# Integral of digamma function

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.

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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. Mar 28 '12 at 22:25
BTW: ISC doesn't seem to have any ideas for a closed form. – J. M. Mar 28 '12 at 22:30
Relatedly: mathoverflow.net/questions/22088/… (the constant you have above is the logarithm of the number I asked about in that MO question) – deoxygerbe Mar 29 '12 at 2:39
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. Mar 29 '12 at 5:37
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

see too his article concerning the related 'Khintchine-Levy constants' (page 62 the expression for $\ln(W)$ at the middle right)
$$S=-\sum_{k=2}^\infty \zeta'(k)$$