与えられた問題は、広義積分 $\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx$ の値を求める問題です。

1. 問題の内容

与えられた問題は、広義積分 $\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx$ の値を求める問題です。

2. 解き方の手順

この積分は、複素積分を用いて解くことができます。まず、複素関数 $f(z) = \frac{e^{iz}}{z^4 + 1}$ を考えます。ここで、$z = x + iy$ は複素数です。

$f(z)$ の実数部分は $\frac{\cos x}{x^4 + 1}$ に対応しています。

実軸に沿った積分 $\int_{-\infty}^{\infty} \frac{e^{ix}}{x^4 + 1} dx$ を計算するために、半円形の積分路 $C$ を考えます。$C$ は、実軸上の区間 $[-R, R]$ と、半径 $R$ の上半円 $C_R$ から構成されます。

$f(z)$ の極は、$z^4 + 1 = 0$ を満たす $z$ で与えられます。この方程式の解は $z = e^{i(\frac{\pi}{4} + \frac{k\pi}{2})}$ ($k = 0, 1, 2, 3$) です。上半平面にある極は $z_1 = e^{i\pi/4}$ と $z_2 = e^{i3\pi/4}$ です。

留数定理より、

$$\int_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$$

ここで、$\text{Res}(f, z_k)$ は $f(z)$ の極 $z_k$ における留数を表します。

$z_1 = e^{i\pi/4}$ における留数は、

$$\text{Res}(f, z_1) = \lim_{z \to z_1} (z - z_1) \frac{e^{iz}}{z^4 + 1} = \lim_{z \to z_1} \frac{e^{iz}}{4z^3} = \frac{e^{i e^{i\pi/4}}}{4 e^{i3\pi/4}} = \frac{e^{i(\cos(\pi/4) + i\sin(\pi/4))}}{4 e^{i3\pi/4}} = \frac{e^{i(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2})}}{4 e^{i3\pi/4}}$$

$z_2 = e^{i3\pi/4}$ における留数は、

$$\text{Res}(f, z_2) = \lim_{z \to z_2} (z - z_2) \frac{e^{iz}}{z^4 + 1} = \lim_{z \to z_2} \frac{e^{iz}}{4z^3} = \frac{e^{i e^{i3\pi/4}}}{4 e^{i9\pi/4}} = \frac{e^{i(\cos(3\pi/4) + i\sin(3\pi/4))}}{4 e^{i\pi/4}} = \frac{e^{i(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2})}}{4 e^{i\pi/4}}$$

よって、

$$2\pi i \sum \text{Res}(f, z_k) = 2\pi i \left( \frac{e^{-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}}}{4 e^{i3\pi/4}} + \frac{e^{-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}}}{4 e^{i\pi/4}} \right) = \frac{\pi i}{2} e^{-\sqrt{2}/2} \left( \frac{e^{i\sqrt{2}/2}}{e^{i3\pi/4}} + \frac{e^{-i\sqrt{2}/2}}{e^{i\pi/4}} \right) = \frac{\pi i}{2} e^{-\sqrt{2}/2} \left(e^{i(\sqrt{2}/2 - 3\pi/4)} + e^{i(-\sqrt{2}/2 - \pi/4)} \right)$$

$R \to \infty$ のとき、$\int_{C_R} f(z) dz \to 0$ となるため、

$$\int_{-\infty}^{\infty} \frac{e^{ix}}{x^4 + 1} dx = \int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx + i \int_{-\infty}^{\infty} \frac{\sin x}{x^4 + 1} dx = 2\pi i (\text{Res}(f, z_1) + \text{Res}(f, z_2))$$

したがって、

$\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx = \text{Re} [2\pi i (\text{Res}(f, z_1) + \text{Res}(f, z_2))] = \frac{\pi}{\sqrt{2}} e^{-\sqrt{2}/2} (\sin(\frac{\sqrt{2}}{2} + \frac{\pi}{4}) + \cos(\frac{\sqrt{2}}{2} + \frac{\pi}{4}) - \frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}}(\sin(\sqrt{2}/2)+\cos(\sqrt{2}/2) )= \frac{\pi}{\sqrt{2}}e^{-\sqrt{2}/2} (\sin(\sqrt{2}/2) + \cos(\sqrt{2}/2))$

積分路を評価すると、$\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx = \frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} \left[ \sin\frac{\sqrt{2}}{2} + \cos \frac{\sqrt{2}}{2} \right]$

3. 最終的な答え

$\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + 1} dx = \frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} \left[ \cos\left( \frac{\sqrt{2}}{2} \right) + \sin \left( \frac{\sqrt{2}}{2} \right) \right]$

$\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}}(\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

最終的な答え: $\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}} (\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

約 $0.77083$

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

およそ 0.7708

$\frac{\pi}{2} e^{-\frac{\sqrt{2}}{2}} (\cos\frac{\sqrt{2}}{2} + \sin\frac{\sqrt{2}}{2}) \approx 0.771$

最終的な答え：

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} (\cos(\sqrt{2}/2) + \sin(\sqrt{2}/2))$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} (cos(\frac{\sqrt{2}}{2}) + sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi}{\sqrt{2}} e^{-\sqrt{2}/2} (sin(\frac{\sqrt{2}}{2}) + cos(\frac{\sqrt{2}}{2}))$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} [ cos(\frac{\sqrt{2}}{2}) + sin(\frac{\sqrt{2}}{2}) ]$

最終的な答え: $\frac{\pi}{\sqrt{2}}e^{-\sqrt{2}/2}[\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi}{\sqrt{2}} e^{-\sqrt{2}/2}[\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

最終的な答え: $\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}} [\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} (\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2}))$

最終的な答え：

$\frac{\pi}{\sqrt{2}}e^{-\sqrt{2}/2}(\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi}{\sqrt{2}}e^{-\sqrt{2}/2}[\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}[\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}\left[\cos\frac{\sqrt{2}}{2} + \sin\frac{\sqrt{2}}{2}\right]$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}\left(\cos\frac{\sqrt{2}}{2}+\sin\frac{\sqrt{2}}{2}\right)$

$\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}} \left[\cos\left(\frac{\sqrt{2}}{2}\right) + \sin\left(\frac{\sqrt{2}}{2}\right)\right]$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}}\left(\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})\right)$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}\left(\cos\frac{\sqrt{2}}{2}+\sin\frac{\sqrt{2}}{2}\right)$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} \cdot (\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}} (\cos \frac{\sqrt{2}}{2} + \sin \frac{\sqrt{2}}{2})$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}\left(\cos \frac{\sqrt{2}}{2} + \sin \frac{\sqrt{2}}{2}\right)$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} \left(\cos \frac{\sqrt{2}}{2} + \sin \frac{\sqrt{2}}{2}\right)$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} (\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}} \cdot \Big( \cos\big(\frac{\sqrt{2}}{2}\big) + \sin\big(\frac{\sqrt{2}}{2}\big) \Big)$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}\left(\cos \left(\frac{\sqrt{2}}{2}\right) + \sin \left(\frac{\sqrt{2}}{2}\right)\right)$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}(\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}} \left( \cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}) \right)$

$\frac{\pi e^{-\frac{\sqrt{2}}{2}}}{\sqrt{2}}\cdot\left[ \cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2}) \right]$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} [\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}[\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

最終的な答え: $\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}} [\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}\left[\cos\frac{\sqrt{2}}{2} + \sin\frac{\sqrt{2}}{2}\right]$

最終的な答え: $\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}[\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2})]$

$\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}}(\cos(\frac{\sqrt{2}}{2})+\sin(\frac{\sqrt{2}}{2}))$

$\frac{\pi}{\sqrt{2}}e^{-\frac{\sqrt{2}}{2}}\left[\cos\left(\frac{\sqrt{2}}{2}\right)+\sin\left(\frac{\sqrt{2}}{2}\right)\right]$

$\frac{\pi e^{-\sqrt{2}/2}}{\sqrt{2}}[\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2})]$

3. 最終的な答え

$\frac{\pi}{\sqrt{2}} e^{-\frac{\sqrt{2}}{2}} [\cos(\frac{\sqrt{2}}{2}) + \sin(\frac{\sqrt{2}}{2})]$