integral from 0 to x^2 of cos(t^3)

解

\int_{0}^{x^{2}} cos (t^{3}) d t

- \frac{x ^{2} ( 3 i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} ) + 3 - i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} ) )}{6 3 ( x ^{2} ) ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}

解答ステップ

\int_{0}^{x^{2}} cos (t^{3}) d t

非初等積分を使用する:

\int cos (x^{3}) d x = - \frac{x ( 3 i x ^{3} Γ ( \frac{1}{3} , - i x ^{3} ) + 3 - i x ^{3} Γ ( \frac{1}{3} , i x ^{3} ) )}{6 3 x ^{6}}

= - \frac{t ( 3 i t ^{3} Γ ( \frac{1}{3} , - i t ^{3} ) + 3 - i t ^{3} Γ ( \frac{1}{3} , i t ^{3} ) )}{6 3 t ^{6}}_{0}^{x^{2}}

限度を計算する:

- \frac{x ^{2} ( 3 i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} ) + 3 - i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} ) )}{6 3 ( x ^{2} ) ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}

= - \frac{x ^{2} ( 3 i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} ) + 3 - i ( x ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} ) )}{6 3 ( x ^{2} ) ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}