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