It is possible to write Fourier transform as:

$$g(\omega)=\frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}f(x)e^{i\omega x}\,dx =\frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}f(x)(\cos\omega x+i\sin\omega x)\,dx.$$

Now, considering an odd function, $f(x) = -f(-x)$,

we can write the Fourier transform as: $$g(\omega)=\frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}f(x)\cos\omega x \, dx+ \frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}f(x) i\sin\omega x\,dx.$$

As $f(x)$ is odd, and the product of an odd function with an even function is odd, then we have that $f(x) \cos \omega x$ is odd, and therefore:

$$ \int^{+\infty}_{-\infty}f(x)\cos\omega x\, dx=0.$$

So we have that the Fourier transform is:

\begin{align*} g(\omega )&=\frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}f(x)i\sin\omega x\, dx\\ &= \frac{2}{\sqrt{2 \pi}}\int^{+\infty}_{0}f(x)i\sin\omega x\, dx\\ &= \sqrt{\frac{2}{\pi}} \int^{+\infty}_{0}f(x)i\sin\omega x\, dx. \end{align*}

The change of the limits and the multiplication of the integral by $2$ is possible since $f(x) \sin\omega x$ is an even function, since both are odd functions.

My problem is that I have seen in some books that the sine transform is: $$ \sqrt{\frac{2}{\pi}}\int^{+\infty}_{0}f(x)\sin\omega x\, dx.$$

My question is: What happens to $i$? Does it disappear?


1 Answer 1


The $i$ doesn't just disappear. What you've done is found a complex vector solution to this problem in some funky function space. What that means is when you're dealing with complex number multiples of functions, you get a real part, and an imaginary part, and these represent directions, so to speak, in a complicated vector space, and you just need to choose the appropriate direction. So $i$ is a direction in this weird viewpoint, not a number, and that's why it disappears when you have finished finding your solution in the complex viewpoint.


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