R^2 - 4$$

Then

$$\vert\frac{1}{(z^2 + 1)(z^2 + 4)}\vert \leq \frac{1}{\vert z^2 + 1\vert\vert z^2 + 4\vert}$$

$$\leq \frac{1}{R^2 -1}\frac{1}{R^2 - 4} \to 0 (R \to \infty)$$

Thus, $\lim_{R \to \infty}\int_{C_R}\frac{1}{(z^2 + 1)(z^2+4)}dx = 0$.

(c) It is a integral of a trigonometric function. So, we let $z = e^{i\theta}$. Then the curve $C$ is a unit circle with the center at the origin. Next we write $\sin{\theta}$ using $z$.

$$\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2i} = \frac{z-z^{-1}}{2i}$$

Then $dz = ie^{i\theta} = izd\theta$. Thus,

$$\int_{0}^{2\pi}\frac{d\theta}{1+\sin{\theta}} = \oint_{C}\frac{1}{2+\frac{z - z^{-1}}{2i}}\frac{dz}{iz} = \oint_{C}\frac{2}{z^2 + 4iz -1}dz$$

The singularities are $z = -2i \pm \sqrt{3}i$. But $z = -2i - \sqrt{3}i$ is outside of the curve $C$. So, we need to find the residue of $z = -2i + \sqrt{3}i$. Since $z = -2i + \sqrt{3}i$ is the pole of the 1st order,

$$Res[-2i + \sqrt{3}i] = \lim_{z \to -2i + \sqrt{3}i}(z - (-2i + \s... ...^2 + 4iz -1}= \lim_{z \to -2i + \sqrt{3}i}\frac{2}{2z+4i} = \frac{1}{\sqrt{3}i}$$

Therefore,

$$\oint_{C}\frac{2}{z^2 + 4iz -1}dz = 2\pi i\frac{1}{\sqrt{3}i} = \frac{2\pi}{\sqrt{3}}$$

and

$$\int_{0}^{2\pi}\frac{d\theta}{2 +\sin{\theta}} = \frac{2\pi}{3}$$

(d) To solve this integral, we conside the curve $C$ represented by $z = e^{i\theta}$. Then $\sin{\theta} = \frac{z - z^{-1}}{2i}, dz = iz d\theta$. Thus

$$\int_{0}^{2\pi}\frac{1}{(2+\sin{\theta})^2}d\theta = \oint_{C}\frac{1}{(2 + \frac{z - z^{-1}}{2i})^2}\frac{1}{iz}dz$$

Using the residue theorem, we find $\oint_{C}\frac{1}{(2 + \frac{z - z^{-1}}{2i})^2}\frac{1}{iz}dz$. Since

$$(2 + \frac{z - z^{-1}}{2i})^2 = (2 + \frac{z^2 - 1}{2iz})^2 = 4 +... ...{2iz} + \frac{z^4 -2z^2 + 1}{-4z^2} = \frac{z^4 + 8z^3i -18z^2 -8zi + 1}{-4z^2}$$

we have

$$\frac{1}{(2 + \frac{z - z^{-1}}{2i})^2} = \frac{-4z^2}{z^4 + 8z^3i -18z^2 -8zi + 1}$$

Note that $z^4 + 8z^3i -18z^2 -8zi + 1 = (z^2 + 4iz -1)^2$. Then the singularities are $z = -2i \pm \sqrt{-3} = -2i \pm \sqrt{3}i$. Here, $-2i - \sqrt{3}i$ is otuside of the curve $C$. So, we only need to find the residue of $-2i + \sqrt{3}i$. Since,

$$\oint_{C}\frac{1}{(2 + \frac{z - z^{-1}}{2i})^2}\frac{1}{iz}dz = \oint_{C}\frac{-4\pi i}{z^4 + 8z^3i -18z^2 -8zi + 1}$$

, $z = -2i + \sqrt{3}i$ is the pole of the 2nd order. Thus,

$$Res[-2i + \sqrt{3}i] = \lim_{z \to -(2-\sqrt{3})i}\left(\frac{4\p... ...{(z -(-2-\sqrt{3}i)^2}\right)' = \frac{32\sqrt{3}i}{48} = \frac{-2\sqrt{3}i}{3}$$

Therefore,

$$\oint_{C}\frac{1}{(2 + \frac{z - z^{-1}}{2i})^2}\frac{1}{iz}dz = 2\pi i(\frac{-2\sqrt{3}i}{3}) = \frac{4\sqrt{3}\pi}{3}$$

(e) Consider the straight line $C_1$ connecting a point $R$ and $-R$ and the curve $C_R$ connecting a point $R$ to $-R$. Let $C$ be the closed curve formed by $C$ and $C_R$. Here, we conside the following integral.

$$\int_{C_1}\frac{e^{ix}}{(x^2 + 1)^2}dx + \int_{C_R}\frac{e^{iz}}{(z^2 + 1)^2}dz = \oint_{C}\frac{iz}{(z^2 + 1) 2}dz$$

First we evaluate $\oint_{C}\frac{e^{iz}}{(z^2 + 1)^2}dz$ using the residue theorem. $z=\pm i$ is the singular points. But $z = -i$ is outside of the curve $C$. Then we find the residue of $z = i$. Since $z = i$ is the pole of the order 2, we have

$$Res[i] = \lim_{z \to i}\left((z-i)^2\frac{e^{iz}}{(z+i)^2 (z-i)^2}\right)' = \frac{-ie^{-1}}{2}$$

Thus,

$$\oint_{C}\frac{e^{iz}}{(z^2 + 1)^2}dz = 2\pi i(\frac{-ie^{-1}}{2}) = \pi e^{-1}$$

Next we integrate on $C_1$.

$$\int_{C_{1}}\frac{e^{ix}}{(x^2 + 1)^2}dx = \int_{-R}^{R}\frac{e^{ix}}{(x^2 + 1)^2}dx = \int_{-R}^{0}\frac{e^{ix}}{(x^2 + 1)^2}dx + \int_{0}^{R}\frac{e^{ix}}{(x^2 + 1)^2}dx$$

$$= \int_{0}^{R}\frac{e^{-ix}}{(x^2 + 1)^2}dx + \int_{0}^{R}\frac{e^{ix}}{(x^2 + 1)^2}dx = 2\int_{0}^{R}\frac{\cos{x}}{(x^2 + 1)^2}dx$$

We integrate on $C_{R}$. We show $\int_{C_R}\frac{e^{iz}}{(z^2 + 1)^2}dx \to 0$ as $r \to \infty$. To do so, we only need to show that there exists $M,k$ such that $\vert\frac{e^{iz}}{(z^2 + 1)^2}\vert \leq \frac{M}{R^k}$.

$$\frac{e^{iz}}{(z^2 + 1)^2}\vert \leq \frac{\vert e^{iz}\vert}{(z^2 + 1)^2} \leq \frac{1}{z^4 - 2z^2} \leq \frac{1}{R^4 - R^2}$$

Thus, since

$$\int_{C_1}\frac{e^{ix}}{(x^2 + 1)^2}dx + \int_{C_R}\frac{e^{iz}}{(z^2 + 1)^2}dz = \oint_{C}\frac{iz}{(z^2 + 1) 2}dz = \pi e^{-1}$$

we have

$$2\int_{0}^{R}\frac{\cos{x}}{(x^2 + 1)^2}dx = \frac{\pi}{2e}$$

(f) Consider the straight line $C_1$ connecting a point $R$ and $-R$ and the curve $C_R$ connecting a point $R$ to $-R$. Let $C$ be the closed curve formed by $C$ and $C_R$. Here, we conside the following integral.

$$\int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx + \int_{C_R}\frac{-ize^{imz}}{z^2 + 1}dz = \oint_{C}\frac{-ize^{imz}}{z^2 + 1}dz$$

First we evaluate $\oint_{C}\frac{-ize^{imz}}{z^2 + 1}dz$. Note that $z=\pm i$ are the singularities. But $z = ^i$ is outside of the curve $C$. Thus we only need to find the residue of $z = i$. Since $z = i$ is the pole of the order 1. Thus,

$$Res[i] = \lim_{z \to i}\frac{-ize^{imz}}{(z+i)(z-i)} = \frac{e^{-m}}{2i}$$

Thus,

$$\oint_{C}\frac{-ize^{imz}}{z^2 + 1}dz = 2\pi i\frac{e^{-m}}{2i} = \pi e^{-m}$$

Next we consider the integral on $C_1$.

$$\int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx = \int_{-R}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx = \int_{-R}^{0}\frac{-ixe^{imx}}{x^2 + 1}dx + \int_{0}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx$$

$$= \int_{0}^{R}\frac{ixe^{-imx}}{x^2 + 1}dx + \int_{0}^{R}\frac{-ixe^{imx}}{x^2 + 1}dx = \int_{0}^{R}\frac{-ix(e^{imx}- e^{-imx})}{x^2 + 1}dx$$

$$= 2\int_{0}^{R}\frac{x\sin{mx}}{x^2 + 1}dx$$

Finally, we consider the integral on $C_{R}$. We show that $\int_{C_R}\frac{-ize^{imz}}{z^2 + 1}dx$ converges to 0 as $R \to \infty$. For $C_R$, $z = Re^{i\theta}$. Thus $dz = Rie^{i\theta}$ and

$$\int_{C_R}\vert\frac{-ize^{imz}}{z^2 +1}\vert dz = \int_{0}^{\pi}\vert\frac{-iRe^{i\theta}e^{iRme^{i\theta}}}{R^2 e^... ...}Rie^{i\theta}\vert d\theta = \int_{0}^{R}\vert e^{iRme^{i\theta}}\vert d\theta$$

$$= \int_{0}^{\pi}\vert e^{iRm(\cos{\theta} + i\sin{\theta})}\vert d\... ...int_{0}^{\pi}\vert e^{iRm(\cos{\theta})}\cdot e^{iRm(\sin{\theta}}\vert d\theta$$

$$= \int_{0}^{\pi}e^{-Rm\sin{\theta}}d\theta$$

Now we consider the graph of $\sin{\theta}$ and divide the above integral for $[0, \frac{\pi}{2}]$ and $[\frac{\pi}{2},\pi]$. Then

$$\int_{0}^{\pi}e^{-Rm\sin{\theta}}d\theta = 2\int_{0}^{\pi/2}e^{-Rmsin{\theta}}d\theta$$

Note that for $[0, \frac{\pi}{2}]$, $\sin{\theta} \geq \frac{2\theta}{\pi}$. Thu

$$2\int_{0}^{\pi/2}e^{-Rmsin{\theta}}d\theta \leq 2 \int_{0}^{\pi/2}e^{-Rm \frac{2\theta}{\pi}} d\theta$$

$$= 2\frac{-\pi}{2Rm}e^{-Rm\frac{2\theta}{\pi}}\mid_{0}^{\pi/2}$$

$$= \frac{\pi}{Rm}(e^{-Rm} -1) \to 0 (R \to \infty)$$

Thus,

$$\int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx + \int_{C_R}\frac{-ize^{imz}}{z^2 + 1}dz = \oint_{C}\frac{-ize^{imz}}{z^2 + 1}dz = \pi e^{-m}$$

and

$$\int_{C_1}\frac{-ixe^{imx}}{x^2 + 1}dx = 2\int_{0}^{R}\frac{x\sin{mx}}{x^2 + 1}dx$$

Therefore,

$$\int_{0}^{\infty}\frac{x\sin{mx}}{x^2 + 1}dx = \frac{\pi}{2e^{m}}$$

(g) To evaluate $\int_{0}^{\infty}\frac{\sin{x}}{x(x^2+1)^2}dx$, we consider the straight line $C_1$ connecting point $\varepsilon$ to $R$, the curve $C_2$ connecting $R$ to $-R$, the straight line $C_3$ connecting $-R$ to $\varepsilon$, the curve $C_4$ connecting $-\varepsilon$ to $\varepsilon$. The curve $C$ is composed of $C_1, C_2, C_3, C_4$. We consider the following integral.

$$\int_{C_1}\frac{-ie^{ix}}{x(x^2+1)^2}dx + \int_{C_2}\frac{-ie^{iz... ...{C_4}\frac{-ie^{iz}}{z(z^2 + 1)^2}dz = \oint_{C}\frac{-ie^{iz}}{z(z^2 + 1)^2}dz$$

First by using the residue theorem, we evaluate $\oint_{C}\frac{-ie^{iz}}{z(z^2 + 1)^2}dz$. The singularities are $z = 0, \pm i$. But $z = -i$ is outside of $C$. So, we need to find the residue of $z = 0, i$.

$$Res[0] = \lim_{z \to 0}\frac{-zie^{iz}}{z(z^2 + 1)^2} = -i$$

$z = i$ is the pole of the 2nd order. So,

$$Res[i] = \lim_{z \to i}\left((z-i)^2\frac{-ie^{iz}}{z(z^2 + 1)^2}\right)' = \frac{-12ie^{-1}}{-16} = \frac{3i}{4e}$$

Thus,

$$\oint_{C}\frac{-ie^{iz}}{z(z^2 + 1)^2}dz = 2\pi i(-i + \frac{3i}{4e}) = 2\pi - \frac{3\pi}{2e}$$

Next we evaluate the integral on $C_1$ and $C_3$.

$$\int_{\varepsilon}^{R}\frac{-ie^{ix}}{x(x^2+1)^2}dx + \int_{-R}^{-\varepsilon}\frac{-ie^{ix}}{x(x^2+1)^2}dx$$

Let $x = -u$. Then $dx = -du$. For $x: -R \to -\varepsilon$, $u:R \to \varepsilon$. Thus,

$$\int_{-R}^{-\varepsilon}\frac{-ie^{ix}}{x(x^2+1)^2}dx = \int_{R}^... ...iu}}{(-u)(u^2+1)^2}(-du) = \int_{\varepsilon}^{R}\frac{-ie^{-ix}}{x(x^2+1)^2}dx$$

Therefore,

$$\int_{-R}^{-\varepsilon}\frac{-ie^{ix}}{x(x^2+1)^2}dx + \int_{\varepsilon}^{R}\frac{-ie^{ix}}{x(x^2+1)^2}dx = \int_{\varepsilon}^{R}\frac{ie^{-ix} - ie^{ix}}{x(x^2+1)^2}dx = 2\int_{\varepsilon}^{R}\frac{e^{ix} - e^{-ix}}{2ix(x^2+1)^2}dx$$

$$= 2\int_{\varepsilon}^{R}\frac{\sin{x}}{x(x^2+1)^2}dx$$

We evaluate the integral on $C_2$. Here we show $\int_{C_2}\frac{-ie^{iz}}{z(z^2 + 1)^2}dz$ converges to 0 as $R \to \infty$. Since $z = Re^{i\theta}$,

$$\int_{C_2}\vert\frac{-ie^{iz}}{z(z^2 + 1)^2}dz\vert \leq \frac{1}... ...vert^2}\vert dz\vert \leq \frac{1}{R(R^4 - 2R^2)}\pi R \leq \frac{1}{R^4 - R^2}$$

Thus, this integral converges to 0 as $R$ goes to $\infty$.

Lastly, we evaluate the integral on $C_4$. Let $f(z) = \frac{-ie^{iz}}{z(z^2 + 1)^2}$. Then Laurent expansion of $f(z)$ around $z=0$ is

$$f(z) = \frac{-i}{z} + 1 + \frac{z}{2} + \cdots$$

Here we integrate the 1st term. Since $z = \varepsilon e^{i\theta}$, $dz = \varepsilon i e^{i\theta}$. Thus,

$$\int_{C_4}\frac{-i}{z}dz = -i\int_{C_4}\frac{\varepsilon i e^{i\theta}}{\varepsilon e^{i\theta}}d\theta = \int_{\pi}^{0}d\theta = \pi$$

Now the integration of the rest of terms are

$$\int_{C_4}\alpha z^{n}dz = \alpha\int_{\pi}^{0}\vert\varepsilon^n... ...)}\theta\vert d\theta = -\alpha \varepsilon^{n+1}\pi \to 0 (\varepsilon \to 0)$$

Putting all integrals together, we have

$$\int_{-\infty}^{\infty}\frac{\sin{x}}{x(x^2+1)^2}dx = 2\pi - \frac{3\pi}{2e} - \pi$$

Therefore,

$$\int_{0}^{\infty}frac{\sin{x}}{x(x^2+1)^2}dx = \frac{\pi}{2} - \frac{3\pi}{4e}$$

(g) To evaluate $\int_{0}^{\infty}\frac{1- \cos{mx}}{x^2} dx$, we consider the straight line $C_1$ connecting point $\varepsilon$ to $R$, the curve $C_2$ connecting $R$ to $-R$, the straight line $C_3$ connecting $-R$ to $\varepsilon$, the curve $C_4$ connecting $-\varepsilon$ to $\varepsilon$. The curve $C$ is composed of $C_1, C_2, C_3, C_4$. We consider the following integral.

$$\int_{C_1}\frac{1-e^{imx}}{x^2}dx + \int_{C_2}\frac{1-e^{imz}}{z^... ...}{x^2}dx + \int_{C_4}\frac{1-e^{imz}}{z^2}dz = \oint_{C}\frac{1-e^{imz}}{z^2}dz$$

First we evaluate $\oint_{C}\frac{1-e^{imz}}{z^2}dz$ using the residue theorem. Expand $f(z) = \frac{1-e^{imz}}{z^2}$ using Laurent expansion about $z=0$. Then

$$f(z) = \frac{1-(1+imz + \frac{1}{2}(imz)^2 + \cdots}{z^2} = \frac{-im}{z} + m^2 + \cdots$$

Thus, $z=0$ is the pole of the 1st order and the residue is $-im$. Therefore,

$$\oint_{C}\frac{1-e^{imz}}{z^2}dz = 2\pi i(-im) = 2\pi m$$

Next we evaluate the integral on $C_1$ and $C_3$.

$$\int_{\varepsilon}^{R}\frac{1-e^{imx}}{x^2}dx + \int_{-R}^{-\varepsilon}\frac{1-e^{imx}}{x^2}dx$$

Here, we set $x = -u$. Then $dx = -du$ and $x: -R \to -\varepsilon$, $u:R \to \varepsilon$. Thus,

$$\int_{\varepsilon}^{R}\frac{1-e^{imx}}{x^2}dx + \int_{\varepsilon}^{R}\frac{1-e^{-imx}}{x^2}dx = 2\int_{\varepsilon}^{R}\frac{1- \cos{mx}}{x^2}dx$$

We evaluate the integral on $C_2$. We show that $\int_{C_2}\frac{1-e^{imz}}{z^2}dz \to 0$ as $R \to \infty$.

$$\int_{C_2}\frac{1-e^{imz}}{z^2}dz \leq \frac{1}{\vert z\vert^2}\vert dz\vert \leq \frac{2\pi R}{R^2} \leq \frac{2\pi}{R} \to 0 (R \to \infty)$$

Finally , we evaluate the integral on $C_4$. Let $f(z) = \frac{1-e^{imz}}{z^2}$. Then expand $f(z)$ using Laurent expansion about $z=0$.

$$f(z) = \frac{1-(1+imz + \frac{(imz)^2}{2} + \cdots)}{z^2} = \frac{-im}{z} + \frac{m^2}{2} + \cdots$$

We let $z = \varepsilon e^{i\theta}$. Then $dz = \varepsilon i e^{i\theta}d\theta$. Also, $\theta$ moves from $\pi$ to 0. We evaluate this integral for the 1st term. Then

$$\int_{C_4}\frac{-im}{z}dz = \int_{\pi}^{0}\frac{-im}{\varepsilon ... ...)d \theta = \int_{\pi}^{0}\frac{-im}{i}d\theta = m\theta \mid_{\pi}^{0} = -m\pi$$

Putting all integrals together, we have

$$\int_{C_1}\frac{1-e^{imx}}{x^2}dx + \int_{C_2}\frac{1-e^{imz}}{z^... ... + \int_{C_4}\frac{1-e^{imz}}{z^2}dz = \oint_{C}\frac{1-e^{imz}}{z^2}dz = 2m\pi$$

Therefore,

$$\lim_{R \to \infty}\int_{-R}^{R}\frac{1-\cos{mx}}{x^2}dx= 2\pi m + m\pi = 3m\pi$$

and

$$\int_{0}^{\infty}\frac{1-\cos{mx}}{x^2}dx = \frac{3m\pi}{2}$$