Definite Integral from 0 to 1 of Power of u over 1 + Power of u

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Theorem

\(\ds \int_0^1 \dfrac {u^{a - 1} \rd u} {1 + u^d}\) \(=\) \(\ds \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {a + j d}\)
\(\ds \) \(=\) \(\ds \frac 1 a - \frac 1 {a + d} + \frac 1 {a + 2 d} - \frac 1 {a + 3 d} + \cdots\)

where $a, d > 0$.


Proof

\(\ds \int_0^1 \frac {u^{a - 1} \rd u} {1 + u^d}\) \(=\) \(\ds \int_0^1 \frac {u^{a - 1} \rd u} {1 - \paren {-u^d} }\)
\(\ds \) \(=\) \(\ds \int_0^1 u^{a - 1} \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{j d} \rd u\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \int_0^1 \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{a - 1 + j d} \rd u\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^\infty \paren {-1}^j \int_0^1 u^{a - 1 + j d} \rd u\) Fubini's Theorem
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^\infty \paren {-1}^j \intlimits {\frac {u^{a + j d} } {a + j d} } 0 1\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {a + j d}\)

$\blacksquare$


Sources