Is an odd/even function multiplied by $i$ still an odd/ even function

even-and-odd-functions

Let's say I have a function that I want to integrate an even function multiplied by $i\sin(x)$ between $-1$ and $1$ and the function is $1-|x|$ then does this integral become zero because integrating an odd function over a symmetric domain is zero or is $i\sin(x)$ not odd anymore when multiplied by $i$.

Best Answer

Multiplying an odd or even function by a constant doesn't change the parity of the function.

The symbol $i$ sometimes represents the imaginary unit, which is a number, a constant.

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[Math] Area under a curve of an odd function from negative infinity to positive infinity

For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $\int_{-\infty}^0 x^3\,dx$ is $-\infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$\lim_{N\to\infty}\int_{-N}^N x^3\,dx=\lim_{N\to\infty}0=0,$$ and in this sense, the limit will always give $0$.

Even/odd signals and their FFT

My code, as a response to your comment:

from numpy.fft import *
import numpy as np

t = np.linspace(-10, 10, 2**11) #2^11 points on [-10, 10]

#making the signal: 5 on (-7.5, -2.5), -5 on (2.5, 7.5) and 0 otherwise
x = np.zeros(len(t))
x[((t < -2.5) & (t > -7.5))] = 5
x[((t > 2.5) & (t < 7.5))] = -5

f = rfft(x) #the rfft
freal = f.real #real part
fimag = f.imag #imaginary part

fq = rfftfreq(len(t), t[1]-t[0]) #the frequencies

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[Math] Area under a curve of an odd function from negative infinity to positive infinity

Even/odd signals and their FFT

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