This is an expression in my Fourier Analysis text book, but I cann't understand it. 
This is an expression in my Fourier Analysis text book, but I cann't understand it.
I have done something,but I can't go on. Is there something wrong with the expression in my text book Or my work can go on? I need your help. Thanks.
 A: As your work shows, the integral becomes
$$\dfrac{2}{\pi}\int_0^v \dfrac{\sin u}{u} \dfrac{u/2(n+1)}{\sin \bigl(  u/2(n+1)\bigr)} du,$$
where $v = 2(n+1)x$.
Now what is missing in what you have posted (but must be in your text book,
either explicitly or implicitly) is that the limit is being taken as $n \to \infty$.  In this limit, the ratio $\dfrac{u/2(n+1)}{\sin \bigl( u/2(n+1)\bigr)}$
goes to $1$, and the quantity $v$ (which equals $2(n+1)x$) goes to $\infty$, and so the infinite series
$$\dfrac{4}{\pi}\sum_{k=0}^{\infty} \dfrac{1}{2k+1} \sin(2k+1)x$$
is equal to
$$\dfrac{2}{\pi}\int_0^{\pm \infty} \dfrac{\sin u}{u} du,$$ where the sign depends on the sign of $x$ (positive or negative), which we can rewrite as
$$\pm \dfrac{2}{\pi}\int_0^{\infty} \dfrac{\sin u}{u} du,$$
since the integrand is an even function of $u$.
Now actually $\displaystyle \dfrac{2}{\pi}\int_0^{\infty} \dfrac{\sin u}{u} du = 1$ (see here for example), and so the value of the Fourier series is equal to $\pm 1$, depending on the sign of $x$.
Now this last limit argument is a bit non-rigorous, because I am taking a limit
in both the argument of the integrand, and the limit of the integrand, at the
same time, so you have to be more careful than I am being.  Also, it is important
that $-\pi < x < \pi,$ otherwise the function $\sin \bigl(u/2(n+1)\bigr)$ (which appears in the denominator of the integrand) will have a zero over the range
of integration, which will completely mess up the analysis.  But this restriction on $x$ is okay, because the Fourier series is obviously periodic with period $2\pi$, and so if we know its value on the interval $[-\pi,\pi],$ 
we know its value everywhere.
In any case, the key thing to notice is that the only way that $x$ affects
the value of $\pm 1$ (i.e. of $\displaystyle\pm\dfrac{2}{\pi}\int_0^{\infty} \dfrac{\sin u}{u} du$) is through its sign.  The conclusion is that the value of the Fourier series only depends on the sign of $x$ for $x \in [-\pi,\pi],$ and so it is the Fourier series for a square wave.  (See here for a discussion from a slightly different view-point, beginning with the square wave and deriving the Fourier series, which is the opposite direction to your question; also, for your question the value of $L$ is $\pi$.)
