Messing around with Infinities

Finding $\Psi$ : Let $k$ denote some positive integer. Then $\begin{aligned} \int_0^\infty \left(\frac{e^{-x/(k+1)}}{x} - \frac{e^{-x/k}}{x}\right)\,dx &= \lim_{\varepsilon\to 0^+} \int_{\varepsilon}^\infty \frac{e^{-x/(k+1)}}{x}\,dx - \int_{\varepsilon}^\infty \frac{e^{-x/k}}{x}\,dx \\ &= \lim_{\varepsilon\to 0^+} \int_{\varepsilon/(k+1)}^\infty \frac{e^{-u}}{u}\,du - \int_{\varepsilon/k}^\infty \frac{e^{-u}}{u}\,du \\ &= \lim_{\varepsilon\to 0^+} \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{e^{-u}}{u}\,du \end{aligned}$ To evaluate this, we note that for any $\varepsilon > 0$ and $0 < u \leq \frac{\varepsilon}{k}$ , we have $\frac{e^{-\varepsilon/k}}{u} \leq \frac{e^{-u}}{u} \leq \frac{1}{u}$ so by integrating, $e^{-\varepsilon/k}\ln\left(\frac{k+1}{k}\right) = \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{e^{-\varepsilon/k}}{u}\,du \leq \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{e^{-u}}{u}\,du \leq \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{1}{u}\,du = \ln\left(\frac{k+1}{k}\right)$ and taking the limit, $\ln\left(\frac{k+1}{k}\right) = \lim_{\varepsilon\to 0^+}e^{-\varepsilon/k}\ln\left(\frac{k+1}{k}\right) \leq \lim_{\varepsilon\to 0^+} \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{e^{-u}}{u}\,du \leq \lim_{\varepsilon\to 0^+} \ln\left(\frac{k+1}{k}\right) = \ln\left(\frac{k+1}{k}\right).$ Therefore, we see $\int_0^\infty \left(\frac{e^{-x/(k+1)}}{x} - \frac{e^{-x/k}}{x}\right)\,dx = \lim_{\varepsilon\to 0^+} \int_{\varepsilon/(k+1)}^{\varepsilon/k} \frac{e^{-u}}{u}\,du = \ln\left(\frac{k+1}{k}\right) = \ln(k+1) - \ln(k)$ Now note that since $\alpha+\beta$ is odd, $\beta - \alpha$ is also odd, so $\begin{aligned} \sum_{n=\alpha}^\beta \frac{\sqrt[n]{e^{-x}}(-1)^{n+1}}{x} &= \sum_{n=0}^{\beta-\alpha} \frac{e^{-x/(\alpha+n)}(-1)^{\alpha+n+1}}{x} \\ &= \sum_{n=0}^{(\beta-\alpha-1)/2} \frac{e^{-x/(\alpha+(2n))}(-1)^{\alpha+(2n)+1}}{x} + \frac{e^{-x/(\alpha+(2n+1))}(-1)^{\alpha+(2n+1)+1}}{x} \\ &= (-1)^\alpha \sum_{n=0}^{(\beta-\alpha-1)/2} \frac{e^{-x/(\alpha+2n+1)}}{x} - \frac{e^{-x/(\alpha+2n)}}{x} \end{aligned}$ From this along with our previous work, we can conclude $\Psi = (-1)^\alpha \sum_{n=0}^{(\beta-\alpha-1)/2} \ln(\alpha+2n+1) - \ln(\alpha+2n) = \sum_{n=\alpha}^\beta (-1)^{n+1} \ln(n)$

Finding $\Phi$ : Once again, we let $k$ denote some positive integer to write $\begin{aligned} \int_0^\infty &\left(\frac{\left(\arctan(1+(k+1)x)\right)^{1/(2+(k+1)x)}}{x} - \frac{\left(\arctan(1+kx)\right)^{1/(2+kx)}}{x}\right)\,dx \\ &= \lim_{\varepsilon\to 0^+}\lim_{N\to\infty} \int_{\varepsilon}^N \frac{\left(\arctan(1+(k+1)x)\right)^{1/(2+(k+1)x)}}{x}\,dx - \int_{\varepsilon}^N \frac{\left(\arctan(1+kx)\right)^{1/(2+kx)}}{x}\,dx \\ &= \lim_{\varepsilon\to 0^+}\lim_{N\to\infty} \int_{(k+1)\varepsilon}^{(k+1)N} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du - \int_{k\varepsilon}^{kN} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du \\ &= \lim_{N\to\infty} \int_{kN}^{(k+1)N} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du - \lim_{\varepsilon\to 0^+} \int_{k\varepsilon}^{(k+1)\varepsilon} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du \end{aligned}$ Since $\lim_{u\to 0^+} \left(\arctan(1+u)\right)^{1/(2+u)} = \arctan(1)^{1/2} = \frac{\sqrt{\pi}}{2} \\ \lim_{u\to \infty} \left(\arctan(1+u)\right)^{1/(2+u)} = 1$ we can once again use the squeeze theorem to show that $\lim_{\varepsilon\to 0^+} \int_{k\varepsilon}^{(k+1)\varepsilon} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du = \frac{\sqrt{\pi}}{2} \lim_{\varepsilon\to 0^+} \int_{k\varepsilon}^{(k+1)\varepsilon} \frac{1}{u}\,du = \frac{\sqrt{\pi}}{2}\ln\left(\frac{k+1}{k}\right)$ $\lim_{N\to\infty} \int_{kN}^{(k+1)N} \frac{\left(\arctan(1+u)\right)^{1/(2+u)}}{u}\,du = \lim_{N\to\infty} \int_{kN}^{(k+1)N} \frac{1}{u}\,du = \ln\left(\frac{k+1}{k}\right)$ and therefore $\begin{aligned} \int_0^\infty \left(\frac{\left(\arctan(1+(k+1)x)\right)^{1/(2+(k+1)x)}}{x} - \frac{\left(\arctan(1+kx)\right)^{1/(2+kx)}}{x}\right)\,dx &= \ln\left(\frac{k+1}{k}\right) - \frac{\sqrt{\pi}}{2}\ln\left(\frac{k+1}{k}\right) \\ &= \left(1 - \frac{\sqrt{\pi}}{2}\right)\left(\ln(k+1) - \ln(k)\right) \end{aligned}$ Now, using the same method as before, we can conclude $\Phi = \left(1 - \frac{\sqrt{\pi}}{2}\right)\sum_{n=\alpha}^\beta (-1)^{n+1} \ln(n)$

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Finally, we have $1 + \frac{\Phi}{\Psi} = 1 + \frac{\left(1 - \frac{\sqrt{\pi}}{2}\right)\displaystyle\sum_{n=\alpha}^\beta (-1)^{n+1} \ln(n)}{\displaystyle\sum_{n=\alpha}^\beta (-1)^{n+1} \ln(n)} = 1 + \left(1 - \frac{\sqrt{\pi}}{2}\right) = \frac{4-\sqrt{\pi}}{2}$ so that $A = 4, B=2 \implies \sqrt[B]{A}-B = \sqrt{4}-2 = \boxed{0}$