What is the integral from 0 to x^4 of sin(t^3) ?

The integral from 0 to x^4 of sin(t^3) is

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integral from 0 to x^4 of sin(t^3)

\int_{0}^{x^{4}} sin (t^{3}) d t

- \frac{1}{2} i \frac{x ^{4} Γ ( \frac{1}{3} , i ( x ^{4} ) ^{3} )}{3 3 i ( x ^{4} ) ^{3}} - \frac{x ^{4} Γ ( \frac{1}{3} , - i ( x ^{4} ) ^{3} )}{3 3 - i ( x ^{4} ) ^{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}}))

Solution steps

\int_{0}^{x^{4}} sin (t^{3}) d t

Use the nonelementary integral:

\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^{4}}

Compute the boundaries:

- \frac{1}{2} i \frac{x ^{4} Γ ( \frac{1}{3} , i ( x ^{4} ) ^{3} )}{3 3 i ( x ^{4} ) ^{3}} - \frac{x ^{4} Γ ( \frac{1}{3} , - i ( x ^{4} ) ^{3} )}{3 3 - i ( x ^{4} ) ^{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 ^{4} Γ ( \frac{1}{3} , i ( x ^{4} ) ^{3} )}{3 3 i ( x ^{4} ) ^{3}} - \frac{x ^{4} Γ ( \frac{1}{3} , - i ( x ^{4} ) ^{3} )}{3 3 - i ( x ^{4} ) ^{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}}))

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What is the integral from 0 to x^4 of sin(t^3) ?
The integral from 0 to x^4 of sin(t^3) is