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In the steps of the proofs highlighted below, how does a multiple integral changes in to multiplication of two integral. This is only possible if V is independent of u, but as it turns out V = t - u, so they are not actually independent.

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The substitution leads to a definite integral $$ I = \int\limits_0^\infty e^{-sv} f(v) \, dv $$ which does not depend on $u$.

The prior integral $$ I_0 = \int\limits_u^\infty e^{-s(t-u)} f(t-u) \, dt $$ seems to depend on $u$, but it is not as $I$ shows.

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  • $\begingroup$ But v is clearly dependent on u $\endgroup$ May 2, 2016 at 9:56
  • $\begingroup$ $v$ is derived from $u$, but that $I$ is not dependent on $u$ anymore, $v$ is just an integration variable. $\endgroup$
    – mvw
    May 2, 2016 at 10:07

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