Not your ordinary integral

Relevant wiki: Gamma Function

$\begin{aligned} I & = \int_0^\infty \sqrt{\frac{9x^7}{e^{x^3}}} \ dx \\ & = \int_0^\infty 3x^{\frac 72}e^{-\color{#3D99F6}{\frac{x^3}2}} \ dx \quad \quad \small \color{#3D99F6}{\text{Let }t = \frac {x^3}2 \implies x = 2^\frac 13t^\frac 13 \implies dx = \frac {2^\frac 13 t^{- \frac 23}}3 \ dt} \\ & = \int_0^\infty 3e^{-t} 2^\frac 76 t^\frac 76 \cdot \frac {2^\frac 13 t^{- \frac 23}}3 \ dt \\ & = \int_0^\infty 2\sqrt 2 e^{-t}t^\frac 12 \ dt \\ & = 2\sqrt 2 \Gamma \left( \frac 32 \right) \\ & = 2 \sqrt 2 \cdot \frac 12 \Gamma \left( \frac 12 \right) \\ & = 2 \sqrt 2 \cdot \frac 12 \cdot \sqrt \pi \\ & = \sqrt{2\pi} \end{aligned}$

$\implies k = \boxed{2}$

You can solve it without the gamma function, but its a bit lengthy -

$\int_{0}^{\infty} \sqrt{\frac{9 x^7}{e^{x^3}}} \, dx$ $\displaystyle =\int_0^\infty 3x^{\frac 72}e^{\frac{x^3}2} \ dx \quad \quad \small \color{#3D99F6}{\text{Let }t^2 = {x^3} \implies t = x^\frac{3}{2} \implies dx = \frac 23 x^{- \frac 12}dt\\ } \\= \displaystyle \int_{0}^{\infty} \frac 23 . 3 x^{\frac 72 - \frac 12} e^{\frac {x^3}2} dt \\ = \displaystyle 2\int_{0}^{\infty} x^3 e^{\frac {x^3}2} dt \\= \displaystyle 2\int_{0}^{\infty} {\color{#D61F06}t^2} e^{- \color{#D61F06}{\frac{t^2}2}} dt$

$\text{Generalise for}$ $\text{f(}{\color{#3D99F6}n}, {\color{#CEBB00}a})\quad = \quad \displaystyle 2\int_{0}^{\infty} t^{\color{#3D99F6}n} e^{{- \color{#CEBB00}a}t^2} dt$

$\text{If we observe, we see that} \\ \displaystyle \text{f(}{\color{#3D99F6}n}, {\color{#CEBB00}a}) = \displaystyle - \frac {\partial}{\partial a} \text{f(}{\color{#3D99F6}n-2}, {\color{#CEBB00}a}) \\ \implies \text{f(}{\color{#3D99F6}n}, {\color{#CEBB00}a}) = \displaystyle - \frac {\partial}{\partial a} 2\int_{0}^{\infty} t^{\color{#3D99F6}n-2} e^{{-\color{#CEBB00}a}t^2} dt \\ \text{In this case, n = 2. Therefore} \\ \implies \text{f(}{\color{#3D99F6}2}, {\color{#CEBB00}a}) = \displaystyle - \frac {\partial}{\partial a} 2\int_{0}^{\infty} t^{\color{#3D99F6}2-2} e^{{-\color{#CEBB00}a}t^2} dt = \displaystyle - \frac {\partial}{\partial a} 2\int_{0}^{\infty} e^{{-\color{#CEBB00}a}t^2} dt \quad \quad \small \color{#3D99F6}{\text{Let}u^2 = {\color{#CEBB00}{a}}t^2 \implies dt = {\color{#CEBB00}a}^{- \frac 12}du} \\ \implies \text{f(}{\color{#3D99F6}2}, {\color{#CEBB00}a}) = \displaystyle - \frac {\partial}{\partial a} 2 {\color{#CEBB00}a}^{- \frac 12} \boxed{\displaystyle \int_{0}^{\infty} e^{{-\color{#D61F06}u^2}} du} \\ \text{The above boxed integral is the famous 'Gaussian Integral' and the result is} \frac{\sqrt{\pi}}2 \\ \implies \text{f(}{\color{#3D99F6}2}, {\color{#CEBB00}a}) = \displaystyle - 2 . -\frac 12 {\color{#CEBB00}a}^{- \frac 32} . \frac{\sqrt{\pi}}2 \quad \small \text{In this case,} {\color{#CEBB00}a} = \frac 12 \\ \implies \text{f(}{\color{#3D99F6}2}, {\color{#CEBB00}\frac 12}) = \displaystyle{\color{#CEBB00}\frac 12}^{-\frac 32} . \frac{\sqrt{\pi}}2 = 2\sqrt 2 . \frac{\sqrt{\pi}}2 = \boxed{\sqrt{2\pi}} \\ \implies \text{k} = \boxed{2}$