Series Expansion An old problem from Whittaker and Watson I'm having issues with. Any guidance would be appreciated.
Show that the function
$$
f(x)=\int_0^\infty \left\{ \log u +\log\left(\frac{1}{1-e^{-u}} \right) \right\}\frac{du}{u}e^{-xu}
$$
has the asymptotic expansion
$$
f(x)=\frac{1}{2x}-\frac{B_1}{2^2x^2}+\frac{B_3}{4^2x^4}-\frac{B_5}{6^2x^6}+\;... \;,
$$
where 
$$B_1, B_3, ...$$ 
are Bernoulli's numbers.
Show also that f(x) can be developed as an absolutely convergent series of the form
$$
f(x)=\sum_{k=1}^\infty\frac{c_k}{(x+1)(x+2)...(x+k)}
$$

Note that:
$$\frac{d}{du} \left\{ \ln u + \ln \left( \frac{1}{1- e^{-u}}\right) \right\} = \frac{1}{u} - \frac{1}{e^u - 1} = -\sum_{n = 1}^{+\infty} \frac{B_n u^{n-1}}{n!}$$
then:
$$\ln u + \ln \left( \frac{1}{1- e^{-u}}\right) = -\sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} u^n $$
so we can rewrite the integral as:
$$-\int_0^{+\infty} \sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} u^n  \cdot \frac{e^{-xu}}{u} \, du = - \sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} \cdot \frac{(n-1)!}{x^n} = - \sum_{n=1}^{+\infty} \frac{B_n}{n^2 x^n}$$
In other words it's equal to:
$$f(x) = \frac{1}{2x} - \frac{B_2}{2^2 \cdot x^2} + \frac{B_4}{4^2 \cdot x^4} - \ldots$$
I've used the same notation for Bernoulli numbers as it is at MathWorld.
