I have an even function with the Fourier series as below:
$f(t) = \frac{a_0}2 + \sum_{n=1}^\infty a_n \cos nt$
with $a_n = \frac{1}\pi $$\int_{-\pi}^{\pi} f(x)\cos nx\ dx$
And I have to demonstrate the Fourier series for $f(\pi (t-2))$ in the interval {1,3} using $a_n$
Here what's I am thinking: Let $f(\pi (t-2)) = \frac{a'_0}2 + \sum_{n=1}^\infty a'_n \cos(n\pi (t-2))$
So $a'_0 = \frac{1}\pi $$\int_{-\pi}^{\pi} f(\pi (x-2))\ dx$
Let $ u = \pi (x-2)$, then $du = \pi dx$, and x: 1 to 3 give u: $-\pi$ to $\pi$
Then $a'_0 = \frac{1}{\pi^2} $$\int_{-\pi}^{\pi} f(u)\ du$ = $\frac{a_0}\pi$
But when I calculate $a_n$ with the same method, it gave $a_n = \frac{1}{\pi^2} $$\int_{-\pi}^{\pi} f(u)\cos n\pi u du$
and I stuck here.
You don't have to evaluate any coefficients. Just change $t$ to $\pi (t-2)$ in the given series and use the identity $\cos (n\pi (t-2))=\cos (n \pi t) \cos 2n\pi) -\sin (n \pi t) \sin 2n\pi)=\cos (n \pi t)$ The Fourier series on $[1,3]$ is $f=\frac {a_0} 2 +\sum_{n=1}^{\infty} a_n \cos (n\pi t)$. Note that $\{\cos (n \pi t)\}$ is orthonormal on $[1,3]$ (and $\{\cos (n t)\}$ is not!). Hence all you have to do is replace $\{\cos (n t)\}$ by $\{\cos (n \pi t)\}$.