Gamma-Zeta Product

Start with the substitution s=nus = nu. Show that Γ(x)ζ(x)=0essx1ds(n=1nx)\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 Γ(x)ζ(x)=0ux1eu1du.\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 Γ(x)ζ(x)=n=10es(sn)x11nds.\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=nus = nu and ds=nduds=ndu, thus

Γ(x)ζ(x)=n=10enuux1du=0(eu+e2u+e3u+...)ux1du=0ux1eu(eueu1)du=0ux1eu1du. \begin{aligned} \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{aligned}

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

Note by Steven Zheng
4 years, 11 months ago

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