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 $|x| < 1$ .

$\displaystyle\sum_{n = 0}^{\infty} (a + nd)x^n = a\displaystyle\sum_{n = 0}^{\infty} x^n + d\displaystyle\sum_{n = 1}^{\infty} nx^n =$ $\dfrac{a}{1 - x} + d\displaystyle\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} x^n =$ $\dfrac{a}{1 - x} + \dfrac{d}{1 - x}\displaystyle\sum_{j = 1}^{\infty} x^{j} = \dfrac{a}{1 - x} + \dfrac{d}{1 - x}(\dfrac{x}{1 - x}) = \dfrac{a}{1 - x} + \dfrac{dx}{(1 - x)^2}$

$\implies$

$f(x) = \displaystyle\sum_{n = 0}^{\infty} \left(\left(\frac{2n + 1}{2}\right)! \left(\frac{-2n - 3}{2}\right)! * (a + nd)x^n\right) = -\pi\displaystyle\sum_{n = 0}^{\infty} (-1)^n (a + nd)x^n = \boxed{-\pi(\dfrac{a}{1 + x} - \dfrac{dx}{(1 + x)^2})}$

and,

$g(x) =\displaystyle\sum_{n = 0}^{\infty} (a^{*} + n^2 d^{*})x^n = a^{*}\displaystyle\sum_{n = 0}^{\infty} x^n + d^{*}\displaystyle\sum_{n = 1}^{\infty} n^2 x^n = \dfrac{a^{*}}{1 - x} + d^{*}\displaystyle\sum_{j = 1}^{\infty}\displaystyle\sum_{n = j}^{\infty} n x^n =$ $(\dfrac{a^{*}}{1 - x} + d^{*}\displaystyle\sum_{j = 1}^{\infty} (j x^{j} + (j + 1)x^{j + 1} + (j + 2)x^{j + 2} + ... ) =$ $\dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{1}{1 - x}\displaystyle\sum_{j = 1}^{\infty} j x^{j} + \dfrac{x}{(1 - x)^2}\displaystyle\sum_{j = 1}^{\infty} x^{j}) =$ $\dfrac{a^{*}}{1 - x} + d^{*}((\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x})) = \boxed{\dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{x^2 + x}{(1 - x)^3})}$

$f(\dfrac{1}{2}) = 2$ and $f(-\dfrac{1}{2}) = \dfrac{2}{9} \implies$

$3a - d = \dfrac{-9}{\pi}$

$a + d = \dfrac{-1}{9\pi}$

$\implies a = \dfrac{-41}{18\pi}$ and $d = \dfrac{13}{6\pi} \implies f(x) = \dfrac{1}{6}(\dfrac{41}{3(1 + x)} + \dfrac{13x}{(1 + x)^2})$

Let $u = 1 + x \implies dx = du \implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) dx =$ $\boxed{\dfrac{1}{18}(80\ln(3) - 52)}$ .

$g(\dfrac{1}{2}) = 2$ and $g(-\dfrac{1}{2}) = \dfrac{2}{9} \implies$

$9a^{*} - d^{*} = 3$

$a^{*} + 3d^{*} = 1$

$\implies a^{*} = \dfrac{5}{14}$ and $d^{*} = \dfrac{3}{14}$ $\implies g(x) = \dfrac{1}{14}(\dfrac{5}{1 - x} + \dfrac{3(x^2 + x)}{(1 - x)^3})$

Let $I(x) = \displaystyle\int g(x) dx$ .

Let $u = 1 - x \implies x = 1 - u$ and $dx = -du$ $\implies I(x) = \dfrac{1}{14}(\displaystyle\int \dfrac{5}{1 - x} dx - 3\int (\dfrac{1}{u} - 3u^{-2} + 2u^{-3}) du) = \dfrac{1}{14}(-5\ln(1 - x) - 3(\ln(u) + \dfrac{3}{u} - \dfrac{1}{u^2})) =$ $\dfrac{1}{14}(-5\ln(1 - x) - 3\ln(1 - x) - \dfrac{9}{1 - x} + \dfrac{3}{(1 - x)^2}) = \dfrac{1}{14}(-8\ln(1 - x) + \dfrac{9x - 6}{(1 - x)^2}) \implies$

$\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} g(x) dx = \boxed{\dfrac{1}{14}(8\ln(3) - \dfrac{12}{9})}$

$\implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) - g(x) dx = \dfrac{4}{63}(61\ln(3) - 44) \approx \boxed{1.461292}$ .