Evaluating a limit similar to the Euler constant In the course of studying a certain complex-valued functional equation, I have had a need to evaluate the following limit:
$$\gamma_\mathcal{T}=\lim_{n\to\infty}\left(-\frac{i}{2}\sum_{k=1}^n \frac1{ik+k^{3/2}}-\log\left(1+\frac{i}{\sqrt n}\right)\right)$$
which is structurally similar to the usual limit definition for the Euler constant $\gamma$.
So far as I can tell, there seems to be no elementary closed form for this limit, so I set about trying for numerical estimation.
The problem is that the convergence of this limit looks to be excruciatingly slow. Even with the help of a sequence extrapolation method, I only managed to produce a few good digits:
$$\gamma_\mathcal{T}\approx-0.5-0.9300125396i$$
I am wondering if there are more efficient, alternative methods for numerically evaluating this limit. Thanks in advance!
 A: To expand on @Fedor's comment: if you rationalize the denominator, you get the general term of the sum to be $f(k)=\frac{k^{3/2} - i k}{k^3 + k^2}.$ The imaginary part of $\sum_{k=1}^\infty f(k)$ is easy to evaluate (it is equal to $-1$). The real part is not so easy, but Mathematica returns immediately with $1.8600250792211903071806\dots,$ so its sequence acceleration techniques are up to the task.
A: As already mentioned by Fedor and Igor, you can ignore the logarithmic term since it zeroes out at $\infty$, and you can just concentrate on the series
$$-\frac{i}{2}\sum_{k=1}^\infty \frac1{ik+k^{3/2}}$$
Using Laplace transform techniques, your sum can be transformed into the integral
$$-\frac12-\frac{i}{\sqrt\pi}\int_0^\infty \frac{F(\sqrt{u})}{\exp\,u-1}\mathrm du$$
where $F(z)$ is Dawson's integral.
I don't know of a closed form for this integral, but Mathematica easily evaluates this numerically:
-1/2 - I NIntegrate[DawsonF[Sqrt[u]]/(E^u - 1), {u, 0, Infinity}, 
    Method -> "DoubleExponential", WorkingPrecision -> 50]/Sqrt[Pi]
-1/2 - 0.93001253961059515359034795785857166233326206076173 I

