Monday, 18 November 2013

limits - How is this series for the Euler-Mascheroni constant derived?.



We are familiar with the classic sum for Euler's constant γ:



γ=lim



But, how is this one derived?:



\gamma=\lim_{n\to \infty}\left(\frac12\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k(2k)!}n^{2k}-\ln(n)\right) .




I thought perhaps it may have something to do with the Bernoulli numbers or even Zeta because there are terms which appear in their identities (such as the formula for \zeta(2n)), but I am not sure.



Thanks for any input.


Answer



Consider \sin^2\left(\frac{n}{2} \right):
\sin^2\left(\frac{n}{2} \right) = \frac{1-\cos(n)}{2} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2 \cdot (2k)!} n^{2k}
Now, since \frac{n^{2k}}{2 k} = \int\limits_0^n t^{2k-1} \mathrm{d} t, we get:
\sum_{k=1}^\infty \frac{(-1)^{k+1}}{(2k)!} \frac{n^{2k}}{2 k} = \int_0^n \frac{1-\cos(t)}{t} \mathrm{d} t
The latter integral gives, by the definition of the cosine integral:
\int_0^n \frac{1-\cos(t)}{t} \mathrm{d} t = \gamma + \log(n) - \operatorname{Ci}(n)



In order to get a sense of why Euler-Mascheroni constant \gamma appears here, consider another definition of cosine integral (in which it is manifest that \lim_{x \to \infty} \operatorname{Ci}(x) = 0):
\operatorname{Ci}(n) = -\int_{n}^\infty \frac{\cos(t)}{t} \mathrm{d} t



Both definitions fulfill \operatorname{Ci}^\prime(x) = \frac{\cos(x)}{x}. In order to establish that the integration constant is indeed the Euler-Mascheroni constant, we consider n=1:
\begin{eqnarray} \gamma &=& \int_0^1 \frac{1 - \cos(t)}{t} \mathrm{d} t - \int_1^\infty \frac{\cos(t)}{t} \mathrm{d} t = \Re\left( \int_0^\infty \frac{[0<|t|<1]-\mathrm{e}^{i t}}{t} \mathrm{d} t \right) \\ &\stackrel{t \to i t}{=}& \Re\left( \int_0^\infty \frac{[0<|t|<1]-\mathrm{e}^{-t}}{t} \mathrm{d} t \right) \stackrel{\text{by parts}}{=} \Re\left( -\int_0^\infty \mathrm{e}^{-t} \ln(t) \mathrm{d} t \right) = \gamma \end{eqnarray}
The last equality follows from the definition of the constant.



No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...