How does this identity so elegantly combine an infinite sum in $\eta$ and an improper integral in $\Gamma$?

Question

This is all well and good, but where did this come from?

In the article on the Gamma function, Wikipedia shows most of its alternate definitions with clear proofs, yet in the article on the Dirichlet Eta function, this bizarre identity is just asserted with the phrase, "we may begin by defining: ". My calculus is not at all good enough to relate a $\Sigma$ sum to a $\int$ sum like this, but here Wikipedia's assertion seems to be that multiplying in the $\eta$ sum to the $\Gamma$ function adds one to the denominator of the integrand and I cannot for the life of me see why. A quick google search avails me nought, either.

Answer 1

We have $\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}\,\mathrm{d}t$ for any $s>0$. From the geometrical series it follows that $$ \frac{1}{1-(-e^x)} = \sum_{n\ge0}(-e^{-x})^{n}.$$ with uniform convergence over any compact subset of $\mathbb{R}^+$. $$\begin{align} \int_0^\infty \frac {x^{s - 1} } {e^x + 1}\,\mathrm{d}x &= \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - ( {-e^{-x} }) } \,\mathrm{d} x \\ &= \int_0^\infty x^{s - 1} e^{-x} \left( {\sum_{n\ge0} ({-e^{-x} })^n} \,\mathrm{d} x\right)\\ &= \sum_{n \ge 0} \left( {( {-1})^n \int_0^\infty x^{s - 1} e^{- {(n + 1)} x} \,\mathrm{d} x}\right) \end{align}$$ Substitute $t=(n+1)x\Rightarrow \,\mathrm{d}x=1/(n+1)\,\mathrm{d}t$. The integral now becomes $\int_0^\infty \left(\frac{t}{n+1}\right)^{s-1} e^{- t}\frac{\mathrm{d}t}{n+1}=\int_0^\infty \frac{1}{(n+1)^s} t^{s-1} e^{- t}\,\mathrm{d} t$. The first two factors can be written outside the integral, which leaves us with the Gamma function. $$ \begin{align} &= \Gamma (s) \sum_{n \ge 0} \frac {( {-1})^n} { ({n + 1})^s} \\ &= \Gamma (s) \sum_{n \ge 1} \frac {( {-1})^{n-1}} { n^s} \\ &= \Gamma(s) \, \eta(s) \end{align}$$

Answer 2

Check out this section of the Wikipedia entry for "Mellin transform"; it shows the derivation of the analogous formula $$ \zeta(s) = \frac1{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx, $$ and the method works great for the desired $\eta(s)$ identity as well.

Answer 3

$\frac{1}{e^x+1}=\frac{e^{-x}}{1+e^{-x}}=\sum\limits_{n=0}^\infty (-1)^ne^{-(n+1)x}$

Leads to $\eta(s)=\frac{1}{\Gamma(s)}\sum\limits_{n=0}^\infty(-1)^n\int\limits_0^\infty x^{s-1}e^{-(n+1)x}dx=\sum\limits_{n=0}^\infty\frac{(-1)^n}{n^s}$