# $\int_0^{\infty}e^{-x^3}\sin(x^3)dx$

definite integralsgamma function

Evaluate$$\int_0^{\infty}e^{-x^3}\sin(x^3)dx$$

I substitute $$x^3=t$$ Integral becomes
$${1\over3}\int_0^{\infty}t^{-2\over3}e^{-t}\sin(t)dt$$

I guess this function has something to do with gamma function but sine function is causing problem here . I tried using integration by parts but it didn't seem to work out . How can I get rid off sine ?

According to Table of Integrals, Series, and Products Seventh Edition, I.S. Gradshteyn and I.M. Ryzhik, 3.326.1,

$$\int_0^{\infty} e^{-x^u}dx = \dfrac1{u}\Gamma(\dfrac1{u}) = \Gamma(1+\dfrac1{u})$$ when $$\Re(u) > 0$$.

Letting $$x = cy$$, this becomes $$c\int_0^{\infty} e^{-c^uy^u}dy = \Gamma(1+\dfrac1{u})$$.

Letting $$c^u=a$$, $$a^{1/u}\int_0^{\infty} e^{-ay^u}dy = \Gamma(1+\dfrac1{u})$$, so $$\int_0^{\infty} e^{-ay^u}dy = \dfrac1{a^{1/u}}\Gamma(1+\dfrac1{u})$$.

Putting $$a = 1+i$$, as Claude Leibovici suggested, and not worrying about this being legal,

$$\begin{array}\\ \dfrac1{(1+i)^{1/u}}\Gamma(1+\dfrac1{u}) &=\int_0^{\infty} e^{-(1+i)y^u}dy\\ &=\int_0^{\infty} e^{-y^u}e^{-iy^u}dy\\ &=\int_0^{\infty} e^{-y^u}(\cos(-y^u)+i\sin(-y^u))dy\\ &=\int_0^{\infty} e^{-y^u}\cos(-y^u)dy+i\int_0^{\infty} e^{-y^u}\sin(-y^u)dy\\ \end{array}$$

Since $$1+i =\sqrt{2}e^{\pi i/4}$$,

$$\begin{array}\\ (1+i)^{-1/u} &=\sqrt{2^{-1/u}}e^{-\pi i/(4u)}\\ &=2^{-1/(2u)}(\cos(-\pi /(4u))-i\sin(-\pi /(4u)))\\ &=2^{-1/(2u)}(\cos(\pi /(4u))+i\sin(\pi /(4u)))\\ \end{array}$$

$$\int_0^{\infty} e^{-y^u}\sin(-y^u)dy =2^{-1/(2u)}\Gamma(1+\dfrac1{u})\sin(\pi /(4u))$$

and

$$\int_0^{\infty} e^{-y^u}\cos(-y^u)dy =2^{-1/(2u)}\Gamma(1+\dfrac1{u})\cos(\pi /(4u))$$.

Putting $$u=3$$, $$\int_0^{\infty} e^{-y^3}\sin(-y^3)dy =2^{-1/6}\Gamma(\frac43)\sin(\pi /12)$$.

Wolfy says $$\int_0^∞ e^{-y^3} \sin(-y^3) dy = -\frac{(\sqrt{3} - 1) Γ(4/3)}{2^{5/3}} ≈-0.205905$$ and $$\sin(\pi/12) =\dfrac{\sqrt{3} - 1}{2 \sqrt{2}}$$ so the results agree.