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?



Evaluating the integral $\int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt$
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


$$j\_k=\left(\frac{\partial^k}{\partial s^k} \left. \frac{\sqrt{\pi}\Gamma(s+1)}{2 \Gamma(s+5/2)}\right)\right|\_{s=-1/2}.$$


