I need to evaluate the integral $\int_2^{\infty} \frac{\ln x}{x^{15}}\,dx$. 
I need to evaluate the integral
  $$\int_2^{\infty} \frac{\ln x}{x^{15}}\,dx$$

I tried integration by substitution but cant get the $x$ term to cancel, so im kind of stuck, any ideas?
 A: Alternatively, one may start with
$$
\int_2^{\infty}  \frac1{x^{s}}\: dx=\frac1{(s-1)\cdot2^{s-1}},\qquad s>1,
$$ then one is allowed to differentiate with respect to $s$ under the integral sign to get
$$
\int_2^{\infty}  -\frac{\ln x}{x^{s}}\: dx=\left(\frac1{(s-1)\cdot2^{s-1}} \right)',\qquad s>1,
$$ that is
$$
\int_2^{\infty}  \frac{\ln x}{x^{s}}\: dx=\frac1{(s-1)^2\cdot2^{s-1}}+\frac{\ln 2}{(s-1)\cdot2^{s-1}},\qquad s>1,
$$ putting $s=15$ gives

$$
\int_2^{\infty}  \frac{\ln x}{x^{15}}\: dx=\frac{1}{3\:211\:264}+\frac{\ln  2}{229\:376}.
$$

A: HINT:Integration by parts gives
$$\left(-\frac{x^{-14}}{14}\cdot\ln x\right)_2^{\infty}+\frac{1}{14}\int_2^{\infty} x^{-15}dx$$
A: $$\int_2^{\infty} \log x  \cdot x^{-15} dx$$
Integration by parts,
$$= \left[ \log x \cdot \frac{x^{-14}}{-14}\right]_2^{\infty} - \int_2^{\infty} \frac 1x \cdot \frac{x^{-14}}{-14}dx$$
$$= \left[ \log x \cdot \frac{x^{-14}}{-14}\right]_2^{\infty} - \int_2^{\infty} \frac{x^{-15}}{-14}dx$$
$$= \left[ \log x \cdot \frac{x^{-14}}{-14}\right]_2^{\infty} -  \left[\frac{x^{-14}}{-14 \cdot -14}\right]_2^{\infty}$$
Can you proceed further?
