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.