Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.


In [my
answer](https://mathoverflow.net/questions/25794/shortest-most-elegant-proof-for-l1-chi-neq-0/25822#25822) to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples
of the real characters modulo 3 and 4. The proof goes as follows: one takes
$$
f(x)=\sum\_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum\_{j=1}^{m-1}\chi(j)x^j
$$
and uses Abel's theorem to write
$$
L(1,\chi)=\int\_0^1f(x)dx;
$$
since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.


Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of
the polynomial
$$
g\_\chi(x)=\sum\_{n=1}^{m-1}\chi(n)x^n
$$
on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works
perfectly further; for example,
$$
g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5
$$
or
$$
g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.
$$
Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is
always holds for small values $m\le30$.


**Question.** Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$,
is it true that $g\_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for
which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed
in the literature?



Positivity of $L(1,\chi)$ for real Dirichlet's character
These are called Fekete polynomials, and you can find out a great deal about them [here](http://www.dms.umontreal.ca/~andrew/PDF/fekete.pdf). Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.


