Question

Prove that :-

${\quad \leadsto \quad \bf \Gamma ( s ) \cdot \zeta ( s ) = \displaystyle \bf \int_{ \bf 0}^{\bf \infty} \bf \dfrac{{u}^{s-1}}{e^{u} - 1} du}$ ​

Answer 1

Question

$\color{red}{ \quad \rm \Gamma ( s ) \cdot \zeta ( s ) = \displaystyle \bf \int_{ \rm0}^{\rm\infty} \rm \dfrac{{u}^{s-1}}{e^{u} - 1} du}$

Answer

$\displaystyle\rm \red{ \Gamma(x) = \int \limits _{t = 0}^{t = \infty } {t}^{x - 1} \: {e}^{ - t} }$

$\displaystyle\rm \red{ \Gamma(x) = \int \limits _{t = 0}^{t = \infty } {t}^{x - 1} \: {e}^{ - t} \: dt}$

$\displaystyle\rm \red{ \int \limits _{u = 0}^{u = \infty } {(nu)}^{x - 1} \: {e}^{ - nu} \: du}$

$\displaystyle\rm \red{ \int \limits _{ 0}^{ \infty } {n}^{x - 1} \cdot {u}^{x - 1} \cdot {e}^{ - nu} \: {n}^{1} \: du}$

$\displaystyle\rm \red{ \frac{\Gamma(x)}{ {n}^x } = \frac{\displaystyle \rm \cancel{{n}^{x} } \int \limits _{0}^{ \infty } u^{x - 1} \cdot {e}^{ - nu} \: du}{ \cancel{{n}^{x} }} }$

$\displaystyle\rm \red{ \sum_{n = 1}^{ \infty } \bigg( \Gamma(x) \cdot \frac{1}{ {n}^{x} }\bigg) = \sum_{n = 1}^{ \infty } \left( \int \limits_{0}^ \infty {u}^{x - 1} \: {e}^{ - nu} \: du \right) }$

$\displaystyle\rm \red{ \Gamma(x) \cdot \sum_{n = 1}^{ \infty } \frac{1}{ {n}^{x} } = \int \limits_{0}^ \infty {u}^{x - 1} \sum_{n = 1}^{ \infty } \: ( {e}^{ - u} {)}^n\: du }$

$\displaystyle\rm \red{ \Gamma(x) \: \zeta(x)= \int \limits_{0}^ \infty {u}^{x - 1} \: \frac{( {e}^{ - u}) {e}^{u} }{(1 - {e}^{ - u} ) {e}^u } \: du\color{red}{ \quad \rm = \displaystyle \bf \int_{ \rm0}^{\rm\infty} \rm \dfrac{{u}^{s-1}}{e^{u} - 1} du} }$

$\color{red}{ \quad \rm \Gamma ( s ) \cdot \zeta ( s ) = \displaystyle \bf \int_{ \rm0}^{\rm\infty} \rm \dfrac{{u}^{s-1}}{e^{u} - 1} du}$