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I saw someone do this (in a youtube video):

$$\sum_{\text{n}=1}^\infty\frac{\Gamma\left(\text{s}\right)}{\text{n}^\text{s}}=\Gamma\left(\text{s}\right)\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}=\Gamma\left(\text{s}\right)\zeta\left(\text{s}\right)=\sum_{\text{n}=1}^\infty\left\{\int_0^\infty\text{u}^{\text{s}-1}e^{-\text{n}\text{u}}\space\text{d}\text{u}\right\}=$$ $$\int_0^\infty\text{u}^{\text{s}-1}\left\{\sum_{\text{n}=1}^\infty e^{-\text{n}\text{u}}\right\}\space\text{d}\text{u}=\int_0^\infty\text{u}^{\text{s}-1}\cdot\frac{1}{e^\text{u}-1}\space\text{d}\text{u}$$

But, I can follow all the steps he did but the last integral does not converge because the geometric series only hold when the real part of $\text{u}$ is bigger then $0$, but the lower bound of the integral equals $0$. So why are those two things equal?

Or can we assign a value to:

$$\lim_{u\to0}\text{u}^{\text{s}-1}\cdot\frac{1}{e^\text{u}-1}$$

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    The last integral exist as Lebesgue integral. – 2017-01-06
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    For small $u$, the integrand behaves as $u^{s-2}$. – 2017-01-06
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    @user2958701 The geometric series works like this: $$\sum_{n=1}^\infty r^n=\frac{1}{r-1}$$ When $|r|<1$ – 2017-01-06
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    @YvesDaoust Yes, but the sum does not exists at $u=0$?! – 2017-01-06
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    @user403351: I didn't say it exists. On the opposite, the behavior is $u^{-1}$. But the integral can very well converge thanks to the factor $u^{s-1}$. – 2017-01-06
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    @YvesDaoust I see that the limit exists for $\Re(s)>1$. So then we can assign a value to $u=0$ – 2017-01-06
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    @user403351: no, values with $1<\Re(s)\le2$ don' t work. But $\Re(s)<1$ works. – 2017-01-06

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Last statement could wrote like this ‎$$\frac{u^{s-1}}{e^u-1}=\frac{e^{-u}u^{s-1}}{1-e^{-u}}=e^{-u}u^{s-1}\sum_{k=0}^{\infty}e^{-ku}=\sum_{k=1}^{\infty}e^{-ku}u^{s-1}$$‎

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    Ok, but then does the integral not exists because of the lower bound $u=0$?! – 2017-01-06
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Usually, we take integrals like this as follows:

$$\int_0^\infty\frac{u^{s-1}}{e^u-1}\ du=\lim_{a\to0}\int_a^1\frac{u^{s-1}}{e^u-1}\ du+\lim_{b\to\infty}\int_1^b\frac{u^{s-1}}{e^u-1}\ du$$

And when $s>1$, there is no problem as $a\to0$.

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    I would mention that $\int_0^\infty u^{s-1} \frac{e^{-u}-e^{-N u}}{1-e^{-u}}du =\Gamma(s)\sum_{n=1}^N n^{-s}$ and the result follows from $\lim_{N \to \infty} \int_0^\infty u^{s-1} \frac{e^{-N u}}{1-e^{-u}}du = 0$ – 2017-01-07