Does it even converge?

Relevant wiki: Gamma Function

$\begin{aligned} I & = \int_0^\infty \sqrt{\frac {x^{10}}{e^{x^3}}} dx \\ & = \int_0^\infty x^5 e^{-\frac {x^3}2} dx & \small \color{#3D99F6} \text{Let }t = \frac {x^3}2, \ dt = \frac 32 x^2 \ dx \\ & = \frac 43 \int_0^\infty t^{2-1} e^{-t} \ dt \\ & = \frac 43 \Gamma (2) & \small \color{#3D99F6} \text{where } \Gamma (\cdot) \text{ denotes the gamma function.} \\ & = \frac {4(1!)}3 \\ & = \frac 43 \end{aligned}$

$\implies a+b = 4+3 = \boxed{7}$

You can also do this integral the conventional way if you did not see the clever Gamma Function substitution .

Start by simplifying the integrand $\sqrt { \frac { { x }^{ 10 } }{ { e }^{ { x }^{ 3 } } } } = \frac{x^5}{\sqrt{e^{x^3}}}.$ Then make the substitution $k = x^3, dk = 3x^2 dx$ . The integral becomes now $\frac{1}{3} \int _{ 0 }^{ \infty }{ k * e^{-\frac{k}{2}} dk}$

Using integration by parts with $u = k, dv = e^{-\frac{k}{2}} \Rightarrow du = dk, v = -2e^{-\frac{k}{2}}$ , we subsequently have $\frac{1}{3}(-2ke^{-\frac{k}{2}}|_{(k=0→\infty)} - \int _{ 0 }^{ \infty }{ -2ke^{-\frac{k}{2}} dk}) =$ $\frac{1}{3} ( -2ke^{-\frac{k}{2}} - 4e^{-\frac{k}{2}} )|_{(k=0→\infty)}$

All terms cancel except for $\frac{1}{3}(-(-4)) = \frac{4}{3}$ , and $4+3=\boxed{7}$ .