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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 $$ f(x)=\sum_{k=1}^\infty\frac{c_k}{(x+1)(x+2)...(x+k)} $$

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1 Answer 1

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.

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Much appreciated. If you have any ideas on the second part of the problem I'd be grateful as well. –  John Doe Jun 25 '12 at 4:27

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