Is this formula for product of integrals correct?

Question

$$\int_0^x f(x)dx \int_0^xg(x)dx=\int_0^xf(-x)*g(-x)dx+\frac12 \int_0^xf(2x)*g(2x)dx $$

Where * means convolution.

Also, the convolution is an operation on two functions rather than two function values. I guess it should be $(f \ast g)(-x)$ and $(f \ast g)(2x)$?
Do you mean: $$ \int_0^x f(t)dt \int_0^xg(s)ds=\int_0^x(f*g)(-u)du+\frac12 \int_0^x(f*g)(2v)dv$$
The left-hand side (however you interpret it) is defined for many functions for which convolution doesn't even make sense...
Since OP didn't bother to make a sensible statement out of the question I voted to close as "not a real question".

pedja · Answer 1 · 2011-09-19 07:20:11Z

up vote 1 down vote

HINT : try to use the following: integral of convolution rule, substitutions $-x=u$ and $2x=v$ with reversing limits of integration.

answered Sep 19 '11 at 7:20

pedja
6,811427

	And what the result? – Anixx Sep 20 '11 at 13:46
	@Anixx,I just pointed to direction of reasoning,it is up to you whether you will accept suggestion or not... – pedja Sep 20 '11 at 14:33
	@ pedja I derived it from that exactly expression, and I wnd to know whether I am correct. – Anixx Apr 12 '12 at 23:46

1 Answer