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I was attempting to evaluate a series $$\sum_{n=1}^\infty \frac{1}{n} \ln\left(1+\frac{1}{n}\right)$$

Since $$\frac{1}{n}\ln\left(1+\frac{1}{n}\right)=\int_0^1 \frac{1}{n(n+t)}dt,$$

I rewrote it as $$\sum_{n=1}^\infty \int_0^1 \frac{1}{n(n+t)} \; dt$$

and switched the sum and integral:

$$\int_0^1 \sum_{n=1}^\infty \frac{1}{n(n+t)}dt$$

The sum is related to digamma. Specifically,

$$\frac{\gamma-\psi(t+1)}{t}=\sum_{n=1}^\infty \frac{1}{n(n+t)}$$

Now, integrating this is the problem:

$$\int_0^1 \frac{\gamma+\psi(t+1)}{t} \; dt=1.257746887\ldots$$

That $t$ in the denominator causes a fit. I tried integration by parts, to no avail.

I ran it through Maple and it gave me $$1.257746887\ldots$$ which is indeed what the sum converges to.

Does anyone know if we can evaluate the above digamma integral? Perhaps a numerical approximation is the best we can do? After all, Maple would not give me a closed form; just the decimal solution. Can it be related to zeta or some other advanced function somehow?

Thanks very much for any input.

J. M. ain't a mathematician
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Cody
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    Not sure about a closed form, but there is yet another pretty expression for your sum: $$\sum_{k=1}^\infty \frac{(-1)^{k+1}\zeta(k+1)}{k}$$ – J. M. ain't a mathematician Mar 28 '12 at 22:25
  • BTW: ISC doesn't seem to have any ideas for a closed form. – J. M. ain't a mathematician Mar 28 '12 at 22:30
  • Relatedly: http://mathoverflow.net/questions/22088/infinite-product-experimental-mathematics-question (the constant you have above is the logarithm of the number I asked about in that MO question) – graveolensa Mar 29 '12 at 02:39
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    Two more "pretty" representations: $$\int_0^1 \frac{u\log,u}{(1-u)\log(1-u)}\mathrm du=\int_0^1 \frac{(1-u)\log(1-u)}{u\log,u}\mathrm du$$ – J. M. ain't a mathematician Mar 29 '12 at 05:37
  • Thanks everyone for your input. Much appreciated. I thought that may be the case, but certainly wasn't sure. I suppose that's why Maple would only give a numerical solution. I am glad I stumbled upon something interesting. Thanks. – Cody Mar 29 '12 at 10:53

1 Answers1

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This formula appeared in earlier work of Coffey 'Series of zeta values, the Stieltjes constants, and a sum S_γ(n)' (see (38a) and following expressions),
in his paper of 2011 'Series representations of the Riemann and Hurwitz zeta functions and series and integral representations of the first Stieltjes constant' (1.23 and next)

as well as in Steven Finch's 'Continued Fraction Transformation III' (page 5)
see too his article concerning the related 'Khintchine-Levy constants' (page 62 the expression for $\ln(W)$ at the middle right)

It seems that it may be written in many ways! For example :

$$S=-\sum_{k=2}^\infty \zeta'(k)$$

Raymond Manzoni
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