5
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ what is $i$, $\sqrt{-1}$? Then the logarithm tends to 0 and the series converges, right? $\endgroup$ Aug 6, 2012 at 11:56

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
2
$\begingroup$

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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.