$\begingroup$

Consider $$\int^{\frac{1}{2}}_{-\frac{1}{2} } e^{i2\pi f} \,df = \int^{\frac{1}{2} }_{-\frac{1}{2} } \cos(2 \pi f)\, df$$

Why do we only look at the real part? What about the imaginary part $i\sin(2\pi f)$?

What is the reasoning behind this?

$\endgroup$ 10

3 Answers

$\begingroup$

$$\int\limits_{-1/2}^{1/2}e^{2\pi ix}\,dx=\left.\frac{1}{2\pi i}e^{2\pi ix}\right|_{-1/2}^{1/2}=\frac{1}{2\pi i}\left(e^{\pi i}-e^{-\pi i}\right)= \frac{1}{\pi}\sin\pi \\ \int\limits_{-1/2}^{1/2}\cos (2\pi x)\,dx=\frac{1}{2\pi}\sin(2\pi x)\Bigg|_{-1/2}^{1/2}=\frac{1}{2\pi}(\sin\pi-\sin(-\pi))=\frac{1}{\pi}\sin\pi $$

$\endgroup$ 1 $\begingroup$

Simply put, the reason is that it's an integral that is symmetric around $f=0$, and because sin is an odd function, the integral of the sin component must be zero.

$\endgroup$ 2 $\begingroup$

As the sine function is odd, the integral over a symmetric range is null.

In this particular case, as you are integrating over a whole period, you can also trade the cosine for a sine, and

$$\int^{\frac{1}{2} }_{-\frac{1}{2} } \cos(2 \pi f)\,df=\int^{\frac{1}{2} }_{-\frac{1}{2} } \sin(2 \pi f)\,df=0\ !$$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy