It's an Area Feast.

Using the gamma function $\Gamma(p) = \int_{0}^{\infty} t^{p - 1} e^{-t} dt$ we obtain:

$\Gamma(\dfrac{1}{2}) = \displaystyle\int_{0}^{\infty} t^{-\frac{1}{2}} e^{-t} dt$

Let $s^2 = t \implies 2s ds = dt \implies$

$\Gamma(\dfrac{1}{2}) = 2\displaystyle\int_{0}^{\infty} e^{-s^2} ds$

Since s is a dummy variable we can write:

$(\Gamma(\dfrac{1}{2}))^2 =$ $4\displaystyle\int_{0}^{\infty} e^{-x^2} dx \int_{0}^{\infty} e^{-y^2} dy =$ $4\displaystyle\int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2)} dx dy$

Let $x = r\cos\theta, y = r\sin\theta$

Using the Jacobian $\implies (\Gamma(\dfrac{1}{2}))^2 = 4\displaystyle\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-r^2} r dr d\theta =$ $-2\displaystyle\int_{0}^{\frac{\pi}{2}} e^{-r^2}|_{0}^{\infty} d\theta = 2\displaystyle\int_{0}^{\frac{\pi}{2}} d\theta = \pi \implies \Gamma(\dfrac{1}{2}) = \sqrt{\pi}$

Now, using integration by parts you can show that $\Gamma(p + 1) = p \Gamma(p)$

Show $\Gamma(1) = 1$ and recursively using $\Gamma(p + 1) = p \Gamma(p) \implies \Gamma(p + 1) = p!$ for any integer $p \geq 0$ and p can be extended to reals so that $\Gamma(\dfrac{1}{2}) = \Gamma(\dfrac{-1}{2} + 1) = \sqrt{\pi} = (\dfrac{-1}{2})!$

Using $\Gamma(p + 1) = p \Gamma(p)$ and $\Gamma(\dfrac{1}{2}) = \sqrt{\pi} = (\dfrac{-1}{2})!$

$\implies$

$S_{0} = \Gamma(\dfrac{3}{2}) = \dfrac{1}{2} * \Gamma(\dfrac{1}{2}) = \dfrac{\sqrt{\pi}}{2} = (\dfrac{1}{2})!$

$$

$S_{1} = \Gamma(\dfrac{5}{2}) = \dfrac{3}{2} * \Gamma(\dfrac{3}{2}) = \dfrac{1 * 3 * \sqrt{\pi}}{2^2} = (\dfrac{3}{2})!$

$S_{2} = \Gamma(\dfrac{7}{2}) = \dfrac{5}{2} * \Gamma(\dfrac{5}{2}) = \dfrac{1 * 3 * 5 * \sqrt{\pi}}{2^3} = (\dfrac{5}{2})!$

In General:

$S_{n} = \Gamma(\dfrac{2n + 3}{2}) = \dfrac{1 * 3 * 5 * * * (2n + 1) * \sqrt{\pi}}{2^{n + 1}} = (n + \frac{1}{2})!$ $$

$\implies (n + \dfrac{1}{2})! = \dfrac{(2n + 1)! * \sqrt{\pi}}{2^{2n + 1} * n!}$ , where $n$ is a non-negative integer. $$

Note: $(2n + 1)! = 1 * 2 * 3 * * * (2n) * (2n + 1) = 2^n * n! * (1 * 3 * 5 * * * (2n + 1)) \implies$ $1 * 3 * 5 * * * (2n + 1) = \dfrac{(2n + 1)!}{2^n * n!}$

and,

Using $\Gamma(p) = \dfrac{\Gamma(p + 1)}{p}$ and $\Gamma(\dfrac{1}{2}) = \sqrt{\pi} = (\dfrac{-1}{2})!$

$\implies$

$T_{0} = \Gamma(-\dfrac{1}{2}) = -2\Gamma(\dfrac{1}{2}) = -2\sqrt{\pi} = (-\dfrac{3}{2})!$

$T_{1} = \Gamma(-\dfrac{3}{2}) = -\dfrac{2}{3}\Gamma(-\dfrac{1}{2}) = \dfrac{2^2}{1 * 3}\sqrt{\pi} = (-\dfrac{5}{2})!$

$T_{2} = \Gamma(-\dfrac{5}{2}) = -\dfrac{2}{5}\Gamma(-\dfrac{3}{2}) = -\dfrac{2^3}{1 * 3 * 5}\sqrt{\pi} = (-\dfrac{7}{2})!$

In General:

$T_{n} = \Gamma(-n -\dfrac{1}{2}) = \pm\dfrac{2^{n + 1}\sqrt{\pi}}{1 * 3 * 5 * * * (2n + 1)} = (-n -\dfrac{3}{2})!$ $$

$\implies (-n - \dfrac{3}{2})! = \pm\dfrac{2^{2n + 1} * n!\sqrt{\pi}}{(2n + 1)!}$ , where $n$ is a non-negative integer

$\implies$

$S_{n} * T_{n} =$ $\begin{cases} \pi & \text{for} \:\ n \:\ \text{odd}\\ -\pi & \text{for} \:\ n \:\ \text{even}\\ \end{cases}$ .

$$

$$

Let $(0 \leq x < 1)$

$\displaystyle\sum_{n = 1}^{\infty} n x^n = \displaystyle\sum_{j = 1}^{\infty} \displaystyle\sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\displaystyle\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

$\implies$

$f(x) = \displaystyle\sum_{n = 1}^{\infty} \left(\left(\frac{2n + 1}{2}\right)! \left(\frac{-2n - 3}{2}\right)! * n x^n\right) = -\pi\displaystyle\sum_{n = 1}^{\infty} (-1)^n n x^n = \dfrac{\pi x}{(1 + x)^2}$

Let $u = 1 + x \implies dx = du \implies \displaystyle\int_{0}^{\frac{1}{2}} f(x) dx = \pi\displaystyle\int_{1}^{\frac{3}{2}} \dfrac{1}{u} - \dfrac{1}{u^2} du = \pi(\ln(u) + \dfrac{1}{u})|_{1}^{\frac{3}{2}} = \boxed{\pi(\ln(\dfrac{3}{2}) - \dfrac{1}{3})}$

Let $g(x) = \pi\displaystyle\sum_{n = 1}^{\infty} \dfrac{x^n}{n}$ on $(0 \leq x < 1)$

$\implies \dfrac{d}{dx}(g(x)) = \pi\displaystyle\sum_{n = 1}^{\infty} x^{n - 1} = \dfrac{\pi}{1 - x} \implies g(x) = \pi\displaystyle\int_{0}^{x} \dfrac{1}{1 - x} dx = \pi\ln(\dfrac{1}{1 - x})$ .

For $\pi\displaystyle\int_{0}^{\frac{1}{2}} \ln(\dfrac{1}{1 - x}) dx$

Let $u = \ln(\dfrac{1}{1 - x}) \implies du = \dfrac{dx}{1 - x}$ and $dv = dx \implies v = x \implies$

$\pi\displaystyle\int_{0}^{\frac{1}{2}} \ln(\dfrac{1}{1 - x}) dx = \pi(x\ln(\dfrac{1}{1 - x}) + \ln(1 - x) + x)|_{0}^{\frac{1}{2}} = \boxed{\dfrac{\pi}{2}(1 - \ln(2))}$

$\implies \displaystyle\int_{0}^{\frac{1}{2}} g(x) - f(x) = \pi(\dfrac{5}{6} - \ln(\dfrac{3}{\sqrt{2}})) =$ $\pi(\dfrac{5}{2 * 3} - \ln(\dfrac{3}{\sqrt{2}})) = \pi(\dfrac{\alpha}{\lambda * \beta} - \ln(\dfrac{\beta}{\sqrt{\lambda}})) \implies \alpha + \beta + \lambda = \boxed{10}$ .