integral from 0 to 1 of (sin(x^4))/(x^4)

解

\int_{0}^{1} \frac{sin ( x ^{4} )}{x ^{4}} d x

- \frac{1}{3} sin (1) - \frac{1}{6} (E_{\frac{3}{4}} (1 i) + E_{\frac{3}{4}} (- 1 i))

解答ステップ

\int_{0}^{1} \frac{sin ( x ^{4} )}{x ^{4}} d x

部分積分を適用する

= [- \frac{sin ( x ^{4} )}{3 x ^{3}} - \int - \frac{4}{3} cos (x^{4}) d x]_{0}^{1}

\int - \frac{4}{3} cos (x^{4}) d x = \frac{1}{6} x (E_{\frac{3}{4}} (- i x^{4}) + E_{\frac{3}{4}} (i x^{4}))

= [- \frac{sin ( x ^{4} )}{3 x ^{3}} - \frac{1}{6} x (E_{\frac{3}{4}} (- i x^{4}) + E_{\frac{3}{4}} (i x^{4}))]_{0}^{1}

限度を計算する:

- \frac{1}{3} sin (1) - \frac{1}{6} (E_{\frac{3}{4}} (1 i) + E_{\frac{3}{4}} (- 1 i))

= - \frac{1}{3} sin (1) - \frac{1}{6} (E_{\frac{3}{4}} (1 i) + E_{\frac{3}{4}} (- 1 i))