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 $1_{0\leq u\leq t}$.
$$ \begin{align}\int_{-\infty }^{\infty } h(t-s)\ \theta (0<s<t) \ \int_{-\infty }^{\infty } \ u \ H(s-u) \ \theta (0<u<s) \, du \, ds \\ = \int_{-\infty }^{\infty } \left(\int_{-\infty }^{\infty }\ u \ h(t-s) \ H(s-u) \ \theta (0<s<t) \ \theta (0<u<s) \, du\right) \, ds \\ = \int_{-\infty }^{\infty } \ u \ \int_{-\infty }^{\infty } \ h(t-s) \ H(s-u) \ \theta (0<u<s<t) \, ds \, du \\ = \int _0^t \ u \ * \ \left(\int _u^t \ h(t-s) \ * \ H(s-u)ds\right)du\end{align}$$