Estimating the rate of convergence of an integral I'm studying the integral $\displaystyle\int_0^w \frac{s\mathrm ds}{(e^s+1)\sqrt{1-(s/w)^2}}$ as $w\to\infty$. The intuition suggests that this integral converges to $\displaystyle\int_0^\infty \frac{s\mathrm ds}{ e^s+1 }=\frac{\pi^2}{12}$, because the singularity in $s=w$ is integrable but the obtained mass goes to zero thanks to the exponential, and everywhere else the integrands are close (none of this is formal, of course).
To the more formal approach. We split the integral:
$\displaystyle\int_{w/2}^w \frac{s\mathrm ds}{(e^s+1)\sqrt{1-(s/w)^2}}\le \frac{w^2}{2\sqrt{2}(e^{w/2}+1)}\int_{1/2}^1 \frac{ \mathrm ds}{ \sqrt{1- s  }}=\frac{w^2}{ 2(e^{w/2}+1)} \to 0$ as $w\to\infty$
and
$\displaystyle\int_0^{w/2} \frac{s\mathrm ds}{(e^s+1)\sqrt{1-(s/w)^2}} $ which converges to $\displaystyle\int_0^\infty \frac{s\mathrm ds}{ e^s+1 }$ by Dominated Convergence Theorem.
Now I'd like to estimate the rate of convergence with all constants in closed form (the knowledge of asymptotic behavior is not sufficient, unfortunately). Is there a method to achieve it? Maybe, with some trickier splitting of the integral?
I'd be glad to hear all suggestions.
 A: Using the identity $\dfrac1{\sqrt{1-t}}=1+\dfrac{t}{(1+\sqrt{1-t})\sqrt{1-t}}$, one gets
$$
\int_0^w \frac{s\mathrm ds}{(\mathrm e^s+1)\sqrt{1-(s/w)^2}}=I_\infty+\frac1{w^2}J(w)-K(w),
$$
with
$$
I_\infty=\int_0^\infty \frac{s\mathrm ds}{ \mathrm e^s+1 },\qquad
K(w)=\int_w^\infty \frac{s\mathrm ds}{ \mathrm e^s+1 }.
$$
and
$$
J(w)=\int_0^w \frac{s^3\mathrm ds}{(\mathrm e^s+1)\left(1+\sqrt{1-(s/w)^2}\right)\sqrt{1-(s/w)^2}}.
$$
Arguments similar to the ones used in the question show that $J(w)\to J_\infty$ with
$$
J_\infty=\int_0^\infty \frac{s^3\mathrm ds}{2(\mathrm e^s+1)}.
$$
Finally, $K(w)\ll\dfrac1{w^2}$ hence
$$
\lim_{w\to\infty}w^2\cdot\left(\int_0^w \frac{s\mathrm ds}{(\mathrm e^s+1)\sqrt{1-(s/w)^2}}-I_\infty\right)=J_\infty.
$$
Edit: To get an upper bound on $J$, note that $\left(1+\sqrt{1-(s/w)^2}\right)\sqrt{1-(s/w)^2}\geqslant\frac32$ and $\mathrm e^s+1\gt\mathrm e^s$ for every $s\leqslant\frac12w$, and that $s^2\lt w^2$, $\mathrm e^s+1\gt\mathrm e^{w/2}$ and $1+\sqrt{1-(s/w)^2}\gt1$ for every $\frac12w\leqslant s\leqslant w$. Hence,
$$
J(w)\leqslant\frac23\int_0^{w/2}s^3\mathrm e^{-s}\mathrm ds+w^2\mathrm e^{-w/2}\int_{w/2}^w\frac{s\mathrm ds}{\sqrt{1-(s/w)^2}},
$$
which, using the change of variable $s\to ws$ in the last integral, implies that 
$$
J(w)\leqslant\frac23\int_0^{+\infty}s^3\mathrm e^{-s}\mathrm ds+\max\{w^4\mathrm e^{-w/2};w\gt0\}\cdot\int_{1/2}^1\frac{s\mathrm ds}{\sqrt{1-s^2}},
$$
that is, $J(w)\leqslant\frac23\cdot6+8^4\mathrm e^{-4}\cdot(1-\frac12\sqrt3)\lt15$.
