Double Integrals: Computing P with Constant Limits

[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.

 

[jvicens]

Hi there,

Thank you for your question. Your initial approach to computing P looks correct. However, when you use the function K1, you are essentially changing the region of integration from an infinite triangle to a finite rectangle. This means that your result will not be equivalent to the original expression, where the region of integration is an infinite triangle.

To see this, consider the case where K(t) = 1 for all positive t. In the original expression, you are integrating over the entire region below the line y=x, which has an infinite area. However, when you use K1, you are only integrating over the finite rectangle with corners at (0,0) and (0,-1). This will result in a different value for P.

In general, when dealing with infinite limits of integration, it is important to consider the region of integration carefully. In this case, using the original K function is the correct approach, as it takes into account the entire region of integration. I hope this helps clarify things for you. Let me know if you have any further questions.