- #1

- 107

- 26

f(t) = t:

$$t=\sum_{n=-\infty }^{\infty } \frac{e^{itn}}{2\pi }\int_{-\pi}^{\pi}t\: e^{-itn} dt$$

$$t=\sum_{n=-\infty }^{\infty } \frac{cos(tn)+i\, sin(tn)}{2\pi }\int_{-\pi}^{\pi}t\: e^{-itn} dt$$

$$t=\sum_{n=-\infty }^{\infty } \frac{cos(tn)+i\, sin(tn)}{2\pi }\: (2i)(\frac{\pi cos(\pi n)}{n}-\frac{sin(\pi n)}{n^{2}})$$

$$t=\sum_{n=-\infty }^{\infty } \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

Because the imaginary part is a odd function only remains the term with n=0

so:

$$t=\sum_{n=-\infty }^{\infty } \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

Because the real part is a even function we can transform it into this:

$$t=2\sum_{n=1 }^{\infty } \left ( \frac{sin(tn)sin(\pi n )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+\lim_{n\rightarrow 0}\, \, \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right)+$$

$$+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

the first limit is 0 and in the sum we can delete the term with contains ##sin(\pi n )## and get:

$$t=-2\sum_{n=1 }^{\infty }\frac{sin(nt)cos(n\pi )}{n}+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

$$t=-2\sum_{n=1 }^{\infty }(-1)^{n}\frac{sin(nt)}{n}+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

this is right if the limit is equal to 0 but is undefined so where is the error?