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I have as a part of a problem the following integrals

$\frac{1}{T}\int_{t-T}^t f(\tau) \, \mathrm{d}\tau = \frac{1}{T}\int_{0}^T f(t-u) \, \mathrm{d}u$, where $T > 0$ and $u \in [0,T]$

I cannot find the right substition. Can someone provide a hint?

Yes
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Carlos
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1 Answers1

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We have $$ \int_{0}^{T}f(t-u)du \underset{\tau := t-u}{=} -\int_{t}^{t-T}f(\tau)d\tau = \int_{t-T}^{t}f(\tau)d\tau. $$

Yes
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  • It doesen't matter if I write $0$ or $t$ for the lower bound, respectively, $T$ or $t-T$ for the upper bound? – Carlos Nov 17 '15 at 10:22
  • @Carlos It matters in the sense that it makes the integral have different signs! – Yes Nov 17 '15 at 10:24
  • @ Gudson Chou The rule that after switching the sign also the limits have to be switched I understand. However, the bounds from the left integral are $0$ and $T$. In the middle integral they are $t$ and $t-T$. – Carlos Nov 17 '15 at 10:27
  • @Carlos Ah, try to prove this $0 \leq u \leq T$ iff $t \leq \tau \leq t-T$. – Yes Nov 17 '15 at 10:29
  • Ok I got it: Since for the bounds of the left integral it holds $0 \leq u \leq T$ it holds for the bounds of the right integral, after substitution with $\tau = t-u$: $0\leq t-\tau\leq T$ $\Rightarrow$ $-t\leq-\tau\leq T-t$ $\Rightarrow$ $t \geq \tau \geq t-T$ – Carlos Nov 17 '15 at 10:47