Does substitution need be bijective when the integrated function is only piecewise continuous ? I guess the formula could be wrong if not, but cannot find a counter-example.
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For Riemann integral in one dimension, the substitution need not be bijective. Specifically: if $u'$ is differentiable with continuous derivative, and $f$ is Riemann integrable, then
$$
\int_a^b f(x)\,dx = \int_c^d f(u(t))u'(t)\,dt
$$
provided that $u(c)=a$ and $u(d)=b$.
Indeed, if $f=F'$ then $f(u(t))u'(t)$ is just the derivative of $F(u(t))$, so integration on the right yields $F(u(d))-F(u(c))$, which is the same as $F(b)-F(a)$ on the left.
The technical reason is that we are really integrating a differential form here, which can be pulled back by any smooth map. In other contexts, when integrating a function on a measure space, the change of variable works differently: it has the absolute value of derivative, and requires a bijective map.
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Excuse me, but how can you set f=F' if f is just piecewise continuous ? In this case, if not continuous, it has no antiderivative. – kikoo lol Jun 20 '15 at 12:40