Evaluating the integral $\int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt$ I am trying to evaluate the integral
$$
I_k(x)=\int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt
$$
with $x$ tending to infinity.
In fact, I wish to have an estimate
$$
\sum_{k=0}^\infty \frac{1}{\log^k x} \int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt = C+E(x),
$$
where $C$ is explicit and $E(x) \to 0$ as $x \to \infty$.
The integral inside may be actually rewritten as
$$
I_k(x) = \int_1^x \log^k t \sqrt{1-\frac{1}{t}} \frac{dt}{t^{3/2}} = c_k \int_{1/x^2}^1 \sqrt{1-t^2} \log^k t dt,
$$
hence can be possibly attacked via multiple zeta values.
Is that possible?
 A: First of all, your series diverges for any $x>1$, because we always have
$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{x}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{x}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$
If you want to get the asymptotics for $I_k(x)$ when $k$ is fixed, then using the substitution $t=1/u$ we get
$$I_k(x)=\int_{1/x}^1 (-1)^k\log^k u \sqrt{(1-u)/u}du=(-1)^k(j_k+R_k(x)),$$
where
$$j_k=\int_0^1 u^{-1/2}(1-u)^{1/2}\log^k u du$$
and
$$R_k(x)=\int_0^{1/x} u^{-1/2}(1-u)^{1/2}\log^k u du=O_k\left(\frac{\log^{k+1} x}{x^{3/2}}\right).$$
Now lets compute $j_k$ in terms of the derivatives of Gamma-function. By the beta-integral, we have for any $s>-1$:
$$\int_0^1 u^s(1-u)^{1/2}du=B(s+1,3/2)=\frac{\sqrt{\pi}\Gamma(s+1)}{2 \Gamma(s+5/2)}.$$
Now differentiate the lhs of this equality $k$ times and set $s=-1/2$. Using $(u^s)'=u^s\log u$, we conclude that
$$j_k=\left(\frac{\partial^k}{\partial s^k} \left. \frac{\sqrt{\pi}\Gamma(s+1)}{2 \Gamma(s+5/2)}\right)\right|_{s=-1/2}.$$
A: In view of the output of the Mathematica code
Table[Series[Integrate[Log[t]^k*Sqrt[-t^2+1],{t, 1/x^2, 1},Assumptions->x> 1],{x,Infinity,2}],{k,1,3}]

$$O\left(\left(\frac{1}{x}\right)^3\right)+\frac{2 \log (x)+1}{x^2}+\left(-\frac{\pi }{8}-\frac{1}{8} \pi  \log (4)\right), $$
$$\frac{1}{48} \left(\pi ^3+6 \pi +3 \pi  \log ^2(4)+3 \pi  \log (16)\right)+\frac{-4 \log ^2(x)-4 \log (x)-2}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right), $$
$$\frac{1}{32} \left(-12 \pi  \zeta (3)-\pi ^3-6 \pi -\pi  \log ^3(4)-3 \pi  \log ^2(4)-\pi ^3 \log (4)-\pi  \log (4096)\right)+\frac{8 \log ^3(x)+12 \log ^2(x)+12 \log (x)+6}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right) $$
I have doubts concerning  simple asymptotics for $I_k(x)$.
