Evaluation of $$\int^{\infty}_{0}(2x+1)e^{-x^{3}}dx$$
What i try::
Let $x^{\frac{3}{2}}=t$. Then $\displaystyle \sqrt{x}dx=\frac{2}{3}dt$ and changing the limits
$$I=\frac{2}{3}\int^{\infty}_{0}\bigg(2t^{\frac{2}{3}}+1\bigg)t^{-\frac{1}{3}}e^{-t^2}dt$$
I did not inderstand How do i solve it after that, Help me please. Thanks
As @AndrewChin said, you can use upper incomplete gamma function as below: $$\int^{\infty}_{0}(2x+1)e^{-x^{3}}dx \underset{dx=1/3t^{-2/3}dt}{\overset{x^3=t, \ x=t^{-1/3}}{======}} \int^{\infty}_{0}(2t^{-1/3}+1)e^{-t}(1/3t^{-2/3}dt) =$$ $$= \int^{\infty}_{0} (2/3t^{-1}e^{-t} + 1/3t^{-2/3}e^{-t}) dt = 2/3\Gamma(0,0) + 1/3\Gamma(1/3,0)$$
Let $x^3=t$, then $x=t^{1/3}$, $dx=t^{-2/3}dt/3$, and $$ I=\int_0^\infty (2t^{1/3}+1)e^{-t} \frac{1}{3}t^{-2/3}dt =\frac23 \int_0^\infty t^{-1/3}e^{-t} +\frac13 \int_0^\infty t^{-2/3}e^{-t} $$ $$ =\frac23 \Gamma(2/3) +\frac13 \Gamma(1/3) \approx \frac23 1.354117... +\frac13 \frac23 \frac{\pi \surd 3}{\Gamma(2/3)} $$ $$ \approx 1.7957248045201828225154229991... $$ See https://oeis.org/A073006 and https://oeis.org/A073005 for more digits.
Generalization:
If $p(x)=\sum_{n=0}^Nc_nx^n$ then $$\large\int_0^\infty p(x)\,e^{-ax^m}dx\\=\sum_{n=0}^N\,\frac{c_n}{m\,a^{\frac{n+1}{m}}}\Gamma(\tfrac{n+1}{m}).$$ With $c_1=2$, $c_0=1$, $a=1$, $m=3$ $$\int_0^\infty(2x+1)e^{-x^3}dx=\frac{c_1}{m}\Gamma(\tfrac{1+1}{m})+\frac{c_0}{m}\Gamma(\tfrac{0+1}{m})=\frac23\Gamma(\tfrac23)+\frac13\Gamma(\tfrac13)$$
