That's plausible

Relevant wiki: Gamma Function

$\displaystyle I = \int_0^{\infty} \dfrac{\sin(x)}{\sqrt{x}} ~~ dx \\ \text{Let, } u = \sqrt[]{x} \\ \dfrac{du}{dx} = \dfrac{d}{dx} \sqrt[]{x} = \dfrac{1}{2\sqrt[]{x}} \\ dx = 2\sqrt[]{x} ~ du = 2u du \\ I = \int_0^{\infty} ~ \dfrac{\sin(u^2)}{u} ~ 2u ~ du \\ I = 2~\int_0^{\infty}~\sin(u^2)~du \\ \text{Now, use the property : } \int_0^{\infty} ~ \sin(x^a) ~ dx ~= ~ \Gamma \left(1 + \dfrac{1}{a}\right)~\cdot ~ \sin\left(\dfrac{\pi}{2a}\right) \text{ [Fresnel\_integral]} \\ I = 2 \left[ \Gamma\left(1 + \dfrac{1}{2}\right)~\sin\left(\dfrac{\pi}{4}\right) \right] \\ I = 2 \left[ \Gamma\left(\dfrac{3}{2}\right) ~ \dfrac{1}{\sqrt[]{2}}\right] \\ I = 2 \left[ \dfrac{\sqrt[]{\pi}}{2} ~ \dfrac{1}{\sqrt[]{2}} \right] \\ I = \sqrt[]{\dfrac{\pi}{2}} \\ \therefore \dfrac{\pi^a}{b^c} = \dfrac{\pi^{\frac{1}{2}}}{2^{\frac{1}{2}}} \\ \boxed{\therefore \dfrac{1}{abc} = \dfrac{1}{\dfrac{1}{2} . \dfrac{1}{2} . 2} = 2}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \Gamma\left(\dfrac{n}{2}\right) = \dfrac{(n-2)!! ~ \sqrt[]{\pi}}{2^{\frac{n-1}{2}}} \\ \Gamma\left(\dfrac{3}{2}\right) = \dfrac{(3-2)!! ~ \sqrt[]{\pi}}{2^{\frac{3-1}{2}}} \\ = \dfrac{\sqrt[]{\pi}}{2^{\frac{2}{2}}} = \dfrac{\sqrt[]{\pi}}{2} \\ \boxed{\therefore \Gamma\left(\dfrac{n}{2}\right) = \dfrac{\sqrt[]{\pi}}{2}}$