I am trying to solve this indefinite integral:
$$\int x^{-1/4} e^{-2/3 x^{3/2}} \mathrm{d}x$$
I'd like to make it $\Gamma$ so substituting $t=\frac 23x^{\frac 32}$ to have $e^{-t}$.
$dt=x^{\frac 12}dx$
$\displaystyle \int x^{-\frac 14}e^{-t}x^{-\frac 12}dt=\int x^{-\frac 34}e^{-t}dt=\sqrt{\frac 23}\int t^{-\frac 12}e^{-t}dt=\sqrt{\frac 23}\ \gamma(\frac 12,t)=\sqrt{\frac {2\pi}3}\operatorname{erf}(\sqrt{t})$
Copyright © 2021 JogjaFile Inc.
I'd like to make it $\Gamma$ so substituting $t=\frac 23x^{\frac 32}$ to have $e^{-t}$.
$dt=x^{\frac 12}dx$
$\displaystyle \int x^{-\frac 14}e^{-t}x^{-\frac 12}dt=\int x^{-\frac 34}e^{-t}dt=\sqrt{\frac 23}\int t^{-\frac 12}e^{-t}dt=\sqrt{\frac 23}\ \gamma(\frac 12,t)=\sqrt{\frac {2\pi}3}\operatorname{erf}(\sqrt{t})$