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

Solution 1:

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.