# Is Fubini’s theorem behind this equality?

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 $$1_{0\leq u\leq t}$$.

• Hint: use an indicator function to make the bounds of integration constant, and then use Fubini. Dec 4, 2023 at 9:44
• @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. Dec 4, 2023 at 10:33
• @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. Dec 4, 2023 at 10:37
• 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. Dec 4, 2023 at 10:42
• I was not aware of this command, thank you. Dec 4, 2023 at 11:12

## 1 Answer

\begin{align}\int_{-\infty }^{\infty } h(t-s)\ \theta (0

• Thank you for the answer ! Dec 4, 2023 at 11:24