Integrals for proof of Fourier series

Contents

Sin(mt)

\displaystyle\int_0^{2\pi}\sin mt\,dt = 0

Proof

We need to find the antiderivative of \int\sin mt\,dt. We know that

\frac{d}{dt} \left(\cos mt\right) = m(-\sin mt) = -m\sin mt

So we don’t change the value of our original integral, we need to find a creative way of multiplying that expression by 1. The way to do it is to multiply our original expression by \left(-\frac{1}{m}\right)(-m). We get:

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt\,dt &=& \left(-\frac{1}{m}\right)\displaystyle\int_0^{2\pi}-m\sin mt\,dt \\ &=& \eval{\left(-\frac{1}{m}\right)(\cos mt)}_0^{2\pi}\\ &=&\left(-\frac{1}{m}\right)\left( \cos (m2\pi) - \cos (m\cdot 0) \right)\\ &=& \left(-\frac{1}{m}\right)\left( 1 - 1 \right)\\ &=& \left(-\frac{1}{m}\right)\cdot 0\\ &=& 0 \end{array}

Cos(mt)

\displaystyle\int_0^{2\pi}\cos mt\,dt = 0

Proof

We need to find the antiderivative of \int\sin mt\,dt. We know that

\frac{d}{dt} \left(\sin mt\right) = m(\cos mt)

So we don’t change the value of our original integral, we need to find a creative way of multiplying that expression by 1. The way to do it is to multiply our original expression by \left(\frac{1}{m}\right)(m). We get:

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\cos mt\,dt &=& \left(\frac{1}{m}\right)\displaystyle\int_0^{2\pi}m\cos mt\,dt \\ &=& \eval{\left(\frac{1}{m}\right)(\sin mt)}_0^{2\pi}\\ &=&\left(\frac{1}{m}\right)\left( \sin (m2\pi) - \sin (m\cdot 0) \right)\\ &=& \left(\frac{1}{m}\right)\left( 0 - 0 \right)\\ &=& \left(\frac{1}{m}\right)\cdot 0\\ &=& 0 \end{array}

Sin(mt)Cos(nt)

\displaystyle\int_0^{2\pi}\sin mt \cos nt\,dt = 0

Proof

We need to make use of the product-to-sum trigonometric identity

2\sin mt \cos nt\=\sin[(m+n)t] + \sin[(m-n)t]

This stems from 2 sum-to-product trigonometric identities, the proofs of which can be found here:

\begin{array}{cccccc} \, & \sin(m+n) & = & \sin m \cos n & + & \cos m \sin n\\ + & \sin(m-n) & = & \sin m \cos n & - & \cos m \sin n\\ \hline \, & \sin(m+n)+\sin(m-n) & = & 2\sin m \cos n &+& 0 \end{array}

Divide both sides of this last equation by 2. We get

\sin m \cos n = \frac12\left[ \sin(m+n)+\sin(m-n)\right]

So,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt \cos nt\,dt &=& \displaystyle\int_0^{2\pi} \frac12\left[ \sin(m+n)+\sin(m-n)\right]\,dt\\ &=& \frac12\displaystyle\int_0^{2\pi}\sin(m+n)\,dt + \frac12\displaystyle\int_0^{2\pi}\sin(m-n)\,dt \end{array}

Since m and n are both integers, then m+n and m-n are integers. But we proved above that

\displaystyle\int_0^{2\pi} \sin [(\text{integer})t] = 0

Therefore,

  • \displaystyle\int_0^{2\pi}\sin(m+n)\,dt=0
  • \displaystyle\int_0^{2\pi}\sin(m-n)\,dt=0

Thus,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt \cos nt\,dt &=&\frac12\displaystyle\int_0^{2\pi}\sin(m+n)\,dt + \frac12\displaystyle\int_0^{2\pi}\sin(m-n)\,dt\\ &=&\frac12\cdot 0 + \frac12\cdot 0\\ &=& 0 \end{array}

Sin(nt) Sin(mt)

We have 2 things to prove here:

  • \displaystyle\int_0^{2\pi}\sin mt \sin nt\,dt = 0 if integers m\neq n or m\neq -n
  • \displaystyle\int_0^{2\pi}\sin mt \sin nt\,dt = \pi if m=n

Proof

To prove the above statements, we have to invoke another product-to-sum trigonometric identity:

2\sin mt \sin nt\=\cos[(m-n)t] - \cos[(m+n)t]

This stems from 2 sum-to-product trigonometric identities, the proofs of which can be found here:

\begin{array}{cccccc} \, & \cos(m-n) & = & \cos m \cos n & + & \sin m \sin n\\ - & \cos(m+n) & = & \cos m \cos n & - & \sin m \sin n\\ \hline \, & \cos(m-n)-\cos(m+n) & = & 0 &+& 2\sin m \sin n \end{array}

Divide both sides of this last equation by 2. We get

\sin m \sin n = \frac12\left[ \cos(m-n)-\cos(m+n)\right]

So,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt \sin nt\,dt &=& \displaystyle\int_0^{2\pi} \frac12\left[ \cos\left((m-n)t)-\cos[(m+n)t\right)\right]\,dt\\ &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt - \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt \end{array}

There are 2 possibilities here. First,

If \lvert m \rvert \neq \lvert n \rvert then both m-n and m+n are integers. But we previously proved that

\displaystyle\int_0^{2\pi} \cos [(\text{integer})t] = 0

Therefore,

  • \displaystyle\int_0^{2\pi}\cos(m-n)\,dt=0
  • \displaystyle\int_0^{2\pi}\cos(m+n)\,dt=0

Thus,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt \sin nt\,dt &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt - \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt\\ &=&\frac12\cdot 0 - \frac12\cdot 0\\ &=& 0 \end{array}

In the second scenario where m=n, m+n will still be an integer so the righthand-most term, \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt will be 0 and can be ignored.

However, if m=n, m-n=0. That means that

  • \cos((m-n)t)=\cos((0)t)=\cos 0 = 1

Then

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\sin mt \sin nt\,dt &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt - \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt\\ &=&\frac12\displaystyle\int_0^{2\pi}\cos\left((0)t\right)\,dt - 0\\ &=&\frac12\displaystyle\int_0^{2\pi} 1\,dt\\ &=&\eval{\frac12 t}_0^{2\pi}\\ &=& \frac12 \cdot 2\pi - \frac12 \cdot 0\\ &=& \pi \end{array}

Cos(mt)Cos(nt)

Again, in this case, we have 2 things to prove:

  • \displaystyle\int_0^{2\pi}\cos mt \cos nt\,dt = 0 if integers m\neq n or m\neq -n
  • \displaystyle\int_0^{2\pi}\cos mt \cos nt\,dt = \pi if m=n

Proof

To prove the above statements, we have to invoke yet another product-to-sum trigonometric identity:

2\cos m \cos n=\cos(m-n) + \cos(m+n)

This stems from 2 sum-to-product trigonometric identities, the proofs of which can be found here:

\begin{array}{cccccc} \, & \cos(m-n) & = & \cos m \cos n & + & \sin m \sin n\\ + & \cos(m+n) & = & \cos m \cos n & - & \sin m \sin n\\ \hline \, & \cos(m-n)+\cos(m+n) & = & 2\cos m \cos n &+& 0 \end{array}

Divide both sides of this last equation by 2. We get

\cos m \cos n = \frac12\left[ \cos(m-n)+\cos(m+n)\right]

So,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\cos mt \cos nt\,dt &=& \displaystyle\int_0^{2\pi} \frac12\left[ \cos\left((m-n)t)+\cos[(m+n)t\right)\right]\,dt\\ &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt + \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt \end{array}

There are 2 possibilities to consider. First,

If \lvert m \rvert \neq \lvert n \rvert then both m-n and m+n are integers. But we previously proved that

\displaystyle\int_0^{2\pi} \cos [(\text{integer})t] = 0

Therefore,

  • \displaystyle\int_0^{2\pi}\cos(m-n)\,dt=0
  • \displaystyle\int_0^{2\pi}\cos(m+n)\,dt=0

Thus,

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\cos mt \cos nt\,dt &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt + \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt\\ &=&\frac12\cdot 0 + \frac12\cdot 0\\ &=& 0 \end{array}

In the second scenario where m=n, m+n will still be an integer so the righthand-most term, \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt will be 0 and can be ignored.

However, if m=n, m-n=0. That means that

  • \cos((m-n)t)=\cos((0)t)=\cos 0 = 1

Then

\begin{array}{rcl} \displaystyle\int_0^{2\pi}\cos mt \cos nt\,dt &=& \frac12\displaystyle\int_0^{2\pi}\cos\left((m-n)t\right)\,dt + \frac12\displaystyle\int_0^{2\pi}\cos\left((m+n)t\right)\,dt\\ &=&\frac12\displaystyle\int_0^{2\pi}\cos\left((0)t\right)\,dt + 0\\ &=&\frac12\displaystyle\int_0^{2\pi} 1\,dt\\ &=&\eval{\frac12 t}_0^{2\pi}\\ &=& \frac12 \cdot 2\pi - \frac12 \cdot 0\\ &=& \pi \end{array}