# Double Integrals

1. Nov 5, 2013

### muzialis

Hi all,

I have a question regarding certain double integrals.
Assume the function $$l(t)$$ is given as well as the function $$K(t)$$, defined only for positive argument. Also the definition $$n(t) = \int_{-\infty}^{t} K(t-\tau) l(\tau) \mathrm{d}\tau$$ is given. if I wish to compute $$P = \int_{-\infty}^{0} n(s) l(s) \mathrm{d}s$$ I obtain $$P = \int_{-\infty}^{0} \int_{-\infty}^{s} K(s-\tau) l(\tau) \mathrm{d}\tau l(s) \mathrm{d}s$$. So far, so good.
If I wish to have constant limits of integration I wrote down
$$P =0.5 * \int_{-\infty}^{0} \int_{-\infty}^{0} K1(s-\tau) l(\tau) \mathrm{d}\tau l(s) \mathrm{d}s$$, where K1(t) = K(t) if t is postive, K(-t) otherwise. I based my "guess" on the fact I am integrating in the region from ngative infinity to zero (on each axis), instead of integrating on an infinite "triangle". Is this correct? If not, can somebody please correct me?
What I particularly fail to grasp is that, if the previous expression were correct, one would expect that the same result would be obtained without resorting to the function K1, extended for negative arguments, but using the original K function in
$$P =\int_{-\infty}^{0} \int_{-\infty}^{0} K(s-\tau) l(\tau) \mathrm{d}\tau l(s) \mathrm{d}s$$, but I know the latter is wrong. Thank you very much for your help, always the most appreciated.