Tricky Area II

Consider $y$ as follows:

$\begin{aligned} y & = \frac 1{x^2+\frac 1{x^2+\frac 1{x^2 + \ddots}}} \\ y & = \frac 1{x^2 + y} \\ y^2 + x^2y - 1& = 0 & \small \color{#3D99F6} \text{Solving the quadratic equation for }y \\ \implies y & = \frac {\sqrt{x^4+4}-x^2}2 \end{aligned}$

We note that $y$ is even as shown in the figure. Now again consider:

$y = \dfrac 1{x^2+y} \implies x^2 = \dfrac 1y - y \implies x = \sqrt{\dfrac 1y-y}$

The area under the curve is given by $A = \displaystyle \int_{-\infty}^\infty y \ dx = 2 \int_0^\infty y \ dx = 2 \int_0^1 x \ dy$ . Therefore,

$\begin{aligned} A & = 2 \int_0^1 \sqrt{\frac 1y - y} dy \\ & = 2 \int_0^1 \sqrt{\frac {1 - y^2}y} dy \\ & = 2 \int_0^1 y^{-\frac 12} \left(1-y^2\right)^\frac 12 dy & \small \color{#3D99F6} \text{Note that beta function }B(m,n) = 2\int_0^1 x^{2m-1}\left(1-x^2\right)^{n-1} dx \\ & = B \left(\frac 14, \frac 32 \right) \\ & = \frac {\Gamma \left(\frac 14 \right)\Gamma \left(\frac 32 \right)}{\color{#D61F06}\Gamma \left(\frac 74\right)} & \small \color{#3D99F6} \text{where }\Gamma (s) \text{ is gamma function.} \\ & = \frac {{\color{#D61F06}4\sqrt 2 \Gamma \left(\frac 94 \right)}\Gamma \left(\frac 14 \right)\Gamma \left(\frac 32 \right)}{\color{#D61F06}\sqrt \pi \Gamma \left(\frac 72 \right)} & \small \color{#D61F06} \text{Using }\Gamma (2z) = \frac {2^{2z-1}\Gamma (z) \Gamma \left(z+\frac 12\right)}{\sqrt \pi} \\ & = \frac {4\sqrt 2 \cdot {\color{#3D99F6}\frac 54 \Gamma \left(\frac 54 \right)} \cdot {\color{#3D99F6} 4 \Gamma \left(\frac 54 \right)} \Gamma \left(\frac 32 \right)}{\sqrt \pi \cdot \color{#3D99F6} \frac 52 \cdot \frac 32 \Gamma \left(\frac 32 \right)} & \small \color{#3D99F6} \text{Using }\Gamma (1+z) = z \Gamma (z) \\ & = \frac {16}3 \sqrt{\frac 2 \pi} {\color{#3D99F6} \Gamma \left(\frac 54 \right)}^2 & \small \color{#3D99F6} \text{Note that }\Gamma (1+z) = z! \\ & = \frac {16}3 \sqrt{\frac 2 \pi} \left({\color{#3D99F6}\frac 14 !}\right)^2 \end{aligned}$

$\implies \alpha + \beta = 16 + 3 = \boxed{19}$

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References:

• Beta function
• Gamma function

Rewriting the function as $y=\frac1{x^2+y}$ And seeing the area, $\textbf{A}$ as

The function can be written in terms of $x$ and then integrated from 0 to 1 just as a form of the beta function.

$\displaystyle x=\sqrt{\frac1{y}-y}=\frac{\sqrt{1-y^2}}{\sqrt{y}}$

$\displaystyle\textbf{A}=2\int_0^1\frac{\sqrt{1-y^2}}{\sqrt{y}}dy\quad\quad\text{Note: }\beta(a,b)=2\int_0^1x^{2a-1}(1-x^2)^{b-1}dx$

$\textbf{A}=\beta(\frac1{4},\frac3{2})=\frac{16}{3}\sqrt{\frac2{\pi}}\Gamma(\frac5{4})^2$