I want to explain this summation to Mathematica,

For $a >0$ define $n_u, n_d \in \mathbb{Z}$ such that if $\{ a \} \leq 0.5$ then $n_u = [a]-1$ and $n_d = - [a]$ and if $\{ a \} > 0.5$ then $n_u = [a]$ and $n_d = -([a]+1)$. Now compute the sum, $\sum _ {n = n_d} ^{n_u} (1+2n)Log[a^2 - (n + \frac{1}{2} )^2]$

Can someone help me write this in Mathematica?

Withe everything that I tried so far, Mathematica produced a complex number as the answer! I don't know how Mathematica is adding up a finite number of real numbers to get a complex number!

-
I'm curious to see the code that you tried. Could you post it? –  Sjoerd C. de Vries May 10 at 22:50
@SjoerdC.deVries I am a noob enough to not know how to put in the conditionals on nu and nd. I just this plain thing, Assuming [ n1 [Element] Integers && n2 [Element] Integers && n1 < n2 && a > 0 && n2 < a && n1 < 0, Sum [ (1 + 2 n) Log [ a^2 - ( n + (1/2 )^2) ] , {n, n1, n2}]] // FullSimplify –  user6818 May 10 at 23:11
@user6818 is $[a]$ intended to be the Gauss style notation for $\operatorname{floor(a)}=\left \lfloor{a}\right \rfloor$? Additionally, how should I interpret $\{ a \}$? Is this just the scalar $a$? Also, note that in your comment you have inside the logarithm $a^2-\left(n+\left(\frac{1}{2}\right)^2\right)$, but in your post it is $a^2-\left( n+\frac{1}{2} \right)^2$. –  J. W. Perry May 11 at 2:59
Is this a homework problem? If so, it should be tagged as such. –  m_goldberg May 11 at 3:01
@J.W.Perry [a] is the floor function and {a} is the fractional part of a. And the comment has the typo - the correct summand is in the question. –  user6818 May 11 at 3:03

Ok, with your last defining comment, this seems quite clear. Your data suggests (avoid $a\leq0$):

f[a_] := If[
FractionalPart[a] <= 1/2,
Sum[(1 + 2 n)* Log[a^2 - (n + 1/2)^2], {n, -Floor[a], Floor[a] - 1}],
Sum[(1 + 2 n)* Log[a^2 - (n + 1/2)^2], {n, -(Floor[a] + 1), Floor[a]}]
]


Now a set of sums for various fixed values of $a>0$,

Table[f[a], {a, 1/10, 2, 1/10}] (*a from .1 to 2 in steps of .1*)


produces

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

-
Note that this precisely satisfies all of the conditions in your actual post, but the "sorta code" you wrote in your first comment where $n1<n2$, is absolutely false for the branch where $0<a\leq.5$. The lower bound is $0$ there, while the upper bound is $-1$ everywhere on that interval by the definition in your actual post ($n_d$ being lower, and $n_u$ being upper). –  J. W. Perry May 11 at 5:59