# Pre-requisite definition

Stieltjes constants can be thought of as a generalization of Euler-Mascheroni constants defined as, \begin{align} \gamma_n &= \lim_{m \rightarrow \infty} \left( \sum_{k=1}^m \frac{(\log k)^n}{k} - \frac{(\log m)^{n+1})}{n+1} \right) \end{align}

# Question

How do I prove the following integral representation of the above expression? \begin{align} \gamma _{n}={\frac {(-1)^{n}n!}{2\pi }}\int _{0}^{2\pi }e^{-nix}\zeta \left(e^{ix}+1\right)dx \end{align}

The above linked Wikipedia page says that this result is achieved by using Cauchy's Differentiation formula, but I'm not sure how to proceed.

Thanks for any and all help!

The wiki page you link to gives the following Laurent series for Riemann zeta:

$$\zeta(s)={\frac1{s-1}}+\sum_{n=0}^{\infty}{\frac {(-1)^{n}}{n!}}\gamma _{n}(s-1)^{n}$$

If you compare this with the usual form of the Taylor series,

$$f(x)=\sum_{n=0}^{\infty}{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}$$

you get the identity

$$(-1)^n\gamma_n=\left.\frac{\mathrm d^n}{\mathrm ds^n}\left(\zeta(s)-\frac1{s-1}\right)\right|_{s=1}$$

If you then look at the Cauchy differentiation formula

$$f^{(n)}(a)={\frac {n!}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{\left(z-a\right)^{n+1}}}\,\mathrm dz$$

you should be able to figure out that

\begin{align*} (-1)^n\gamma_n&={\frac {n!}{2\pi i}}\oint _{\gamma}{\frac{1}{\left(s-1\right)^{n+1}}\left(\zeta(s)-\frac1{s-1}\right)}\,\mathrm ds\\ &={\frac {n!}{2\pi i}}\left(\oint _{\gamma}{\frac{\zeta(s)}{\left(s-1\right)^{n+1}}}\,\mathrm ds-\oint _{\gamma}{\frac{1}{\left(s-1\right)^{n+2}}}\,\mathrm ds\right) \end{align*}

Since the second contour integral is $$0$$ (exercise: why?), we have

$$(-1)^n\gamma_n={\frac {n!}{2\pi i}}\oint _{\gamma}{\frac{\zeta(s)}{\left(s-1\right)^{n+1}}}\,\mathrm ds={\frac {n!}{2\pi i}}\oint _{\gamma}{\frac{\zeta(1+s)}{s^{n+1}}}\,\mathrm ds$$

Make the substitution $$s=\exp(ix),\; \mathrm ds=i\exp(ix)$$ (i.e. taking a unit circle contour), and you should get the expression you want.