Fourier and integral

Question

Given the below trigonometric series:

$1 + \sum_{n=1}^{\infty} \frac{2}{1+n^{2}}\cos (nt)$

Where $f(t)$ is the value of the series.

Can I then deduce that $\int_{-\pi}^{\pi} f(x) dx$ is $2\pi$? I ask because the series for $f(t)$ looks like a fourier series and I can then recognize that $1 = \frac{1}{2} \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx$.

?

Answer 1

The series converges uniformly by Weierstrass's $M$-test, so it's ok to interchange integration and summation: $$\int_{-\pi}^{\pi} \left( 1+ \sum_{n=1}^\infty \frac{2}{1+n^2}\,\cos nt \right)\,dt = \int_{-\pi}^{\pi} 1\,dt + \sum_{n=1}^\infty \left(\frac{2}{1+n^2}\int_{-\pi}^{\pi}\cos nt \,dt\right) = 2\pi + \sum_{n=1}^\infty 0$$

Answer 2

Note that $\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,\mathrm{d}x$ is the constant coefficient of the Fourier series, which is $1$. Thus, the integral is $2\pi$.