与えられた二つの定積分を計算します。 (1) $I = \int_0^{2\pi} \frac{\cos\theta}{(5+4\cos\theta)^2} d\theta$ (2) $I = \int_{-\infty}^{+\infty} \frac{dx}{x^4+x^2+1}$

1. 問題の内容

与えられた二つの定積分を計算します。

(1) $I = \int_0^{2\pi} \frac{\cos\theta}{(5+4\cos\theta)^2} d\theta$

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

2. 解き方の手順

(1) 複素積分を用いて計算します。$z = e^{i\theta}$とおくと、$dz = i e^{i\theta} d\theta = iz d\theta$なので、$d\theta = \frac{dz}{iz}$となります。また、$\cos\theta = \frac{1}{2}(z + \frac{1}{z})$です。したがって、

$$I = \oint_{|z|=1} \frac{\frac{1}{2}(z+\frac{1}{z})}{(5+4\cdot \frac{1}{2}(z+\frac{1}{z}))^2} \frac{dz}{iz} = \oint_{|z|=1} \frac{\frac{1}{2}(z+\frac{1}{z})}{(5+2(z+\frac{1}{z}))^2} \frac{dz}{iz}$$

$$= \oint_{|z|=1} \frac{\frac{1}{2}(\frac{z^2+1}{z})}{(5+2(\frac{z^2+1}{z}))^2} \frac{dz}{iz} = \oint_{|z|=1} \frac{\frac{z^2+1}{2z}}{(\frac{5z+2z^2+2}{z})^2} \frac{dz}{iz} = \oint_{|z|=1} \frac{\frac{z^2+1}{2z}}{\frac{(2z^2+5z+2)^2}{z^2}} \frac{dz}{iz}$$

$$= \oint_{|z|=1} \frac{z^2+1}{2z} \cdot \frac{z^2}{(2z^2+5z+2)^2} \cdot \frac{1}{iz} dz = \oint_{|z|=1} \frac{z(z^2+1)}{2i(2z^2+5z+2)^2} dz$$

$$= \frac{1}{2i} \oint_{|z|=1} \frac{z(z^2+1)}{(2z^2+5z+2)^2} dz = \frac{1}{2i} \oint_{|z|=1} \frac{z(z^2+1)}{(2(z+2)(z+\frac{1}{2}))^2} dz = \frac{1}{2i} \oint_{|z|=1} \frac{z(z^2+1)}{4(z+2)^2(z+\frac{1}{2})^2} dz$$

$$= \frac{1}{8i} \oint_{|z|=1} \frac{z(z^2+1)}{(z+2)^2(z+\frac{1}{2})^2} dz$$

$f(z) = \frac{z(z^2+1)}{(z+2)^2}$とおくと、被積分関数は$\frac{f(z)}{(z+\frac{1}{2})^2}$となります。留数定理より、

$$\oint_{|z|=1} \frac{f(z)}{(z+\frac{1}{2})^2} dz = 2\pi i f'(-\frac{1}{2})$$

$$f'(z) = \frac{(3z^2+1)(z+2)^2 - z(z^2+1) \cdot 2(z+2)}{(z+2)^4} = \frac{(3z^2+1)(z+2) - 2z(z^2+1)}{(z+2)^3} = \frac{3z^3+6z^2+z+2 - 2z^3-2z}{(z+2)^3} = \frac{z^3+6z^2-z+2}{(z+2)^3}$$

$$f'(-\frac{1}{2}) = \frac{-\frac{1}{8} + 6(\frac{1}{4}) - (-\frac{1}{2}) + 2}{(\frac{3}{2})^3} = \frac{-\frac{1}{8} + \frac{3}{2} + \frac{1}{2} + 2}{\frac{27}{8}} = \frac{\frac{-1+12+4+16}{8}}{\frac{27}{8}} = \frac{31}{27}$$

$$I = \frac{1}{8i} \oint_{|z|=1} \frac{z(z^2+1)}{(z+2)^2(z+\frac{1}{2})^2} dz = \frac{1}{8i} (2\pi i f'(-\frac{1}{2})) = \frac{1}{8i} 2\pi i \frac{31}{27} = \frac{\pi}{4} \cdot \frac{31}{27} = \frac{31\pi}{108}$$

(2) $x^4+x^2+1 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$と因数分解できます。

$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$

$1 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$

$1 = Ax^3 - Ax^2 + Ax + Bx^2 -Bx + B + Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D$

$1 = (A+C)x^3 + (-A+B+C+D)x^2 + (A-B+C+D)x + (B+D)$

$A+C = 0$, $-A+B+C+D = 0$, $A-B+C+D = 0$, $B+D = 1$

$C=-A$, $-2A + B + D = 0$, $2A - B + D = 0$, $B+D = 1$

$-2A + 1 = 0$, $2A+1 = 0$なので、$A = \frac{1}{2}$, $C=-\frac{1}{2}$.

$-\frac{1}{2} + B + D = 0$より、$B+D=\frac{1}{2}$だが、$B+D=1$なので矛盾。

$x^4+x^2+1 = (x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)$と因数分解できる。

$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+\sqrt{3}x+1} + \frac{Cx+D}{x^2-\sqrt{3}x+1}$

$1 = (Ax+B)(x^2-\sqrt{3}x+1) + (Cx+D)(x^2+\sqrt{3}x+1)$

$1 = Ax^3 - \sqrt{3}Ax^2 + Ax + Bx^2 - \sqrt{3}Bx + B + Cx^3 + \sqrt{3}Cx^2 + Cx + Dx^2 + \sqrt{3}Dx + D$

$1 = (A+C)x^3 + (-\sqrt{3}A + B + \sqrt{3}C + D)x^2 + (A - \sqrt{3}B + C + \sqrt{3}D)x + (B+D)$

$A+C = 0$, $-\sqrt{3}A + B + \sqrt{3}C + D = 0$, $A - \sqrt{3}B + C + \sqrt{3}D = 0$, $B+D = 1$

$C = -A$, $-2\sqrt{3}A + B + D = 0$, $2A - \sqrt{3}(B-D) = 0$, $B+D=1$

$-2\sqrt{3}A + 1 = 0$, $2A = \sqrt{3}(B-D)$, $A = \frac{1}{2\sqrt{3}}$, $C = -\frac{1}{2\sqrt{3}}$

$2(\frac{1}{2\sqrt{3}}) = \sqrt{3}(B-D)$より$\frac{1}{\sqrt{3}} = \sqrt{3}(B-D)$

$B-D = \frac{1}{3}$, $B+D = 1$より、$2B = \frac{4}{3}$、$B = \frac{2}{3}$、$D = \frac{1}{3}$.

$$\int_{-\infty}^{\infty} \frac{dx}{x^4+x^2+1} = \int_{-\infty}^{\infty} \frac{\frac{1}{2\sqrt{3}}x + \frac{2}{3}}{x^2+\sqrt{3}x+1} + \frac{-\frac{1}{2\sqrt{3}}x + \frac{1}{3}}{x^2-\sqrt{3}x+1} dx$$

積分が難しいので、留数定理を使う。

$z^4+z^2+1 = 0$の解を求める。

$z^2 = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$

$z^2 = e^{\pm i \frac{2\pi}{3}}$

$z = e^{i (\pm \frac{\pi}{3} + \pi k)}$ ($k=0, 1$)

$z_1 = e^{i\frac{\pi}{3}} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$, $z_2 = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,

$z_3 = e^{-i\frac{\pi}{3}} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$, $z_4 = e^{-i\frac{2\pi}{3}} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.

上半平面にあるのは$z_1, z_2$。

$f(z) = \frac{1}{z^4+z^2+1}$

$I = 2\pi i (\text{Res}(f, z_1) + \text{Res}(f, z_2))$

$\text{Res}(f, z_1) = \lim_{z \to z_1} \frac{z-z_1}{z^4+z^2+1} = \lim_{z \to z_1} \frac{1}{4z^3+2z} = \frac{1}{4e^{i\pi}+2e^{i\frac{\pi}{3}}} = \frac{1}{-4+1+i\sqrt{3}} = \frac{1}{-3+i\sqrt{3}} = \frac{-3-i\sqrt{3}}{9+3} = \frac{-3-i\sqrt{3}}{12} = -\frac{1}{4} - i \frac{\sqrt{3}}{12}$

$\text{Res}(f, z_2) = \lim_{z \to z_2} \frac{1}{4z^3+2z} = \frac{1}{4e^{2\pi i}+2e^{i\frac{2\pi}{3}}} = \frac{1}{4+2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})} = \frac{1}{4-1+i\sqrt{3}} = \frac{1}{3+i\sqrt{3}} = \frac{3-i\sqrt{3}}{12} = \frac{1}{4} - i \frac{\sqrt{3}}{12}$

$\text{Res}(f, z_1) + \text{Res}(f, z_2) = -\frac{i\sqrt{3}}{6}$

$I = 2\pi i (-\frac{i\sqrt{3}}{6}) = \frac{\pi \sqrt{3}}{3}$

3. 最終的な答え

(1) $\frac{31\pi}{108}$

(2) $\frac{\pi \sqrt{3}}{3}$