Change of Variables and Integration by Parts

$\begin{aligned} I & = \int_0^\infty e^{-\color{#3D99F6}{t^\frac{1}{5}}} dt \quad \quad \small \color{#3D99F6}{\text{Let }x = t^\frac{1}{5} \implies x^5 = t \implies 5x^4 \ dx = dt} \\ & = \int_0^\infty 5x^4 e^{-x} \ dx \\ & = 5 \int_0^\infty x^{\color{#3D99F6}{5}-1} e^{-x} \ dx \\ & = 5 \Gamma (\color{#3D99F6}{5}) = 5 \cdot 4! = \boxed{120} \end{aligned}$

By computing definite integral for $\int_{0}^{\infty} e^{-t^{\frac{1}{5}}}\,dt$

$\begin{aligned} \int_{0}^{\infty} e^{-t^{\frac{1}{5}}}\,dt &= 5\left( 4\left( -3\left( e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}}-e^{-t^{\frac{1}{5}}} \right) -e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}} \right)-e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}} \right)+\ce{C} \quad\quad\quad{\text{Indefinite}} \\&=\left[5\left( 4\left( -3\left( e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}}-e^{-t^{\frac{1}{5}}} \right) -e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}} \right)-e^{-t^{\frac{1}{5}}}t^{\frac{1}{5}} \right) \right]_{0}^{\infty} \\&=120 \square \end{aligned}$

---

You can use @Chew-Seong Cheong 's method by using gamma function $\Gamma(x)$ .

$\LARGE \text{ADIOS muchacha!!!}$