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

Where * means convolution.

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Erm, $\int_0^xf(x)dx$.... –  anon Sep 15 '11 at 23:34
Also, the convolution is an operation on two functions rather than two function values. I guess it should be $(f \ast g)(-x)$ and $(f \ast g)(2x)$? –  TMM Sep 16 '11 at 0:30
Do you mean: $$ \int_0^x f(t)dt \int_0^xg(s)ds=\int_0^x(fg)(-u)du+\frac12 \int_0^x(fg)(2v)dv$$ –  GEdgar Sep 16 '11 at 0:31
The left-hand side (however you interpret it) is defined for many functions for which convolution doesn't even make sense... –  Hans Lundmark Sep 16 '11 at 4:44
Since OP didn't bother to make a sensible statement out of the question I voted to close as "not a real question". –  t.b. Oct 19 '11 at 8:56

1 Answer 1

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

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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

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