I have the following question: I want to show, that the improper integral exists:
$$\int_{0}^{b}\tfrac{\sqrt{1+y'(x)^2}}{\sqrt{y(x)}}$$
I stated: For $$[t,b] \subset(0,b]$$, all t must be in the integral. So if we take a sequence $$t_{k}, k \in \mathbb{N} \ and \lim_{k \rightarrow \infty}=0$$, we must show, that
$$\Big| \int_{t_{k}}^{b}f(x)dx - \int_{t_{l}}^{b} f(x)dx\Big|=\int_{t_{k}}^{t_{l}}\tfrac{\sqrt{1+y'(x)^2}}{\sqrt{y(x)}}\leq \tfrac{1}{\varepsilon}$$
I used substitution for the integral with
$$b=x(\beta)=A(\beta-\sin\beta) , \ y(\beta)=R \cdot (1 - cos(\beta)), \ \frac{dx}{d\beta}=A(1-\cos\beta) \ and \ \frac{dy}{dθ}=A\sin(\beta)$$, also $$y'(x) =\tfrac{dy}{dx}=\tfrac{dy}{d \beta} \cdot \tfrac{d \beta}{dx}=\tfrac{Rsin(\beta)}{R\cdot (1-cos(\beta))}$$ so that $$\int_{t_{k}}^{t_{l}}\tfrac{\sqrt{1+y'(x)^2}}{\sqrt{y(x)}}=\Big| \int_{\beta_{k}}^{\beta_{l}} \sqrt{\tfrac{1+(y'(x))^2}{y(\beta)}}\tfrac{dx}{d \beta} d \beta\Big| = \Big| \int_{\beta_{k}}^{\beta_{l}} \sqrt{(1 + \tfrac{R^2sin^2(\beta)}{R^2(1-cos(\beta))^2}) \cdot \tfrac{R^2(1-cos(\beta))^2}{R(1-cos(\beta))}} d \beta \Big| \\ =\Big| \int_{\beta_{k}}^{\beta_{l}} \sqrt{\tfrac{R^2(1-cos(\beta))^2+R^2sin^2(\beta)}{R^2(1-cos(\beta))^2} \cdot R(1-cos(\beta))} d \beta \Big| \\ =\Big| \sqrt{R} \int_{\beta_{k}}^{\beta_{l}} \sqrt{\tfrac{(1-cos(\beta))^2+sin^2(\beta)}{1-cos(\beta)}} d \beta \Big| \\ = \Big| \sqrt{R} \int_{\beta_{k}}^{\beta_{l}} \sqrt{\tfrac{1-2cos(\beta) + cos^2(\beta)+sin^2(\beta)}{1-cos(\beta)}} d \beta \Big| = \Big| \sqrt{2R}\int_{\beta_{l}}^{\beta_{k}} \sqrt{\tfrac{1-cos(\beta)}{1-cos(\beta)}} d \beta \Big| $$ and with the factor $$\tfrac{1}{2g}$$ we have $$= \sqrt{\tfrac{R}{g}} (\beta_{l}-\beta_{k})=\sqrt{\tfrac{R}{g}}R\big(\beta_{l}-\underbrace{sin(\beta_{l})}_{\leq 1}-\beta_{k}+\underbrace{sin(\beta_{k})}_{\leq 1}\bigr) \leq \sqrt{\tfrac{R}{g}}R(\beta_{l}-\beta_{k})$$
My questions is: How do I continue from here? Is my technique even right? Thanks for all answers!