A calculus problem by Kushal Bose

A generalized Gaussian integral states that $\displaystyle \int_0^\infty e^{-\dfrac{x^2}{a}}\; dx = \dfrac{\sqrt{a\pi}}{2}$ , so the integration turns after evaluating the inner integral as $\displaystyle \int_{y=0}^1 \dfrac{\sqrt{y^3 \pi}}{2}\; dy = \dfrac{\sqrt{\pi}}{5}$ . This makes the answer $\boxed{8}$

Let

$I = \int_0^1 \underbrace{\left( \int_0^{\infty} e^{- {x^2}/{y^3}} \,dx \right)}_{I_1} \,dy$

We shall compute $I_1$ first.

Let $\dfrac{x^2}{y^3} = t \implies \dfrac{2x}{y^3} \,dx = \,dt$ and $\displaystyle \int_0^{\infty} \mapsto \int_0^{\infty}$ and so we have

$\begin{aligned} I_1 &= \int_0^{\infty} e^{- {x^2}/{y^3}} \,dx \\ &= \int_0^{\infty} e^{- {x^2}/{y^3}} \cdot \dfrac{y^3}{2x} \cdot \dfrac{2x}{y^3} \,dx \\ &= \dfrac{y^3}{2} \int_0^{\infty} e^{-t} \cdot \dfrac{1}{\sqrt{ty^3}} \,dt \\ &= \dfrac{\sqrt{y^3}}{2} \int_0^{\infty} t^{-{1}/{2}} e^{-t} \,dt \\ &= \dfrac{\sqrt{y^3}}{2} \cdot \Gamma \left( \dfrac 12 \right) \\ &= \dfrac{\sqrt{y^3}}{2} \cdot \sqrt{\pi} \\ &= \dfrac{\sqrt{\pi}}{2} \cdot y^{{3}/{2}} \end{aligned}$

Thus

$\begin{aligned} I &= \int_0^1 \dfrac{\sqrt{\pi}}{2} \cdot y^{{3}/{2}} \,dy \\ &= \dfrac{\sqrt{\pi}}{2} \times \dfrac 25 \times 1 \\ &= \dfrac{\sqrt{\pi}}{5} \end{aligned}$

So we have

$a = 1 \\ b=2 \\ c=5 \\ \implies a+b+c = \boxed{8}$