Leading term asymptotic integral containing logarithm I want to get the leading term in the asymptotic of the integral $$\int_{a}^{+\infty}\frac{x^3}{(1+x^2)^{\alpha}}\ln\frac{x+a}{x-a} \,dx$$ for $a\to +\infty$. Here $a>0$ and $\alpha>\frac{3}{2}$ to ensure integrability at infinity.
My main concern is the integrable singularity in $a$. Usually I do an appropriate change of variables to remove it, but I can't seem to find an appropriate one.
 A: First, we make a change $x=ta$. The integral becomes $I(a, \alpha)=a^4\int_{1}^{\infty}\frac{t^3}{(1+a^2t^2)^{\alpha}}\ln\frac{t+1}{t-1} \,dt=a^{4-2\alpha}\int_{1}^{\infty}\frac{t^{3-2\alpha}}{(1+\frac{1}{a^2t^2})^{\alpha}}\ln\frac{t+1}{t-1} \,dt$.
Integral converges; the only (integrable) singularity is at $t=1$.
We can expand the integrand into the series
$$I(a,\alpha)=a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}{(1-\frac{\alpha}{a^21!}\frac{1}{t^2}+\frac{\alpha(\alpha+1)}{a^42!}\frac{1}{t^4}+...)} \,dt$$ where all integrals converge ($3-2\alpha<1$ and $\ln\frac{t+1}{t-1}\sim\frac{1}{t}$ at $t\to\infty$)
The mian asymptotics term is $a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}dt=a^{4-2\alpha}\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt$
where $b=3-2\alpha$.
It seems that the easiest way to evaluate the integral $J(b)=\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt=\int_{0}^{1}x^{-b-2}\ln\frac{1+x}{1-x}dx$  ($t=\frac{1}{x}$) is to expand the integrand into the series:
$J(b)=2\int_{0}^{1}x^{-b-2}\left(x+\frac{x^3}{3}+\frac{x^5}{5}+...\right)dx=2\left(\frac{1}{-b}+\frac{1}{3(-b+2)}+\frac{1}{5(-b+4)}+...\right)=$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{k+\frac{1}{2}}\frac{1}{k-\frac{b}{2}}=\frac{1}{-b-1}\left(\Psi(-\frac{b}{2})-\Psi(\frac{1}{2})\right)$,
where $\Psi$ is digamma function: $\Psi(1+s)=-\gamma-\sum_{k=1}^{\infty}\left(\frac{1}{k+s}-\frac{1}{k}\right)$
Finally, we get the main asymptotics term$$I(a, \alpha)\sim{a}^{4-2\alpha}\frac{1}{2(\alpha-2)}\left(\Psi(\alpha-\frac{3}{2})-\Psi(\frac{1}{2})\right)$$
At $\alpha\to2$ we have to disclose the uncertainty
$I(a,\alpha\to2)\sim\frac{{a}}{2}^{0}\lim_{\epsilon\to0}\frac{1}{\epsilon}\left(\Psi(\frac{1}{2}+\epsilon)-\Psi(\frac{1}{2})\right)=\frac{1}{2}\Psi'(\frac{1}{2})=\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2}$
$$I(a,\alpha=2)\sim\int_{1}^{\infty}t^{-1}\ln\frac{t+1}{t-1}dt=\frac{3}{2}\zeta(2)=\frac{\pi^2}{4}$$
Other asymptotics terms can be evaluated in the same way.
