Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$ 
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from
  $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$.

For some $\theta$ between $a-1$ and $a$ and odd $n\in\mathbb{N}$ we have equality:
$$\log a - \int_a^{a+1} \log x dx=\log a - \int_{a-1}^atdt+\int_{a-1}^a \frac{t^2}{2}dt-\int_{a-1}^a \frac{t^3}{3}dt+ \ldots -\int_{a-1}^a \frac{\theta^{n+1}}{n+1}dt$$
I don't know how to move further. Any hint please?
 A: We have
$$I=\log(a) - \int_a^{a+1}\log(x)dx = \int_a^{a+1} \left(\log(a)-\log(x)\right)dx =- \int_a^{a+1} \log(x/a)dx$$
Setting $x=at$, we obtain
$$I = - a\int_{t=1}^{1+1/a}\log(t)dt = - a\int_0^{1/a}\log(1+t)dt$$
We have $t-\dfrac{t^2}2 \leq \log(1+t) \leq t$ for all $t \in [0,1]$. Hence, we have
$$-\int_0^{1/a} atdt \leq I \leq -\int_0^{1/a} atdt + \int_0^{1/a}\dfrac{at^2}2dt$$
This gives us
$$-\dfrac1{2a} \leq I \leq -\dfrac1{2a} + \dfrac1{6a^2} \implies 0 \leq I-\left(-\dfrac1{2a}\right) \leq \dfrac1{6a^2}$$
which proves what you want.
A: Direct solution
$\int_a^{a+1} \ln(x)\ dx = [ x \ln(x) - x ]_a^{a+1}$
$ = (a+1) \ln(a+1) - a \ln(a) - 1$
$ = \ln(a) + (a+1) \ln(1+\frac{1}{a}) - 1$
$ = \ln(a) + (a+1) ( \frac{1}{a} - \frac{1}{2a^2} + \frac{1}{3a^3} - \frac{1}{4a^4} + \cdots ) - 1$
$ = \ln(a) + \frac{1}{2a} - \frac{1}{6a^2} + \frac{1}{12a^3} - \cdots$ [which is clearly an alternating series since $a \ge 1$].
Thus you get the answer immediately, and much more, since the general term in the asymptotic expansion is trivial to find.
