Question: A periodic function $f(t)$, with period $2\pi$ is defined as, $$ f(t) = \begin{cases} 0 & \text{ if } -\pi<t<0, \\ \pi & \text{ if } 0<t<\pi. \end{cases} $$ Find the Fourier series expansion of $f$.
Below is my working: Since $f$ is neither odd nor even, we need to calculate both $a_n$ and $b_n$, the coefficients of cosine and sine respectively. Note that for $n\geq 1$, we have \begin{align*} a_n & = \frac{2}{\pi}\int_0^\pi f(t)\cos(nt)dt \\ & = \frac{2}{\pi}\int_0^\pi \pi \cos(nt)dt \\ & = 2 \int_0^\pi \cos(nt)dt \\ & = 0. \end{align*} Also, \begin{align*} a_0 & = \frac{1}{\pi}\int_0^\pi f(t)dt \\ & = \frac{1}{\pi}\int_0^\pi \pi dt \\ & = \int_0^\pi dt \\ & = \pi. \end{align*} On the other hand, note that \begin{align*} b_n & = \frac{2}{\pi}\int_0^\pi f(t)\sin(nt)dt \\ & = \frac{2}{\pi}\int_0^\pi \pi \sin(nt)dt \\ & = 2\int_0^\pi \sin(nt)dt \\ & = 2 \left[ \frac{1-(-1)^n}{n} \right] \\ & = \frac{2[1-(-1)^n]}{n}. \end{align*} Therefore, the Fourier series of $f$ is $$ f(t) = \pi + \sum_{n=1}^\infty \frac{2[1-(-1)^n]}{n} \sin(nt). $$ However, based on Wolfram alpha, it seems that I am missing the factor $\frac12$ throughout the fourier expansion. Wolfram alpha gives $$2 \sin(t) + \frac23 \sin(3t) + \frac25 \sin(5t) + \frac27 \sin(7t) + \frac{\pi}{2}.$$ I am not able to fathom why this is the case. Any explanation would be greatly appreciated.
To see that the coefficients are wrong, The Fourier series for $f(x)$ may be written \begin{aligned} f(x) &= a_0 + \sum_{n=1} a_n \cos n x + \sum_{n=1} b_n \sin n x \end{aligned} The coefficients may be calculated \begin{aligned} a_0 &= \frac{\int_{-\pi}^{\pi} f(x) dx}{\int_{-\pi}^{\pi} 1 dx} \\ a_n &= \frac{\int_{-\pi}^{\pi} f(x) \cos nx dx}{\int_{-\pi}^{\pi} \cos^2 nx dx}, \quad n \geqslant 1 \\ b_n &= \frac{\int_{-\pi}^{\pi} f(x) \sin nx dx}{\int_{-\pi}^{\pi} \sin^2 nx dx}, \quad n\geqslant 1 \\ \end{aligned} giving \begin{aligned} a_0 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x~dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x~dx. \end{aligned} You can easily verify the integrals of $\sin^2nx$ and $\cos^2nx$.