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Let $f$ be a function that is 0 everywhere except in the interval $(0,1)$. Let $$ P_n(x)=\int_{-1}^1 f(x+t) Q_n(t) dt $$ and $0\leq x\leq 1$.

Can someone explain me this substitution, somehow i don't get it: $$ P_n(x)=\int_{-x}^{1-x} f(x+t) Q_n(t) dt = \int_{0}^{1} f(t) Q_n(t-x) dt $$ So I assume the first step is just restricting it because it is zero outside anyway. But the rest is unclear.

calculatormathematical
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  • The second step is a substitution $u = x + t$. You get $\int_{-x}^{1-x} f(x + t)Q_n(t)dt = \int_0^1 f(u) Q_n(u - x) du$. – Tob Ernack Mar 25 '23 at 16:22

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just take $u=x+t$ then the bounds will change to $1-x+x=1$ and $-x+x=0$. finally just replace $u$ by $t$.

Milan Rashed
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