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