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Let f ( t ) = { 1 x ∈ [ − π , π ] 0 otherwise {\displaystyle f(t)={\begin{cases}1&x\in \left[-\pi ,\pi \right]\\0&{\mbox{otherwise}}\end{cases}}}
( f ∗ f ) ( t ) = ∫ − ∞ ∞ f ( τ ) f ( t − τ ) d τ = ∫ t − π t + π f ( τ ) d τ {\displaystyle {\begin{aligned}(f*f)(t)&=\int _{-\infty }^{\infty }f(\tau )\,f(t-\tau )\,\mathrm {d} \tau \\&=\int _{t-\pi }^{t+\pi }f(\tau )\,\mathrm {d} \tau \\\end{aligned}}}
Let G ( τ ) = τ θ ( τ + π ) − τ θ ( τ − π ) {\displaystyle G(\tau )=\tau \,\theta (\tau +\pi )-\tau \,\theta (\tau -\pi )}
Observe that G ′ ( τ ) = f ( τ ) {\displaystyle G'(\tau )=f(\tau )} , so
( f ∗ f ) ( t ) = ∫ t − π t + π G ′ ( τ ) d τ = G ( t + π ) − G ( t − π ) {\displaystyle (f*f)(t)=\int _{t-\pi }^{t+\pi }G'(\tau )\,\mathrm {d} \tau =G(t+\pi )-G(t-\pi )} by the fundamental theorem of calculus
In this case, ( f ∗ f ) ( t ) = ( t + π ) θ ( t + 2 π ) − ( t + π ) θ ( t ) − ( ( t − π ) θ ( t ) − ( t − π ) θ ( t − 2 π ) ) = ( t + π ) θ ( t + 2 π ) + ( t − π ) θ ( t − 2 π ) {\displaystyle (f*f)(t)=(t+\pi )\,\theta (t+2\pi )-(t+\pi )\,\theta (t)-((t-\pi )\,\theta (t)-(t-\pi )\,\theta (t-2\pi ))=(t+\pi )\,\theta (t+2\pi )+(t-\pi )\,\theta (t-2\pi )}
![{\displaystyle {\begin{aligned}{\mathcal {F}}\left[f*g\right]&={\sqrt {2\pi }}{\hat {f}}(\lambda )\,{\hat {g}}(\lambda )\\(f*g)(t)&={\frac {1}{\sqrt {2\pi }}}\,{\mathcal {F}}^{-1}\left[{\hat {f}}(\lambda )\,{\hat {g}}(\lambda )\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3eb82bb6e44484521ae352e18b590b1b17452d89)
![{\displaystyle f(t)={\begin{cases}1&x\in \left[-\pi ,\pi \right]\\0&{\mbox{otherwise}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a39ccc1b0016809c9655c5ee999932fa54410c25)
, so
by the fundamental theorem of calculus
