One Cycle of a Trigonometric Function

Integral of a Complex Exponential

Consider the integral

$ \int_0^{2 \pi} d\theta e^{i \theta} $.

Upper Limit of $2 \pi$ Radians

The upper limit is the number, $2 \pi$. For a circle with radius $r=1$, the circumference of the circle is $2 \pi r = 2 \pi$. Even though $2 \pi$ is a number, it is convenient to construct a unit called the radian. In particular, for a unit circle centered at the origin of a Cartesian coordinate system, 0 radians corresponds to zero angular displacement with respect to the positive $x$-axis, and $+2 \pi $ radians is the counterclockwise angular displacement from zero radians back to the starting position on the positive $x$-axis. Therefore, there are $2 \pi$ radians in one counterclockwise revolution  about a unit circle. Explicitly, this relationship is:

$2 \pi$ radians $ $~ $ 1$ revolution.

This relation allows for converting between the units of radians and revolution. It is easy to remember that there are $2 \pi$ radians in one revolution because the circumference of a unit circle is $2 \pi$ radians. But, regardless of the radius of the circular path, there are always $2 \pi $ radians in one revolution. Something to be careful about is that a revolution or a fraction of a revolution refers to the trajectory of a point along a path whereas the angular unit of a radian refers to the angular displacement corresponding to a particular trajectory along a circular path. Since these units are describing fundamentally different geometrical aspects, a tilde is put  equal sign, to avoid confusion.

Using Euler’s Formula

Now back to the integral. Using without proof Euler’s formula of $e^{i \theta} = \cos\theta + i \sin\theta$, it is possible to rewrite the integral as

$ \int_0^{2 \pi} d\theta \cos\theta + i \int_0^{2 \pi} d\theta \sin\theta$ .

Repetition of the Cosine and Sine Functions

It is a well-known fact that the cosine function $cos\theta$ repeats every $2\pi$ radians, regardless of the starting point. Similarly, the sine function $\sin\theta$ repeats every $2\pi$ radians, regardless of the starting angle or initial angle $\theta_i$. Mathematically, one has

$\sin(\theta_i + 2 \pi) = \sin(\theta_i)$.

However, the angles are different: $\theta_i + 2 \pi \ne \theta_i$. Even though the value of the sine function is the same on both sides of the previous equation, the argument of the sine function on the left is different from the argument of the sine function on the right. Similarly,

$\cos(\theta_i + 2 \pi) = \cos(\theta_i)$.

Using the Unit Circle to Imagine the Plots of the Sine and Cosine Functions

To evaluate the integral $ \int_0^{2 \pi} d\theta \cos\theta + i \int_0^{2 \pi} d\theta \sin\theta$, it suffices to plot the sine and cosine functions versus the angle $\theta$. In the first integral as well as the second integral, the angle $\theta$ ranges from $0$ to $2 \pi$ radians. The unit of radians is implied because $\theta$ is an angle instead of an arbitrary number–the sine and cosine functions require angles as arguments, not dimensionless numbers.

relation to cartesian

The sine function $\sin(\theta)$ can be thought of as the $y$-value on the unit circle for a certain $\theta$. Similarly, $\cos(\theta)$ can be thought of as the $x$-value on the unit circle for a certain $\theta$. This is an extension or generalization of the geometric definitions of the sine and cosine functions based on triangles. This extension is associated with identities for the sine and cosine functions, but these identities are omitted for the purposes of keeping this post a reasonable length.

Sine Function

If one plots the sine function from $0$ radians to $2 \pi$ radians, what one is really doing is plotting the $y$-value on the unit circle as $\theta$ ranges from $0$ radians to $2 \pi$ radians. One can imagine the $y$-value starting at 0, reaching 1, decreasing to 0, decreasing to $-1$, and returning to $y=0$ on the positive $x$-axis. This behavior of the $y$-coordinate constitutes one rudimentary cycle of the sine function with respect to an angular coordinate, since the behavior of the sine function repeats as $\theta$ changes, for example, from $2 \pi$ radians to $4 \pi$ radians. If these $y$-values are plotted versus the angle $\theta$, the net area from $0$ radians to $2 \pi$ radians is zero, with positive area defined as area above the $\theta$-axis and negative area defined as area below the $\theta$ axis,

Cosine Function

A similar situation occurs for the cosine function. If one plots the cosine function from $0$ radians to $2 \pi$ radians, what one is really doing is plotting the $x$-value on the unit circle as $\theta$ ranges from $0$ radians to $2 \pi$ radians. As the value of $\theta$ increases from $0$ to $2 \pi$, one can imagine the $x$-value starting at 1, becoming 0, reaching -1, becoming 0, and returning to the starting value of 1. If these $x$-values are plotting versus the angle $\theta$, the net area is zero. Since an integral is supposed to give the net area between a curve and an axis, it can be concluded that

Evaluating the Integral of a Complex Exponential from $0$ Radians to $2 \pi$ Radians

$ \int_0^{2 \pi} d\theta \cos\theta + i \int_0^{2 \pi} d\theta \sin\theta = 0$.

Inserting Euler’s formula,

$ \int_0^{2 \pi} d\theta e^{i \theta}  = 0.$

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