How to identify quick ways of evaluating integrals of complex exponentials?

Question

For example, on the interval $-\pi\le x\le \pi$ consider $$\int_{x=-\pi}^{\pi}\Big(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\Big)dx$$

One answer states that:

$$\int_{x=-\pi}^{\pi}\Big(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\Big)dx=0\tag{1}$$ The equality follows as the complex exponentials are integrated over $3$ and $1$ full period, respectively, and thus average out to zero.

I'm confused by what the author means by 'integrated over $3$ full periods'. It was my understanding that one full period is $2\pi$. So $3$ full periods must be $6\pi$ so the limits should be $-3\pi\le x\le 3\pi$.

Could someone please explain how I can effectively 'see' that this integral is zero, without going through the calculation?

Using Euler's formula I write $$e^{3ix}=\cos(3x)+i\sin(3x)$$ in order to better understand what the author is talking about, but I'm still confused.

Does anyone have any idea what the author means by this part:

and thus average out to zero

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After an online search I found that another answer gave the following reasoning from this website:

For an odd function $$f(x)=-f(-x):\int_a^af(x)dx=0$$

Now I have no idea whatsoever what this is supposed to mean.

If someone could shed some light on why equation $(1)$ is zero it would be greatly appreciated.

Thank you.

Answer 1

Your internet research brought you on the right track. Yet the formula is somewhat different.

For an odd, integrable function $f:[-a,a]\rightarrow\mathbb{R}$ the following is valid \begin{align*} \int_{-a}^a f(x)\,dx = 0 \end{align*}

Recall a function $f$ is odd iff \begin{align*} f(x)=-f(-x)\tag{1} \end{align*}

On the other hand we know the integration rules \begin{align*} \int_a^b f(x)\,dx&=-\int_b^a f(x)\,dx\tag{2}\\ \int_{a}^b f(x)\,dx&=\int_{a}^c f(x)\,dx+\int_{c}^bf(x)\,dx\tag{3}\\ \int_{a}^b f(x)\,dx&=-\int_{-a}^{-b}f(-x)\,dx\tag{4} \end{align*}

The last rule (4) is due to substitution \begin{align*} x&\rightarrow -x\\ dx&\rightarrow -dx \end{align*}

Therefore we obtain for odd functions $f:[-a,a]\rightarrow\mathbb{R}$

\begin{align*} \int_{-a}^a f(x)\,dx&=\int_{-a}^0 f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (3)\\ &=-\int_{0}^{-a}f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (2)\\ &=\int_{0}^{a}f(-x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (4)\\ &=-\int_{0}^{a}f(x)\,dx + \int_{0}^a f(x)\,dx\qquad\qquad&\longrightarrow (1)\\ &=0 \end{align*}

Since $e^{3ix}-e^{-3ix}$ is odd as well as $e^{ix}-e^{-ix}$ we can see at a glance \begin{align*} \int_{-\pi}^\pi\left(\left(e^{3ix}-e^{-3ix}\right)-\left(e^{ix}-e^{-ix}\right)\right)=0 \end{align*}

$$ $$

Regarding integration over $3$ full periods: Let's consider \begin{align*} &e^{ix}\qquad\qquad &-6\pi\leq x\leq 6\pi\\ &e^{3ix}\qquad\qquad &-2\pi \leq x\leq 2\pi \end{align*}

• In the first case we have three full periods by increasing the length of the interval by a factor three from $2\pi$ to $6\pi$.

• In the second case we have three full periods by increasing the speed of rotation by a factor three from $e^{ix}$ to $e^{3ix}$.

In both cases we have three full periods.

Answer 2

You were on the right track using Euler's formula. Note that $\cos{3x}$ and $\sin{3x}$ both have period $\frac{2\pi}{3}$, so 3 full periods is indeed $2\pi$.

As for the second part, notice $f(x) = e^{aix} - e^{-aix}$ is odd since $-f(-x) = -(e^{ai(-x)} - e^{-ai(-x)}) = -(e^{-aix} - e^{aix}) = f(x)$. In fact, using Euler's formula as you gave shows $f(x) = 2i \sin{ax}$, from which you can deduce that $f$ is odd directly.

So we have the sum of two odd functions, which is also an odd function. Does this all make sense?

EDIT:

Your integral now comes to this.

$$ \int_{-\pi}^{\pi} 2i \sin{3x} - 2i \sin{x} \: dx = \int_{-\pi}^{\pi} 2i \sin{3x} \: dx - \int_{-\pi}^{\pi} 2i \sin{x} \: dx$$

Both of those functions are odd, so each individual integral is $0$.

Here's the other way to think about it. Consider the integral we just looked at, but pull out the constant $2i$ (just for clarity).

$$ 2i \int_{-\pi}^{\pi} \sin{3x} \: dx - 2i \int_{-\pi}^{\pi} \sin{x} \: dx $$

Think of the interpretation of the integral as the (signed) area under the curve. The overall area under the curve, for one full period of the $\sin$ function is $0$. Since the two intgrands are making 3 and 1 full periods respectively, the values of the integrals are $3*0 = 0$ and $0$ respectively.