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I am reading Lorenzo's Bergomi's book Stochastic Volatility Modeling, and I have come to this passage.enter image description here

I just would like to understand the derivation between the first and the second equality. I guess I just have to "correctly" re-express the integral and then use Fubini's theorem so as to obtain an integral with just a $dt$/$du$/whatever term that turns into the $T - \tau$ term, but I can't figure how to do the right change of variables as $t - u$ is a function of $t$ and $u$. Any idea over there?

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Note that the function $f$ only depends on $|t-u|$, meaning it is actually symmetric: $f(x)=f(-x)$. Doing the change of variable $\tau:=t-u$: $$\begin{align} \int_0^Tdu\int_0^Tf(t-u)dt &=\int_0^Tdu\int_{-u}^{T-u}f(\tau)d\tau \\ &=\int_0^Tdu\left(\int_0^{T-u}f(\tau)d\tau+\int_0^uf(\tau)d\tau\right) \\ &=\int_0^T{du \left(\int_0^T{f(\tau) \textbf{1}_{\tau \leq T - u} d\tau} + \int_0^T{f(\tau) \textbf{1}_{\tau \leq u}d\tau}\right)} \\ &=\int_0^T{f(\tau)d\tau \left(\int_0^T{ \textbf{1}_{u \leq T - \tau} du} + \int_0^T{\textbf{1}_{u \geq \tau}du}\right)} \\ &=\int_0^Tf(\tau)d\tau\left(\int_0^{T-\tau}du+\int_\tau^Tdu\right) \\ &=2\int_0^T(T-\tau)f(\tau)d\tau \end{align}$$ For the second equality, note that $0\leq u\leq T$ hence $-u\leq0$ and $0\leq T-u$.

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    $\begingroup$ @siou0107 the indicators are a nice touch, thanks, even clearer now. $\endgroup$ Jun 10 '20 at 12:38

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