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Setup: I need to do a convolution with the function $\cfrac{i}{x}$, and I would like to get rid of the $i$. My functions to be convolved are all real valued.

According to the ever-failable Wikipedia, and I quote:

Associativity with scalar multiplication

$$ a (f*g) = (af)*g $$

for any real (or complex) number $a$.

Is this true for complex numbers? I get why it's true for real numbers, but I'm not sure this is valid for complex ones, as complex integration is not the same as integration over the reals.

Dear forum, can you confirm or deny? And if you confirm, could you point me towards a source, so I can refer to it?

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  • $\begingroup$ Why do you say all your functions are real valued when you're convolvoing with $i/x$? Are your functions defined on the imaginary axis? $\endgroup$ – Harald Hanche-Olsen Jan 31 '14 at 17:48
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Confirm:

$$ f*g(x) = \int_{-\infty}^\infty f(y) g(x-y) \, dy .$$ So $$ a(f*g(x)) = a \int_{-\infty}^\infty f(y) g(x-y) \, dy = \int_{-\infty}^\infty a f(y) g(x-y) \, dy = (af)*g(x) .$$

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Some books define the integral of a complex-valued function $f(x)$ of a real variable $x$ by separating the real and imaginary parts: $$ \begin{align} & \int_{\mathbb R} f(x)\,dx = \int_{\mathbb R} u(x)\,dx + i\int_{\mathbb R} v(x) \, dx \\[10pt] & \text{ where }f(x)=u(x)+iv(x)\text{ and } u,v\text{ are real-valued functions.} \end{align} $$

\begin{align} \text{So }\int_{\mathbb R} if(x)\,dx & = \int_{\mathbb R} i(u(x)+iv(x))\,dx \\[8pt] & = \int_{\mathbb R} (-v(x) + iu(x)) \,dx \\[8pt] & = \int_{\mathbb R} -v(x)\,dx +i\int_{\mathbb R}u(x)\,dx \tag 1 \\[8pt] & = i\left(\int_{\mathbb R} u(x)\,dx + i\int_{\mathbb R} v(x) \, dx \right) \tag 2 \\[8pt] & = i\int_{\mathbb R} f(x)\,dx. \end{align} In $(1)$ and $(2)$, only real-valued functions are integrated, and that's a concept that's already defined before this.

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