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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!

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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.

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