1
$\begingroup$

During my studies of generating functions for expressions composed of harmonic numbers and its asyptotic approximations I found a simple sum which seems to have been neglected so far

$$s171219a=\sum _{k=1}^{\infty } x^k \left(H_k-(\log (k)+\gamma )\right){}^2\tag{1}$$

The tough part for me is the sum of the mixed term

$$s171219b=\sum _{k=1}^{\infty } x^k H_k \log (k)\tag{2}$$

Can you please help me with the tough part?

What I did up to now is:

Using the integral representation $H_k = \int_{0}^{1} \frac{1-u^k}{1-u} \, du$ the sum (2) under the integral becomes

$$A = \frac{1}{1-u} \sum _{k=1}^{\infty } (1-u^k) x^k \log (k)\tag{3}$$

Using the clever trick to generate $\log$-factors

$$\log(k) = \frac{\partial k^t}{\partial t}|_{t=0} \tag{4}$$

we get $A = R-S$ where

$$R =\frac{1}{1-u} \sum _{k=1}^{\infty } x^k k^t = \frac{\text{Li}_{-t}(x)}{1-u} \tag{5}$$

$$S = \frac{\text{Li}_{-t}(x\;u)}{1-u} $$

so that

$$A(t,x) = \int_{0}^1 \frac{\text{Li}_{-t}(x)-\text{Li}_{-t}(x\;u)}{1-u} \, du \tag{6}$$

and the sum (2) becomes

$$ \frac{\partial A(t,x)}{\partial t}|_{t=0}$$

I'm stuck, however, with the integral (6) over the polylog function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.