Integral of Generating Function

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Theorem

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Then:

\(\ds \int_0^z \map G t \rd t\) \(=\) \(\ds \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k\)
\(\ds \) \(=\) \(\ds a_0 z + \dfrac {a_1 z^2} 2 + \dfrac {a_2 z^3} 3 + \dfrac {a_3 z^4} 4 + \cdots\)


Proof

\(\ds \int_0^z \map G t \rd t\) \(=\) \(\ds \int_0^z \paren {\sum_{k \mathop \ge 0} a_k t^k} \rd t\) Definition of Generating Function
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {\int_0^z a_k t^k \rd t}\)
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\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {\intlimits {a_k \dfrac {t^{k + 1} } {k + 1} } 0 z}\) Primitive of Power
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {a_k \dfrac {z^{k + 1} } {k + 1} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k\) Translation of Index Variable of Summation

$\blacksquare$


Sources

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