↑ index§3 Notes

math464

Revised 2026-09-17

Lecture 1

Proposition:

Let ${\theta }_1,{\theta }_2\in \mathbb{R}$. Then $e^{i{\theta }_1}\cdot e^{i{\theta }_2}=e^{i({\theta }_1+{\theta }_2)}$.

Proof:

$$\begin{array}{rl}{e}^{i{\theta }_1}\cdot e^{i{\theta }_2}& =(\cos {\theta }_1+i\sin {\theta }_1)(\cos {\theta }_2+i\sin {\theta }_2)\\ & =(\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)+i(\cos {\theta }_1\sin {\theta }_2+\sin {\theta }_1\cos {\theta }_2)\end{array}$$ $$e^{i({\theta }_1+{\theta }_2)}=\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)$$

By $\sin$ and $\cos$ angle sum identities, as desired.

Corollary:

$$e^{i{\theta }_1}\cdot e^{i{\theta }_2}\cdot e^{i{\theta }_3}\cdots =e^{i({\theta }_1+{\theta }_2+{\theta }_3+\cdots )}$$

(Induction on the number of factors: the base case is the Proposition, and the inductive step applies the Proposition to $e^{i({\theta }_1+\cdots +{\theta }_{n-1})}\cdot e^{i{\theta }_n}$.) Taking every ${\theta }_j=\theta$, $$\Rightarrow (e^{i\theta }{)}^n=e^{in\theta }\text{(De Moivre’s Formula)}$$

Note:

One can derive the exponential representations of $\sin$ and $\cos$: $$e^{i\theta }=\cos \theta +i\sin \theta ,e^{-i\theta }=\overline{e^{i\theta }}=\cos \theta -i\sin \theta$$ Adding the two kills the $\sin$ term, subtracting them kills the $\cos$ term: $$\Rightarrow \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2},\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$$

Lecture 2

Periodic functions.

Definition:

We say a function $f$ is periodic of period $T$ if $$f(t+T)=f(t)\forall t\in \mathbb{R}$$

Fact: If $f$ is periodic of period 1, then $f$ can be “represented” as $$\sum _{n=-\infty }^{\infty }{c}_n{e}^{2\pi int},c_n\in \mathbb{C}$$

Note that not all infinite sums of exponentiations are convergent, but in most cases that we will encounter in this class, this is an exact representation.

The collection of functions $\{e^{2\pi int}:n\in \mathbb{Z}\}$ can be used to represent any function of period 1.

Now suppose $f$ is of period 1, and we are told $$f(t)=\sum _{n\in \mathbb{Z}}{c}_n{e}^{2\pi int}$$

How can we compute $c_n$? $$\begin{array}{rl}{e}^{-2\pi ikt}f(t)& =\sum _{n\in \mathbb{Z}}{c}_n{e}^{-2\pi ikt}{e}^{2\pi int}\\ \Rightarrow {\int }_0^1{e}^{-2\pi ikt}f(t)dt& =\sum _{n\in \mathbb{Z}}{c}_n{\int }_0^1{e}^{2\pi i(n-k)t}dt\end{array}$$

Examine ${\int }_0^1{e}^{2\pi i(n-k)t}dt$. Antidifferentiating, we see that this must be split into two cases: $n=k$ and $n\ne k$. $${\int }_0^1{e}^{2\pi i(n-k)t}dt=\{\begin{array}{ll}{\int }_0^{1}1dt=1& n=k\\ {\frac{e^{2\pi i(n-k)t}}{2\pi i(n-k)}|}_0^1=\frac{1-1}{2\pi i(n-k)}=0& n\ne k\end{array}$$ (The numerator vanishes because $e^{2\pi i(n-k)}=1$ for integer $n-k$.) So every term of the sum dies except $n=k$: $$\Rightarrow {\int }_0^{1}f(t)e^{-2\pi ikt}dt=c_k$$

Thus, if $f$ is of period 1, then $$\boxed{c_n={\int }_0^{1}f(t)e^{-2\pi int}dt}$$

Now suppose $f$ is of period $T$. Let $g(t)=f(Tt)$. Then $g(t)$ is of period 1. We can observe this fact $$g(t+1)=f(T(t+1))=f(Tt+T)=f(Tt)=g(t)$$

We already know how to represent $g(t)$: $$g(t)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{2\pi int},c_n={\int }_0^1{e}^{-2\pi int}g(t)dt$$$$\Rightarrow f(Tt)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{2\pi int}$$

Let $s=Tt\Rightarrow t=\frac{s}{T}$. We can change variables to obtain: $$\boxed{f(s)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi in}{T}s}}$$

The same substitution in the coefficient formula ($s=Tt$, so $ds=Tdt$, and $t:0\to 1$ becomes $s:0\to T$): $$c_n={\int }_0^{1}g(t)e^{-2\pi int}dt={\int }_0^{1}f(Tt)e^{-2\pi int}dt$$$$\boxed{\Rightarrow c_n=\frac{1}{T}{\int }_0^{T}f(s)e^{\frac{-2\pi in}{T}s}ds}$$

Useful result:

If $f\in C^k$ ($k$-times differentiable), then $c_n$ decays like $c_n\sim \frac{1}{|n{|}^k}$.

Why: integrate by parts once. The boundary term cancels because $f$ and $e^{-2\pi int/T}$ are both $T$-periodic, so $$c_n(f^{\prime })=\frac{1}{T}\left({f(t)e^{\frac{-2\pi int}{T}}|}_0^T+\frac{2\pi in}{T}{\int }_0^{T}f(t)e^{\frac{-2\pi int}{T}}dt\right)=\frac{2\pi in}{T}{c}_n(f)$$ Iterating $k$ times gives $c_n(f)={\left(\frac{T}{2\pi in}\right)}^k{c}_n(f^{(k)})$, and $|c_n(f^{(k)})|\le \max |f^{(k)}|$ is bounded, so $|c_n|\le C/|n{|}^k$.

Example:

Suppose $f$ is periodic of period $T$ and even.

$$\begin{array}{rl}{c}_n& =\frac{1}{T}{\int }_0^T{e}^{\frac{-2\pi int}{T}}f(t)dt=\frac{1}{T}{\int }_0^T{e}^{\frac{-2\pi int}{T}}f(-t)dt\\ \text{(Let s:=−t) }& =\frac{-1}{T}{\int }_0^{-T}{e}^{\frac{2\pi ins}{T}}f(s)ds=\frac{1}{T}{\int }_{-T}^0{e}^{\frac{2\pi ins}{T}}f(s)ds\\ & =\frac{1}{T}{\int }_0^T{e}^{\frac{2\pi ins}{T}}f(s)ds=c_{-n}\end{array}$$

The last step shifts the interval $[-T,0]$ over to $[0,T]$, which is legal by the Proposition below.

Proposition:

If $g$ is periodic of period $T$, then $${\int }_a^{a+T}g(t)dt={\int }_0^{T}g(t)dt$$

Proof:

Let $F(x)={\int }_x^{x+T}g(t)dt$. $$F^{\prime }(x)=g(x+T)-g(x)=0\Rightarrow F\text{ is constant.}$$ In particular $F(a)=F(0)$, which is the claim.

Example:

Consider the square wave $f$ of period 1 defined on $[0,1]$ by $$f(t)=\{\begin{array}{ll}1,& 0\le t<\frac{1}{2}\\ -1,& \frac{1}{2}\le t<1\end{array}$$

One can easily find the Fourier coefficients by using casework on the integral between 0 and 1, as this is a piecewise function. Doing it: $$c_0={\int }_0^{1/2}1dt+{\int }_{1/2}^1(-1)dt=\frac{1}{2}-\frac{1}{2}=0$$ For $n\ne 0$, with antiderivative $\frac{e^{-2\pi int}}{-2\pi in}$, $$\begin{array}{rl}{c}_n& ={\int }_0^{1/2}{e}^{-2\pi int}dt-{\int }_{1/2}^1{e}^{-2\pi int}dt\\ & =\frac{e^{-\pi in}-1}{-2\pi in}-\frac{e^{-2\pi in}-e^{-\pi in}}{-2\pi in}=\frac{\left[(-1{)}^n-1\right]-\left[1-(-1{)}^n\right]}{-2\pi in}\\ & =\frac{2(-1{)}^n-2}{-2\pi in}=\boxed{\frac{1-(-1{)}^n}{\pi in}}=\{\begin{array}{ll}0,& n\text{ even}\\ \frac{2}{\pi in},& n\text{ odd}\end{array}\end{array}$$ using $e^{-\pi in}=(-1{)}^n$ and $e^{-2\pi in}=1$. This is the series quoted in Lecture 4.

Lecture 3

Suppose $f(t)$ is periodic of period $T$.

When $f$ is even and real-valued then $c_n=c_{-n}=\overline{c_n}$, or that $c_n$ are real.

Example:

The triangle wave is the following. Let $f(t)$ be a periodic function with period 1 $$f(t)=\{\begin{array}{ll}\frac{1}{2}+t,& -\frac{1}{2}\le t<0\\ \frac{1}{2}-t,& 0\le t<\frac{1}{2}\end{array}$$

Find the Fourier series of $f$. $$c_n={\int }_{-\frac{1}{2}}^{\frac{1}{2}}{e}^{-2\pi int}f(t)dt$$$$c_0={\int }_{-\frac{1}{2}}^{\frac{1}{2}}f(t)dt=2{\int }_0^{1/2}\left(\frac{1}{2}-t\right)dt=\frac{1}{4}$$

For $n\ne 0$: split $e^{-2\pi int}=\cos (2\pi nt)-i\sin (2\pi nt)$. Since $f$ is even, $f(t)\sin (2\pi nt)$ is odd and integrates to $0$, while $f(t)\cos (2\pi nt)$ is even, so the integral is twice the half-range one: $$c_n={\int }_{-\frac{1}{2}}^0{e}^{-2\pi int}\left(\frac{1}{2}+t\right)dt+{\int }_0^{\frac{1}{2}}{e}^{-2\pi int}\left(\frac{1}{2}-t\right)dt=2{\int }_0^{\frac{1}{2}}\left(\frac{1}{2}-t\right)\cos (2\pi nt)dt$$ Integrate by parts with $u=\frac{1}{2}-t$, $dv=\cos (2\pi nt)dt$, so $du=-dt$ and $v=\frac{\sin (2\pi nt)}{2\pi n}$. The boundary term dies at both ends ($u=0$ at $t=\frac{1}{2}$, $v=0$ at $t=0$): $$\begin{array}{rl}{c}_n& =2\left({\left(\frac{1}{2}-t\right)\frac{\sin (2\pi nt)}{2\pi n}|}_0^{1/2}+\frac{1}{2\pi n}{\int }_0^{1/2}\sin (2\pi nt)dt\right)\\ & =\frac{1}{\pi n}\left({\frac{-\cos (2\pi nt)}{2\pi n}|}_0^{1/2}\right)=\frac{1-\cos (\pi n)}{2{\pi }^2{n}^2}\\ & =\boxed{\frac{1-(-1{)}^n}{2{\pi }^2{n}^2}}=\{\begin{array}{ll}0,& n\text{ even}\\ \frac{1}{{\pi }^2{n}^2},& n\text{ odd}\end{array}\end{array}$$

Thus, the Fourier series of $f$ is $$\frac{1}{4}+\sum _{n\text{ odd},n\in \mathbb{Z}}\frac{1}{{\pi }^2{n}^2}{e}^{2\pi int}$$

However, since odd numbers come in $\pm$ pairs, we can write $$\begin{array}{rl}& \frac{1}{4}+\sum _{n\text{ odd},n\ge 1}\frac{1}{{\pi }^2{n}^2}\mathrm{cis}(2\pi nt)+\sum _{n\text{ odd},n\le -1}\frac{1}{{\pi }^2{n}^2}\mathrm{cis}(2\pi nt)\\ & =\frac{1}{4}+\frac{2}{{\pi }^2}\sum _{n\text{ odd},n\ge 1}\frac{\cos (2\pi nt)}{n^2}=\boxed{\frac{1}{4}+\frac{2}{{\pi }^2}\sum _{k=0}^{\infty }\frac{\cos (2\pi (2k+1)t)}{(2k+1{)}^2}}\end{array}$$ pairing $n$ with $-n$ and using $e^{2\pi int}+e^{-2\pi int}=2\cos (2\pi nt)$, then substituting $n=2k+1$.

Rayleigh’s identity (Parseval’s Theorem)

Let $f$ be periodic of period $T$ with Fourier series $$\sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi int}{T}}$$

Assume $f$ has only a finite number of jump discontinuities on the interval $[0,T]$. Then $$\sum _{n=-\infty }^{\infty }|c_n{|}^2=\frac{1}{T}{\int }_0^T|f(t){|}^{2}dt$$

Proof:

$$\begin{array}{rl}|f(t){|}^2=f(t)\overline{f(t)}& =\left(\sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi int}{T}}\right)\left(\sum _{k=-\infty }^{\infty }\overline{c_k}{e}^{-\frac{2\pi ikt}{T}}\right)\\ & =\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{\infty }{c}_n\overline{c_k}{e}^{\frac{2\pi i(n-k)t}{T}}\end{array}$$ Integrate over one period. By the same two-case computation as in Lecture 2, ${\int }_0^T{e}^{2\pi i(n-k)t/T}dt$ equals $T$ when $n=k$ and $0$ otherwise, so only the diagonal survives: $$\Rightarrow {\int }_0^T|f(t){|}^{2}dt=\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{\infty }{c}_n\overline{c_k}{\int }_0^T{e}^{\frac{2\pi i(n-k)t}{T}}dt=T\sum _{n=-\infty }^{\infty }{c}_n\overline{c_n}$$ Divide by $T$ and use $c_n\overline{c_n}=|c_n{|}^2$.

Lecture 4

Example:

Refer back to the square wave diagram. We know that the Fourier series is $$f(t)\sim \sum _{n\mathrm{odd}}\frac{2}{\pi in}{e}^{2\pi int}$$

Using Parseval’s theorem, and noting $|f(t)|=1$ everywhere so the left side is just 1: $${\int }_0^1|f(t){|}^{2}dt=1=\sum _{n\text{ odd}}\frac{4}{{\pi }^2{n}^2}=\frac{8}{{\pi }^2}\sum _{n\text{ odd},n\ge 1}\frac{1}{n^2}$$$$=\frac{8}{{\pi }^2}\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}\Rightarrow \sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}$$ (The 8 is 4 from ${\left|\frac{2}{\pi in}\right|}^2=\frac{4}{{\pi }^2{n}^2}$, doubled because $n$ and $-n$ contribute equally.)

Example:

Refer back to the triangle wave. We know that $$f(t)=\frac{1}{4}+\sum _{n\text{ odd}}\frac{1}{{\pi }^2{n}^2}{e}^{2\pi int}$$ The left-hand side of Parseval is, using evenness, $${\int }_{-\frac{1}{2}}^{\frac{1}{2}}|f(t){|}^{2}dt=2{\int }_0^{1/2}{\left(\frac{1}{2}-t\right)}^{2}dt=2{\frac{-{\left(\frac{1}{2}-t\right)}^3}{3}|}_0^{1/2}=\frac{2}{24}=\frac{1}{12}$$ so $$\frac{1}{12}=\frac{1}{16}+\frac{1}{{\pi }^4}\sum _{n\text{ odd}}\frac{1}{n^4}\Rightarrow \frac{1}{48}=\frac{2}{{\pi }^4}\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^4}$$$$\Rightarrow \sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^4}=\frac{{\pi }^4}{96}.$$ Here $\frac{1}{16}=|c_0{|}^2$, and the $\pm n$ pairing again supplies the factor of 2.

Theorem: (Pointwise convergence of Fourier series)

Let $f$ be a periodic function of period $T$. Suppose $f$ and $f^{\prime }$ are continuous on $[0,T]$ except for a finite number of jump discontinuities. Then the Fourier series converges to $\frac{f(a+)+f(a-)}{2}$ at every number $a$. That is, $$\sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi ina}{T}}=\frac{f(a+)+f(a-)}{2}$$$$f(a+)=\underset{x\to a+}{\lim }f(x),f(a-)=\underset{x\to a-}{\lim }f(x)$$

Example:

Return to the square wave series. Pairing $n$ with $-n$ and using $e^{2\pi int}-e^{-2\pi int}=2i\sin (2\pi nt)$, $$\frac{2}{\pi in}{e}^{2\pi int}+\frac{2}{\pi i(-n)}{e}^{-2\pi int}=\frac{2}{\pi in}\cdot 2i\sin (2\pi nt)=\frac{4}{\pi n}\sin (2\pi nt)$$$$\Rightarrow f(t)\sim \frac{4}{\pi }\sum _{k=0}^{\infty }\frac{1}{2k+1}\sin (2\pi (2k+1)t)$$

At $t=0$, we have $\frac{f(0+)+f(0-)}{2}=0$. The Fourier series returns the same result—since $\sin (0)=0$.

At $t=\frac{1}{4}$, we get $f(t)=1$. In the Fourier series: $$\frac{4}{\pi }\sum _{k=0}^{\infty }\frac{\sin \left(\frac{(2k+1)\pi }{2}\right)}{2k+1}=\frac{4}{\pi }\sum _{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}$$

Hence, we get a nice sum from an alternating series: $$\sum _{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}=\frac{\pi }{4}$$

Example:

Let’s return to the triangle wave again. At $t=0$, we get $f(t)=\frac{1}{2}$. Test in the Fourier series: $$\frac{1}{4}+\frac{1}{{\pi }^2}\sum _{n\text{ odd}}\frac{1}{n^2}=\frac{1}{4}+\frac{2}{{\pi }^2}\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}$$$$\frac{1}{2}=\frac{1}{4}+\frac{2}{{\pi }^2}\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}\Rightarrow \sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}$$

Do the same thing at $t=\frac{1}{2}$. The periodic extension is continuous there with $f\left(\frac{1}{2}\right)=0$, and $e^{\pi in}=(-1{)}^n=-1$ for odd $n$, so $$0=\frac{1}{4}+\sum _{n\text{ odd}}\frac{(-1{)}^n}{{\pi }^2{n}^2}=\frac{1}{4}-\frac{1}{{\pi }^2}\sum _{n\text{ odd}}\frac{1}{n^2}\Rightarrow \sum _{n\text{ odd}}\frac{1}{n^2}=\frac{{\pi }^2}{4}$$

This sum runs over all odd $n\in \mathbb{Z}$, so it is twice the sum over $n\ge 1$: $$\frac{{\pi }^2}{4}=2\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}\Rightarrow \sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}$$ the same result as before.

The geometric Fourier Series.

Motivation:

Suppose $f$ is periodic of period $T=2l$. Then $$\begin{array}{rl}f(x)& \sim \sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{2\pi inx}{T}}=\sum _{n=-\infty }^{\infty }{c}_n{e}^{\frac{\pi inx}{l}}\\ & =\sum _{n=-\infty }^{\infty }{c}_n\cos \left(\frac{n\pi x}{l}\right)+i\sum _{n=-\infty }^{\infty }{c}_n\sin \left(\frac{n\pi x}{l}\right)\end{array}$$

Lecture 5

Trigonometric Fourier Series

Let $f(x)$ be periodic of period $T=2l$. For example, we could have $\cos \left(\frac{k\pi x}{l}\right)$.

Then $f$ can be represented as $$f(x)\sim \frac{a_0}{2}+\sum _{k=1}^{\infty }{a}_k\cos \left(\frac{k\pi x}{l}\right)+\sum _{k=1}^{\infty }{b}_k\sin \left(\frac{k\pi x}{l}\right)$$

Now, how do we compute $a_k,b_j$, the coefficients?

Recall:

Trig identities: $\cos (A+B)=\cos A\cos B-\sin A\sin B$, $\sin (A+B)=\sin A\cos B+\sin B\cos A$. From these, we can derive identities for $\sin A\cos B,\sin B\cos A,\sin A\sin B,\cos A\cos B$. Writing down the $(A-B)$ versions too and adding/subtracting: $$\cos A\cos B=\frac{1}{2}\left[\cos (A-B)+\cos (A+B)\right],\sin A\sin B=\frac{1}{2}\left[\cos (A-B)-\cos (A+B)\right]$$$$\sin A\cos B=\frac{1}{2}\left[\sin (A+B)+\sin (A-B)\right]$$

Return to $$\begin{array}{c}f(x)\sim \frac{a_0}{2}+\sum _{k=1}^{\infty }{a}_k\cos \left(\frac{k\pi x}{l}\right)+\sum _{k=1}^{\infty }{b}_k\sin \left(\frac{k\pi x}{l}\right)\end{array}\text{\hspace{1em}(∗)}$$

Let $m\ge 1$. Multiply $(\ast )$ by $\cos \left(\frac{m\pi x}{l}\right)$ and integrate over $[-l,l]$ (swapping $\sum$ and $\int$ is fine here, since the series converges uniformly under our assumptions): $$\begin{array}{rl}{\int }_{-l}^{l}f(x)\cos \left(\frac{m\pi x}{l}\right)dx& =\frac{a_0}{2}{\int }_{-l}^l\cos \left(\frac{m\pi x}{l}\right)dx\\ & +\sum _{k=1}^{\infty }{a}_k{\int }_{-l}^l\cos \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)dx\\ & +\sum _{k=1}^{\infty }{b}_k{\int }_{-l}^l\sin \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)dx\end{array}$$

We know that $${\int }_{-l}^l\cos \left(\frac{m\pi x}{l}\right)dx={\frac{l}{m\pi }\sin \left(\frac{m\pi x}{l}\right)|}_{-l}^l=\frac{l}{m\pi }\left[\sin (m\pi )-\sin (-m\pi )\right]=0$$$${\int }_{-l}^l\cos \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)dx=\frac{1}{2}\left({\int }_{-l}^l\cos \left(\frac{(k+m)\pi x}{l}\right)dx+{\int }_{-l}^l\cos \left(\frac{(k-m)\pi x}{l}\right)dx\right)$$$$=\frac{1}{2}\left({\frac{l}{(m+k)\pi }\sin \left(\frac{(m+k)\pi x}{l}\right)|}_{-l}^l+{\int }_{-l}^l\cos \left(\frac{(k-m)\pi x}{l}\right)dx\right)$$ The first piece vanishes since $k+m\ge 2$ is a nonzero integer, leaving $$=\frac{1}{2}{\int }_{-l}^l\cos \left(\frac{(k-m)\pi x}{l}\right)dx=\{\begin{array}{ll}{\frac{l}{2\pi (k-m)}\sin \left(\frac{(k-m)\pi x}{l}\right)|}_{-l}^l=0,& k\ne m\\ \frac{1}{2}{\int }_{-l}^{l}1dx=l,& k=m\end{array}$$

Now, return to the sum from before. Only the single term $k=m$ survives: $$\sum _{k=1}^{\infty }{a}_k{\int }_{-l}^l\cos \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)dx=l{a}_m$$

So we have filled in one nasty sum/integral pair, let’s look at the last one: $${\int }_{-l}^l\sin \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)dx=\frac{1}{2}\left({\int }_{-l}^l\sin \left(\frac{(k+m)\pi x}{l}\right)dx+{\int }_{-l}^l\sin \left(\frac{(k-m)\pi x}{l}\right)dx\right)$$$$=\frac{1}{2}\left({\frac{-l}{(k+m)\pi }\cos \left(\frac{(k+m)\pi x}{l}\right)|}_{-l}^l+{\int }_{-l}^l\sin \left(\frac{(k-m)\pi x}{l}\right)dx\right)$$ Since $\cos$ is even, the evaluations at $x=l$ and $x=-l$ agree and the first piece is 0: $$=\frac{1}{2}{\int }_{-l}^l\sin \left(\frac{(k-m)\pi x}{l}\right)dx=\{\begin{array}{ll}{\frac{-l}{2(k-m)\pi }\cos \left(\frac{(k-m)\pi x}{l}\right)|}_{-l}^l=0,& k\ne m\\ \frac{1}{2}{\int }_{-l}^{l}0dx=0,& k=m\end{array}$$

Shortcut: that whole computation was unnecessary. $\sin \left(\frac{k\pi x}{l}\right)\cos \left(\frac{m\pi x}{l}\right)$ is odd in $x$ (odd times even), and we integrate over the symmetric interval $[-l,l]$, so it is 0 immediately.

Now, combine all the sums to obtain $${\int }_{-l}^{l}f(x)\cos \left(\frac{m\pi x}{l}\right)dx=0+l{a}_m+0=l{a}_m$$$$\Rightarrow a_m=\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \left(\frac{m\pi x}{l}\right)dx$$

Now look for $a_0$. Integrate $(\ast )$ itself over $[-l,l]$: $${\int }_{-l}^{l}f(x)dx=\frac{a_0}{2}{\int }_{-l}^{l}dx+\sum _{k=1}^{\infty }{a}_k{\int }_{-l}^l\cos \left(\frac{k\pi x}{l}\right)dx+\sum _{k=1}^{\infty }{b}_k{\int }_{-l}^l\sin \left(\frac{k\pi x}{l}\right)dx$$ The first integral is $2l$; the $\cos$ integrals vanish as computed above; the $\sin$ integrals vanish since $\sin \left(\frac{k\pi x}{l}\right)$ is odd. So $$=l{a}_0+0+0=l{a}_0\Rightarrow a_0=\frac{1}{l}{\int }_{-l}^{l}f(x)dx$$ This is exactly the $m=0$ case of the formula for $a_m$, which is why the constant term of $(\ast )$ is written $\frac{a_0}{2}$ and not $a_0$: it lets one formula cover every $m\ge 0$.

Using the same methods, we can find $b_k$: multiply $(\ast )$ by $\sin \left(\frac{m\pi x}{l}\right)$ instead. We need one more orthogonality relation, $$\begin{array}{rl}{\int }_{-l}^l\sin \left(\frac{k\pi x}{l}\right)\sin \left(\frac{m\pi x}{l}\right)dx& =\frac{1}{2}\left({\int }_{-l}^l\cos \left(\frac{(k-m)\pi x}{l}\right)dx-{\int }_{-l}^l\cos \left(\frac{(k+m)\pi x}{l}\right)dx\right)\\ & =\{\begin{array}{ll}0,& k\ne m\\ l,& k=m\end{array}\end{array}$$ by the identical case analysis. The $\int \cos \left(\frac{k\pi x}{l}\right)\sin \left(\frac{m\pi x}{l}\right)dx$ terms vanish by the oddness shortcut, and ${\int }_{-l}^l\sin \left(\frac{m\pi x}{l}\right)dx=0$ kills the $\frac{a_0}{2}$ term, so $${\int }_{-l}^{l}f(x)\sin \left(\frac{m\pi x}{l}\right)dx=l{b}_m\Rightarrow b_k=\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \left(\frac{k\pi x}{l}\right)dx$$

Example:

Let $f$ be periodic of period $2\pi$, and $$f(x)=\{\begin{array}{ll}0,& -\pi \le x\le 0\\ x,& 0\le x\le \pi \end{array}$$

what is the Fourier series of $f$?

Solution: here $2l=2\pi$, so $l=\pi$ and every $\frac{k\pi x}{l}$ is just $kx$. $$a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)dx=\frac{1}{\pi }{\int }_0^{\pi }xdx=\frac{1}{\pi }\cdot \frac{{\pi }^2}{2}=\frac{\pi }{2}$$ For $a_n$, integrate by parts with $u=x$, $dv=\cos (nx)dx$, $v=\frac{\sin (nx)}{n}$: $$\begin{array}{rl}{a}_n& =\frac{1}{\pi }{\int }_0^{\pi }x\cos (nx)dx=\frac{1}{\pi }\left({\frac{x\sin (nx)}{n}|}_0^{\pi }-\frac{1}{n}{\int }_0^{\pi }\sin (nx)dx\right)\\ & =\frac{1}{\pi }\left(0+{\frac{\cos (nx)}{n^2}|}_0^{\pi }\right)=\frac{1}{\pi n^2}[(-1{)}^n-1]=\{\begin{array}{ll}\frac{-2}{\pi n^2},& n\text{ odd}\\ 0,& n\text{ even}\end{array}\end{array}$$ For $b_n$, take $u=x$, $dv=\sin (nx)dx$, $v=\frac{-\cos (nx)}{n}$: $$\begin{array}{rl}{b}_n& =\frac{1}{\pi }{\int }_0^{\pi }x\sin (nx)dx=\frac{1}{\pi }\left({\frac{-x\cos (nx)}{n}|}_0^{\pi }+\frac{1}{n}{\int }_0^{\pi }\cos (nx)dx\right)\\ & =\frac{1}{\pi }\left(\frac{-\pi (-1{)}^n}{n}+{\frac{\sin (nx)}{n^2}|}_0^{\pi }\right)=\frac{-(-1{)}^n}{n}=\frac{(-1{)}^{n+1}}{n}\end{array}$$ Assembling, with constant term $\frac{a_0}{2}=\frac{\pi }{4}$: $$\Rightarrow f(x)\sim \frac{\pi }{4}-\frac{2}{\pi }\sum _{n\text{ odd},n\ge 1}\frac{\cos (nx)}{n^2}+\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}\sin (nx)$$$$=\frac{\pi }{4}-\frac{2}{\pi }\sum _{k=0}^{\infty }\frac{\cos ((2k+1)x)}{(2k+1{)}^2}+\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}\sin (nx).$$