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Relevant wiki: Gamma Function

Similar solution with @Niraj Sawant 's

Using the identity $\Gamma (n) = (n-1)!$ , where $\Gamma (\cdot)$ denotes the gamma function, we have:

$\begin{aligned} 2.5! & = \color{#3D99F6} \Gamma \left(\frac 72\right) & \small \color{#3D99F6} \text{Note that }\Gamma (s+1) = s\Gamma(s) \\ & = \frac 52 \times \Gamma \left(\frac 52\right) \\ & = \frac 52 \times \frac 32 \times \Gamma \left(\frac 32\right) \\ & = \frac 52 \times \frac 32 \times \frac 12 \times \color{#3D99F6} \Gamma \left(\frac 12\right) & \small \color{#3D99F6} \text{also that }\Gamma (z) \Gamma (1-z) = \frac \pi{\sin (\pi z)} \\ & = \frac {15\color{#3D99F6} \sqrt \pi}8 & \small \color{#3D99F6} \text{putting }z=\frac 12 \end{aligned}$

Therefore, $a-b= 15-8 = \boxed 7$ .

The factorial of any real number is defined by the Pi Function.

$\boxed{(n)! = \prod (n) = \displaystyle \int_0^\infty t^{n} e^{-t} dt = \Gamma(n +1)}$

$\Gamma( x)$ is called the Gamma Function

$\boxed{(2.5)! =\displaystyle (\dfrac {5}{2})! = \displaystyle \int_0^\infty t^{\frac{5}{2}} e^{-t} dt}$

Substitute, $t = r^{2} \Rightarrow dt =2rdr$

$\displaystyle \int_0^\infty r^{5} e^{-r^{2}} (2rdr) = \int_0^\infty r^{5} (2re^{-r^2} )dr$

$=\displaystyle [r^{5}]_0^\infty \int_0^\infty 2re^{-r^2} dr - \int_0^\infty (\dfrac {d}{dr}(r^5) \int_0^\infty (2re^{-r^2}dr))dr$

$= [r^5 (-e^{-r^2})]_0^\infty - \displaystyle \int_0^\infty 5r^4(-e^{-r^2})dr = (0) +\displaystyle \int_0^\infty 5r^3 e^{-r^2} dr$

Now,

$\displaystyle \int_0^\infty \dfrac {5r^3}{2}(2re^{-r^2}) dr = [\dfrac {5r^3}{2}]_0^\infty \displaystyle \int_0^\infty (2re^{-r^2})dr - \displaystyle \int_0^\infty (\dfrac {15r^2}{2} \displaystyle \int_0^\infty (2re^{-r^2})dr)dr = [\dfrac {5r^3}{2}(-e^{-r^2})]_0^\infty - \displaystyle \int_0^\infty \dfrac {15r^2}{2} (-e^{-r^2})dr = (0) + \displaystyle \int_0^\infty \dfrac {15r^2}{2} (e^{-r^2})dr$

Now,

$\displaystyle \int_0^\infty \dfrac {15r^2}{2} (e^{-r^2})dr = \displaystyle \int_0^\infty \dfrac {15r}{4} (2re^{-r^2})dr = [\dfrac {15r}{4}]_0^\infty \displaystyle \int_0^\infty (2re^{-r^2})dr - \displaystyle \int_0^\infty (\dfrac {15}{4} \displaystyle \int_0^\infty (2re^{-r^2})dr) dr = [\dfrac {15r}{4}(-e^{-r^2})]_0^\infty - \displaystyle \int_0^\infty \dfrac {15}{4}(-e^{-r^2})dr$

$=(0) + \dfrac {15}{4}\displaystyle \int_0^\infty e^{-r^2} dr = \dfrac {15}{4}\times \dfrac {\sqrt{π}}{2} = \dfrac {15\sqrt{π}}{8} = \dfrac {a\sqrt{π}}{b}$

.Since, $\color{#20A900}{(\displaystyle \int_{-\infty}^\infty e^{-x^2}dx = \sqrt{π} } )$

$\boxed{(2.5)! = \dfrac {15\sqrt{π}}{8} = 3.32335097044784}$

$ANSWER: a - b = 15 - 8 =\boxed{7}$

For more detailed explanation watch this