Continuity of parametric integral $I(\alpha)=\int_0^\infty \frac{\ln{(1-\alpha^2+\alpha^2x^2)}}{x^2-1}dx$

Question

How to prove that parametric integral $I(\alpha)=\int_0^\infty \frac{\ln{(1-\alpha^2+\alpha^2x^2)}}{x^2-1}dx$ is differentiable on $(1,\infty)$? I wrote $I(\alpha)$ as $\int_0^1 \frac{\ln{(1-\alpha^2+\alpha^2x^2)}}{x^2-1}dx+\int_1^\infty \frac{\ln{(1-\alpha^2+\alpha^2x^2)}}{x^2-1}dx=I_1(\alpha)+I_2(\alpha)$, but I'm having difficulties to show that $I_2(\alpha)$ is differentiable on $(1,\infty)$. Since $f(x,\alpha)= \begin{cases} \frac{\ln{(1-\alpha^2+\alpha^2x^2)}}{x^2-1}, 0<x<1\\ \alpha^2, x=1 \end{cases}$ and $f'(x,\alpha)$ are continuous on $[0,1)\times (1,\infty)$, $I_1(\alpha)$ is continuous on $(1,\infty)$, am I right? Any help is welcome. Thanks in advance.

Answer 1

There's actually a standard complex analytic procedure to show an even stronger result; $I_2(\alpha)$ is analytic over $D=\{\alpha\in\mathbb{C}:\mathfrak{R}(\alpha)>1\}$. It hinges on Morera's theorem.

Let $\gamma$ be a closed $C^1$ contour in $D$. Since for every $\alpha\in D$ the absolute value of the integrand grows asymptotically as $O(x^{-2}\ln(x))$) and thus is integrable over $\gamma\times(1,\infty)$, then through Fubini's theorem, we may exchange the order of integration in the following double integral as follows:

\begin{equation} \begin{split} \oint_\gamma I_2(\alpha)d\alpha &=\oint_\gamma\int_1^\infty\frac{\ln(1-\alpha^2+\alpha^2x^2)}{x^2+1}dxd\alpha\\ &=\int_1^\infty\oint_\gamma\frac{\ln(1-\alpha^2+\alpha^2x^2)}{x^2+1}d\alpha dx\\ &=0 \end{split} \end{equation} where the last equality is due to Cauchy's integral theorem, since the integrand is analytic over $D$ for all $x>1$. We have just proven that \begin{equation} \oint_\gamma I_2(\alpha)d\alpha=0 \end{equation} for all closed piecewise $C^1$ contours on $D$, and thus by Morera's theorem $I_2(\alpha)$ is analytic on $D$. This immediately implies that it is (real) differentiable on $(1,\infty)$.