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It is known that $$\Gamma (s) \zeta (s)=\int_0^{\infty} \frac{x^{s-1}}{e^x-1}dx$$ this function is valid only for $\Re{s}>1$.

However, if we ignore this restriction, and integrate by using $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$ we could get $$\Gamma (s) \zeta (s)=\frac{1}{s-1}-\frac{1}{2s}+\sum_{n=2}^{\infty}\frac{B_n}{n!(s+n-1)}+\int_1^{\infty} \frac{x^{s-1}}{e^x-1}dx$$ Surprisly, this expression coincides with the true value of $\zeta(s)$ precisly in the compelx region $$0\leq\Re(s)\leq 1,-10\leq\Im(s)\leq10$$ Any explanations?

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  • $\begingroup$ Yes, it is well-known that for $\Re(s) > 1$, $\int_0^1 x^{s-2} \frac{x}{e^x-1} dx =\int_0^1 x^{s-2}(\sum_{n=0}^\infty \frac{B_n}{n!} x^n )dx$ $=\sum_{n=0}^\infty \frac{B_n}{n!} \int_0^1 x^{n+s-2}dx=\sum_{n=0}^\infty \frac{B_n}{n!} \frac1{s+n-1}$, where exchanging $\sum$ and $\int$ is justified by absolute convergence. The RHS converges for every $s$. The term $\int_1^\infty x^{s-2} \frac{x}{e^x-1}dx$ is entire and converges for every $s$. This is indeed a good way to find the analytic continuation and poles of $\Gamma(s)\zeta(s)$ and the particular values of $\zeta(s)$ at negative integers $\endgroup$
    – reuns
    Aug 28 '18 at 22:35
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The reason is simple. The functions on the two sides of your identity are holomorphic in the punctured half-plane $\{s:\ \Re(s)>0,\ s\neq 1\}$, and they agree in the half-plane $\{s:\ \Re(s)>1\}$. Hence, by the uniqueness principle, they agree in the in the punctured half-plane $\{s:\ \Re(s)>0,\ s\neq 1\}$.

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  • $\begingroup$ But this relation seems only valid in this small region $\endgroup$
    – Milin
    Jul 17 '18 at 12:03
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    $\begingroup$ @Milin: If two holomorphic functions on a connected open set $D$ agree on infinitely many distinct points with a limit point in $D$, then they agree everywhere in $D$. This is one of the basic theorems in complex analysis. $\endgroup$
    – GH from MO
    Jul 17 '18 at 12:05
  • $\begingroup$ I've checked, the relation gives out too large values for large $t$ in this region $\endgroup$
    – Milin
    Jul 17 '18 at 12:09
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    $\begingroup$ @Milin: Your relation is valid in $\{s:\ \Re(s)>0,\ s\neq 1\}$. Checking it numerically is a different thing, but you don't need computers to do complex analysis. Holomorphicity in my response follows from the exponential decay of $B_n/n!$. $\endgroup$
    – GH from MO
    Jul 17 '18 at 12:09

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