Function Integrable in an improper sense that does not satisfy Riemann's Theorem I need some help over the subject of Fourier series...
Do you know if there's  a function $g(t)$ integrable in a improper sense over an interval $[a,b]$ and such that 
$\lim\limits_{p\rightarrow \infty}\int_a^b g(t)sin(pt)\,dt $ is not cero
Or in less words does not satisfy Riemann?s Theorem
I would be really thankfull if you can help me... Good night
 A: This question is often asked by teacher who hopes to get an answer from the students.  It is not an easy question.  Very often answers are given for infinite interval but very few would give an example for a finite interval.
I shall give an example derived from a Lebesgue function.
First function to be defined is the Lebesgue function in the interval $[0, \pi ]$. 
For each integer $ k  ≥ 0 $, let  ${a_k} = {3^{{k^4}}}$ and let 
${c_r} = \frac{1}{{{r^2}}}$ .
  Subdivide the interval $(0, \pi] $ by the interval 
  ${I_k} = \left[ {\frac{\pi }{{a{}_k}},\;\frac{\pi }{{a{}_{k - 1}}}} \right]{\rm{ ,  }}k = 1,2, \cdots $ 
Define  the function $f$ in  $[0, \pi]$ by 
$f(x) = {c_k}\sin ({a_k}x){\rm{  for }}x \in {I_k} = \left[ {\frac{\pi }{{{a_k}}},\;\frac{\pi }{{{a_{k - 1}}}}} \right]$ and $ f(0) = 0 $ .
Then we have:
(1)  The function  $f$  is continuous in $[0, \pi]$ .
(2)  The Fourier cosine series of $f $ diverges at $x = 0 $.  That is to say, if we extend  $f$  to an even function $F (x)$ in $[ -\pi,\pi ] $, so that  $ F(x) = f(x)$  and $ F(-x) = f(x)$ for $x ≥ 0$, then the Fourier series of $ F $ diverges at $ x = 0 $.
(3) The function $g(x)$  defined by
$g(x) = \left\{ \begin{array}{l}
\frac{{f(x)}}{x},{\rm{  }}0 < x \le \pi ,\\
0{\rm{ ,   }}x = 0
\end{array} \right.$  is improperly Riemann integrable on $[0, \pi] $. 
(4)$ g$ is not absolutely Riemann integrable on $[0, \pi]$ and so it is not Lebesgue integrable on $[0, \pi]$.
(5) The sequence 
$\left( {\int_0^\pi  {g(t)\sin (nt)dt} } \right) = \left( {\int_0^\pi  {f(t)\frac{{\sin (nt)}}{t}dt} } \right)$ 
diverges.  Hence $\int_0^\pi  {g(t)\sin (nt)dt} $ does not tend to 0 as $n$ tends to infinity.  That is to say g does not satisfy the conclusion of the Riemann Lebesgue Theorem.
(2) and (5) are equivalent.
The function you are looking for is g  defined in $[0, \pi]$ a finite interval.
From this example you can make up other examples of the same type.
The verification or proof of (3) (4) and (5) is rather technical. You may access the detail in my article 
Counter example Riemann Lebsgue Lemma
