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

解

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

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

解答ステップ

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

非初等積分を使用する:

\int sin (x^{3}) d x = - \frac{1}{2} i (\frac{x Γ ( \frac{1}{3} , i x ^{3} )}{3 3 i x ^{3}} - \frac{x Γ ( \frac{1}{3} , - i x ^{3} )}{3 3 - i x ^{3}})

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

限度を計算する:

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

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