$\int_0^{\infty}e^{-x^3}\sin(x^3)dx$ 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 ?
 A: 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.
