6
$\begingroup$

I consider $H$ and $h$ two non negative functions.

I had hard time to understand this equality

$$ \int_{0}^{t}h(t-s)\left(\int_{0}^{s}H(s-u)udu\right)ds = \int_{0}^{t}u\left(\int_{u}^{t}h(t-s)H(s-u)ds\right)du $$

I think it is just an application of Fubini theorem, however I don’t succeed to get the same expression.

Indeed I get

$$ \int_{0}^{t}h(t-s)\left(\int_{0}^{s}H(s-u)udu\right)ds = \int_{0}^{s}u\left(\int_{0}^{t}h(t-s)H(s-u)ds\right)du $$

I tried some change of variable ( introduce $v = s - u$), it was not successful.

Am I missing something ?

Thank you a lot


Edit : use of the hint given by Bruno.B

\begin{split} = & \int_{0}^{t}h(t-s)\left(\int_{0}^{s}H(s-u)udu\right)ds \\ = & \int_{0}^{\infty}h(t-s)1_{u\leq s\leq t}\left(\int_{0}^{\infty}H(s-u)u1_{0\leq u \leq s\leq t}du\right)ds \\ = & \int_{0}^{\infty}u1_{0\leq u \leq t}\left(\int_{0}^{\infty}H(s-u)h(t-s)1_{u\leq s\leq t}ds\right)du \\ = & \int_{0}^{t}u\left(\int_{u}^{t}h(t-s)H(s-u)ds\right)du \end{split}

Where the second equality follows because we have $0\leq u\leq s\leq t$ but since $s$ varies between $u$ and $t$ we must have that $u$ varies between $0$ and $t$ that is $1_{0\leq u\leq t}$.

$\endgroup$
5
  • 1
    $\begingroup$ Hint: use an indicator function to make the bounds of integration constant, and then use Fubini. $\endgroup$
    – Bruno B
    Dec 4, 2023 at 9:44
  • 1
    $\begingroup$ @BrunoB Thank you a lot for the hint. I added an attempt to prove this using your hint and making clear all my step. Let me know please if it seems correct. $\endgroup$
    – coboy
    Dec 4, 2023 at 10:33
  • $\begingroup$ @BrunoB Also, does it mean that in general when we want to use Fubini it is more practical to work with constant bounds ? Especially in the case where an order linked the two variables of integration ? Thank you. $\endgroup$
    – coboy
    Dec 4, 2023 at 10:37
  • 1
    $\begingroup$ You should use the syntax \begin{split} ... &[thing you want to align with] \\ &[thing] \\ ... \end{split} for longer calculations like this, otherwise it'll just fly off the screen... But yeah, it's not just more practical, it's how you need to use Fubini's theorem. $\endgroup$
    – Bruno B
    Dec 4, 2023 at 10:42
  • $\begingroup$ I was not aware of this command, thank you. $\endgroup$
    – coboy
    Dec 4, 2023 at 11:12

1 Answer 1

2
$\begingroup$

$$ \begin{align}\int_{-\infty }^{\infty } h(t-s)\ \theta (0<s<t) \ \int_{-\infty }^{\infty } \ u \ H(s-u) \ \theta (0<u<s) \, du \, ds \\ = \int_{-\infty }^{\infty } \left(\int_{-\infty }^{\infty }\ u \ h(t-s) \ H(s-u) \ \theta (0<s<t) \ \theta (0<u<s) \, du\right) \, ds \\ = \int_{-\infty }^{\infty } \ u \ \int_{-\infty }^{\infty } \ h(t-s) \ H(s-u) \ \theta (0<u<s<t) \, ds \, du \\ = \int _0^t \ u \ * \ \left(\int _u^t \ h(t-s) \ * \ H(s-u)ds\right)du\end{align}$$

$\endgroup$
1
  • $\begingroup$ Thank you for the answer ! $\endgroup$
    – coboy
    Dec 4, 2023 at 11:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .