Question: $$f(x)= \sum_{n=0}^\infty \frac {e^{inx}}{1+n^2}$$ if $x\ne 2\pi k$ and $f(x)=0$ if $x=0 , x=2\pi k$
- Find $\hat f(n)$
- Find the Fourier series of $\displaystyle g(x)=\int _0^xf(t)dt$
Thoughts
Our solutions were:
$\displaystyle \sum_{n=0}^\infty \frac 1{1+n^2}$ for (1)
and $\displaystyle i\sum_{n=-\infty}^\infty \sum_{m=1}^\infty \frac 1{n+n^3}e^{inx}$ for (2).
Would love to verify this as we don't have the solutions and the way was very long.
$F\hat(n) $ = $\frac {f\hat(n)}{in}= \frac{1}{in(n^2+1}$ $g(x)=F(x)-F(0)=\sum_{-\infty}^{\infty}F\hat(n)(e^{inx}-1)= \sum_{-\infty}^{\infty} \frac{(e^{inx}-1)}{in(n^2+1)}$ therefore $g\hat (n)=\frac 1{2\pi}\int _0^{2\pi}$$ (\sum_{m=-\infty}^{\infty}\frac 1{im(m^2+1)}e^{imx}-1)e^{-inx}=$ $\frac 1{i2\pi} \sum_{m=-\infty}^{\infty}\frac 1{im(m^2+1)}\int _0^{2\pi}(e^{imx}-1)e^{-inx}$ for m=n the middle first integral is $2\pi$, otherwise it's 0, the second integral is always 0. therefore we get the answer we got.
The part "$f(x)=0$ if $x=0$, $x=2\pi k$" is irrelevant, since Fourier series is not affected by changing $f$ on a set of measure zero.
Your answer to 1) is incorrect: should be $$ \hat f(n)=\begin{cases} 1/(1+n^2),\quad &\text{if }\ n\ge 0 \\ 0 &\text{if } \ n<0\end{cases} $$ because $\hat f(n)$ is the coefficient of $e^{inx}$ in the series.
To answer 2), recall that Fourier series can be integrated term-by-term, and for $n\ne 0$ the function $(in)^{-1} e^{inx}$ is an antiderivative of $e^{inx}$. Hence, $$ \hat g(n)=\begin{cases} (in)^{-1}/(1+n^2),\quad &\text{if }\ n> 0 \\ 0 &\text{if } \ n<0\end{cases} $$ The coefficient with $n=0$ can be found by evaluating $g$ at $x=0$. This gives $$ \hat g(0)= - \sum_{n=1}^\infty \frac{(in)^{-1}}{1+n^2} $$ (there is no simple closed form for this sum).