What is the integral from 0 to 1 of (sin(x^4))/(x^4) ?

The integral from 0 to 1 of (sin(x^4))/(x^4) is -1/3 sin(1)-1/6 (\E_{3/4}(1i)+\E_{3/4}(-1i))

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

\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))

Solution steps

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

Apply Integration By Parts

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

= [- \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}

Compute the boundaries:

- \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))

\int_{0}^{3} (6 x - x^{2}) - x^{2} d x

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What is the integral from 0 to 1 of (sin(x^4))/(x^4) ?
The integral from 0 to 1 of (sin(x^4))/(x^4) is -1/3 sin(1)-1/6 (\E_{3/4}(1i)+\E_{3/4}(-1i))