Start with the substitution \(s = nu\). Show that \[\Gamma(x)\zeta(x) = \int _{ 0 }^{ \infty }{ { e }^{ -s } } { s }^{ x-1 }ds\left( \sum _{ n=1 }^{ \infty }{ { n }^{ -x } } \right) \] is equivalent to the integral \[\Gamma(x)\zeta(x) = \int _{ 0 }^{ \infty }{ \frac { { u }^{ x-1 } }{ { e }^{ u }-1 } du } .\]

**Solution**

Since the gamma function is a real number, we may treat the product as \[\Gamma(x)\zeta(x) = \sum _{ n=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -s } } {\left( \frac{s}{n} \right)}^{ x-1 }\frac{1}{n}ds } . \]

We let \(s = nu\) and \(ds=ndu\), thus

\[ \begin{align*} \Gamma(x)\zeta(x) &= \sum _{ n=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -nu } } {u}^{ x-1 }du } \\ &= \int _{ 0 }^{ \infty }{ ({ e }^{ -u} +{ e }^{ -2u} +{ e }^{ -3u} +... ) } {u}^{ x-1 }du \\ &=\int _{ 0 }^{ \infty }{ { u }^{ x-1 }{ e }^{ -u } } \left( \frac { { e }^{ u } }{ { e }^{ u }-1 } \right) du \\ &=\int _{ 0 }^{ \infty }{ \frac { { u }^{ x-1 } }{ { e }^{ u }-1 } } du. \end{align*}\]

Check out my other notes at Proof, Disproof, and Derivation

## Comments

There are no comments in this discussion.