Asymptotics for $\log$ integral So I have the following integral:
$$
\int_{k - 1/2}^{k + 1/2} \log t \ \mathrm{d}t
$$
And I would like to show it can be written as:
$$
\int_{k - 1/2}^{k + 1/2} \log t \ \mathrm{d}t = \log k + \mathcal{O}(1/k^2)
$$
How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with:
$$
\left(x+\frac{1}{2}\right)\left(\log\left(x+\frac{1}{2}\right)\ -\ 1\right)\ -\left(x-\frac{1}{2}\right)\left(\log\left(x-\frac{1}{2}\right)\ -\ 1\right)\ 
$$ 
which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks
 A: For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $t\in\left(k-1/2,k+1/2\right)$,
$$
\log t=\log\left(\frac{t}{k}k\right)=\log k+\log\frac{t}{k}.
$$
Denote
$$
s=\frac{t}{k}-1\in\left(-\frac{1}{2k},\frac{1}{2k}\right)\subseteq\left(-1,1\right).
$$
Thus Taylor's theorem applies, and
$$
\log t=\log k+\log\left(1+s\right)=\log k+s-\frac{s^2}{2}+O(s^3).
$$
Thanks to this result,
\begin{align}
\int_{k-1/2}^{k+1/2}\log t{\rm d}t&=\log k+\int_{k-1/2}^{k+1/2}\left[s-\frac{s^2}{2}+O(s^3)\right]{\rm d}t\\
&=\log k+k\int_{-1/2k}^{1/2k}\left[s-\frac{s^2}{2}+O(s^3)\right]{\rm d}s,
\end{align}
which suffices to lead to your expected estimate.
A: You properly wrote$$\left(x+\frac{1}{2}\right)\left(\log\left(x+\frac{1}{2}\right)\ -\ 1\right) -\left(x-\frac{1}{2}\right)\left(\log\left(x-\frac{1}{2}\right)\ -\ 1\right)$$
Now rewrite
$$\log(x+a)=\log(x)+\log\left(1+\frac a x\right)=\log(x)+\frac{a}{x}-\frac{a^2}{2 x^2}+\frac{a^3}{3
   x^3}+O\left(\frac{1}{x^4}\right)$$ So,
$$(x+a)\, (\log (x+a)-1)-(x-a)\,(\log (x-a)-1)=2a \log(x)-\frac{a^3}{3 x^2}+O\left(\frac{1}{x^4}\right)$$ and $a=\frac 12$.
