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First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-3/2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

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

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

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

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

First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-3/2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

If you want to get the asymptotics for $I_k(x)$ when $k$ is fixed, then using the substitution $t=1/u$ we get

$$I_k(x)=\int_{1/x}^1 (-1)^k\log^k u \sqrt{u(1-u)}du=(-1)^k(j_k+R_k(x)),$$

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

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

First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-3/2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

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

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

First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-3/2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

If you want to get the asymptotics for $I_k(x)$ when $k$ is fixed, then using the substitution $t=1/u$ we get

$$I_k(x)=\int_{1/x}^1 (-1)^k\log^k u \sqrt{u(1-u)}du=(-1)^k(j_k+R_k(x)),$$

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 (small enough) $s$:

$$\int_0^1 u^s(1-u)^{1/2}du=B(s+1,3/2)=\frac{2 \Gamma(s+5/2)}{\sqrt{\pi}\Gamma(s+1)}.$$

Now differentiate the lhs of this equality $k$ times and set $s=1/2$. Using $(u^s)'=u^s\log u$, we conclude that

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

First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-3/2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

If you want to get the asymptotics for $I_k(x)$ when $k$ is fixed, then using the substitution $t=1/u$ we get

$$I_k(x)=\int_{1/x}^1 (-1)^k\log^k u \sqrt{u(1-u)}du=(-1)^k(j_k+R_k(x)),$$

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

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