Hot Integral - 10

Since $x^4 + 4x^3 + (5-t)x^2 + 2(1-t)x \,=\, \big[(x+1)^2-1\big]\big[(x+1)^2 - t\big]$ , it follows that $\begin{array}{rcl} f(-t) & = & \displaystyle\int_0^\infty \frac{dx}{\sqrt{((x+1)^2 - 1)((x+1)^2+t)}} \; = \; \int_1^\infty \frac{dx}{\sqrt{(x^2-1)(x^2+t)}} \\ & = & \displaystyle \int_0^{\frac12\pi} \frac{\sec\theta \tan\theta\,d\theta}{\tan\theta \sqrt{\sec^2\theta + t}} \; = \; \int_0^{\frac12\pi} \frac{d\theta}{\sqrt{1 + t\cos^2\theta}} \end{array}$ using the substitution $x = \sec\theta$ . Hence $\begin{array}{rcl} I(\alpha) & = & \displaystyle\int_0^\infty t^{\alpha-1}f(-t)\,dt \; = \; \int_0^\infty \left(\int_0^{\frac12\pi} \frac{d\theta}{\sqrt{1 = t\cos^2\theta}}\right)\,dt \\ & = & \displaystyle\int_0^{\frac12\pi} \left(\int_0^\infty \frac{t^{\alpha-1}}{\sqrt{1 + t\cos^2\theta}}\,dt\right)\,d\theta \\ & = & \displaystyle\int_0^{\frac12\pi} \cos^{-2\alpha}\theta B(\alpha,\tfrac12-\alpha)\,d\theta \; = \; \tfrac12 B(\alpha,\tfrac12-\alpha) B(\tfrac12,\tfrac12-\alpha) \\ & = & \displaystyle\tfrac12 \frac{\Gamma(\alpha)\Gamma(\frac12-\alpha)}{\Gamma(\frac12)} \frac{\Gamma(\frac12)\Gamma(\frac12-\alpha)}{\Gamma(1-\alpha)} \; = \; \frac{\Gamma(\alpha)\Gamma(\frac12-\alpha)^2}{2\Gamma(1-\alpha)} \end{array}$ Differentiating gives $\int_0^\infty t^{\alpha-1}\ln t\,f(-t)\,dt \; = \; I'(\alpha) \; = \; \frac{\Gamma(\alpha)\Gamma(\frac12-\alpha)^2}{2\Gamma(1-\alpha)}\big[-2\psi(\tfrac12-\alpha) + \psi(1-\alpha) + \psi(\alpha)\big]$ This certainly all works for $\tfrac{1}{64} < \alpha < \tfrac14$ , as implied by the question, but I think that it holds for a greater range of $\alpha$ ; between $0$ and $\tfrac12$ .

Whatever: the answer is $1+2+2+1+2+1+2+1+1+2+1+1 = 17$ .