Proving that an Integral is Zero does someone have any idea about how to prove that
\begin{equation}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1} dx = 0,
\end{equation}
for $n=0,1,2,...$?
I met this integral while I was solving a PDE.
Thank you very much in advance for your invaulable help.
Best Regards,
Maurizio Barbato
 A: I discovered that the formula I tried to prove is incorrect!
Take e.g. n=0. Then setting $u=x^{4/3}$ the integral becomes
$
\int_{0}^{\infty}  e^{-u}u^{1/2}(cos u)  du,
$
which is the real part of
\begin{equation}
\int_{0}^{\infty} e^{u(i-1)} u^{1/2} du,
\end{equation}
which is equal to $\Gamma(3/2)=\frac{\sqrt{\pi}}{2}$ times the characteristic function of the gamma distribution with parameters $\lambda=1$ and $\nu=3/2$, computed in 1, which is $(1-i)^{-3/2}$. So the real part of the integral is not zero!
Thank you very much for your attention.
A: I guess you already found out that this integral is not equal to zero. I hope 
Actually you can cope with this integral and I hope those derivations will be of any help.
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1}\;\mathrm dx &=& \int_{0}^{\infty} e^{-x^{4/3}} \left(\frac{e^{\mathrm ix^{4/3}}+e^{-\mathrm ix^{4/3}}}{2}\right) x^{4n+1}\;\mathrm dx=\\
&=&\frac{1}{2}\int_{0}^{\infty} e^{-x^{4/3}(1+\mathrm i)}  x^{4n+1}\;\mathrm dx+\frac{1}{2}\int_{0}^{\infty} e^{-x^{4/3}(1+\mathrm i)}  x^{4n+1}\;\mathrm dx=\\
\end{eqnarray}
$$
Multiplying anf dividing by $\frac{4}{3}x^{1/3}$, changing the variable to $t=x^{4/3}$:
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}(1\pm\mathrm i)}  x^{4n+1}\;\mathrm dx&=&\int_{0}^{\infty} e^{-x^{4/3}(1\pm\mathrm i)}  x^{4n+1}\left(\frac{3}{4}x^{-1/3}\right)\;\mathrm dx^{4/3}=\\
&=&\frac{3}{4}\int_{0}^{\infty} e^{-t(1\pm\mathrm i)}  t^{3n+\frac{1}{2}}\;\mathrm dt
\end{eqnarray}
$$
Changing the variable to $z=(1\pm \mathrm i)t$:
$$
\begin{eqnarray}
\frac{3}{4}\int_{0}^{\infty} e^{-t(1\pm\mathrm i)}  t^{3n+\frac{1}{2}}\;\mathrm dt&=&
\frac{3}{4}(1\pm\mathrm i)^{-3n-\frac{3}{2}}\int_{0}^{\infty} e^{-z}  z^{3n+\frac{1}{2}}\;\mathrm dt=\\
&=&\frac{3}{4}(1\pm\mathrm i)^{-3n-\frac{3}{2}}\Gamma\left(3n+\frac{3}{2}\right)
\end{eqnarray}
$$
So
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1}\;\mathrm dx &=& 
\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)\left((1+\mathrm i)^{-3n-\frac{3}{2}}+(1-\mathrm i)^{-3n-\frac{3}{2}}\right)=\\
&=&\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)2^{-\frac{3n}{2}-\frac{3}{4}}\left(e^{\mathrm i \frac{\pi}{4}(-3n-\frac{3}{2})}+e^{-\mathrm i \frac{\pi}{4}(-3n-\frac{3}{2})}\right)=\\
&=&\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)2^{-\frac{3n}{2}+\frac{1}{4}}\cos\left(\frac{3\pi}{4}\left(n+\frac{1}{2}\right)\right)
\end{eqnarray}
$$
