与えられた問題は、次の定積分の値を求めることです。 $\int_{0}^{\infty} \frac{dx}{x^4 + 4}$

1. 問題の内容

与えられた問題は、次の定積分の値を求めることです。

$\int_{0}^{\infty} \frac{dx}{x^4 + 4}$

2. 解き方の手順

まず、$x^4 + 4$ を因数分解します。

$x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)$

次に、被積分関数を部分分数分解します。

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

両辺に $x^4 + 4$ を掛けると、

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

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

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

係数を比較すると、

$A+C = 0$

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

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

$2B + 2D = 1$

$A = -C$, $B+D = \frac{1}{2}$

$-2A + B - 2A + D = 0 \implies -4A + B + D = 0 \implies -4A + \frac{1}{2} = 0 \implies A = \frac{1}{8}$

$2A - 2B - 2A + 2D = 0 \implies -2B + 2D = 0 \implies B = D$

$B = D = \frac{1}{4}$

$A = \frac{1}{8}$, $C = -\frac{1}{8}$, $B = \frac{1}{4}$, $D = \frac{1}{4}$

$\frac{1}{x^4 + 4} = \frac{\frac{1}{8}x + \frac{1}{4}}{x^2 + 2x + 2} + \frac{-\frac{1}{8}x + \frac{1}{4}}{x^2 - 2x + 2}$

$\frac{1}{x^4 + 4} = \frac{1}{8} \left( \frac{x + 2}{x^2 + 2x + 2} + \frac{-x + 2}{x^2 - 2x + 2} \right)$

$\frac{1}{x^4 + 4} = \frac{1}{8} \left( \frac{x + 1}{x^2 + 2x + 2} + \frac{1}{(x+1)^2 + 1} + \frac{-x + 1}{x^2 - 2x + 2} + \frac{1}{(x-1)^2 + 1} \right)$

$\int_{0}^{\infty} \frac{1}{x^4 + 4} dx = \frac{1}{8} \int_{0}^{\infty} \left( \frac{x + 1}{(x+1)^2 + 1} + \frac{-x + 1}{(x-1)^2 + 1} \right) dx$

$\int \frac{x+1}{x^2 + 2x + 2} dx = \frac{1}{2} \ln(x^2 + 2x + 2)$

$\int \frac{-x+1}{x^2 - 2x + 2} dx = -\frac{1}{2} \ln(x^2 - 2x + 2)$

$\int_{0}^{\infty} \frac{dx}{x^4 + 4} = \frac{1}{8} \left[ \frac{1}{2} \ln(x^2 + 2x + 2) - \frac{1}{2} \ln(x^2 - 2x + 2) \right]_{0}^{\infty}$

$= \frac{1}{16} \left[ \ln \left( \frac{x^2 + 2x + 2}{x^2 - 2x + 2} \right) \right]_{0}^{\infty}$

$\lim_{x \to \infty} \ln \left( \frac{x^2 + 2x + 2}{x^2 - 2x + 2} \right) = \ln 1 = 0$

$\ln \left( \frac{0 + 0 + 2}{0 - 0 + 2} \right) = \ln 1 = 0$

$\int \frac{1}{(x-1)^2 + 1} dx = \arctan(x-1)$

$\int \frac{1}{(x+1)^2 + 1} dx = \arctan(x+1)$

$\int_{0}^{\infty} \frac{dx}{x^4 + 4} = \frac{1}{8} \left[ \arctan(x+1) - \arctan(1-x) \right]_{0}^{\infty}$

$= \frac{1}{8} \left[ \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) - \left( \frac{\pi}{4} - \frac{\pi}{4} \right) \right] = \frac{1}{8} \left( \pi - 0 \right)$

$= \frac{1}{8} \left[ \frac{1}{2} \arctan(x+1) + \frac{1}{2}\arctan(x-1)\right]_0^\infty$

$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$

Let $x = 2u$ then $dx = 2du$.

$\int_0^{\infty} \frac{2du}{16u^4 + 4} = \frac{1}{2} \int_0^{\infty} \frac{du}{4u^4 + 1}$

We have the formula $\int_0^\infty \frac{dx}{x^4 + a^4} = \frac{\pi}{2\sqrt{2} a^3}$

$\int_0^\infty \frac{dx}{x^4 + 4} = \int_0^\infty \frac{dx}{x^4 + 2^2} = \frac{\pi}{2\sqrt{2}(2)^{\frac{3}{2}}} = \frac{\pi}{4\sqrt{2}} = \frac{\pi \sqrt{2}}{8\sqrt{2}\sqrt{2}} = \frac{\pi \sqrt{2}}{16}$

Here $a^4 = 4$ so $a = \sqrt{2}$.

$\int_0^{\infty} \frac{dx}{x^4 + 4} = \frac{\pi}{4 \sqrt{2}}$

3. 最終的な答え

$\frac{\pi}{8}$