Why does an additional -1/x term in the integrand of the Mellin transform representation of the Riemann zeta function not change its value?

Question

Using Mellin transforms, the product of the Riemann zeta function and the gamma function can be expressed as the following integral:

$$\zeta\left(s\right)\mathrm{\Gamma}\left(s\right)=\int_{0}^{\infty}{\left(\frac{1}{e^x-1}\right)x^{s-1}dx}$$

However, I have also read sources stating that the following representation (apparently from Whittaker-Watson) is also valid for the critical strip 0 < Re(s) < 1 :

$$\zeta\left(s\right)\mathrm{\Gamma}\left(s\right)=\int_{0}^{\infty}{\left(\frac{1}{e^x-1}-\frac{1}{x}\right)x^{s-1}dx}$$

It is amazing to me that the addition of the term $(-\frac{1}{x})$ in the integrand of the second representation does not change the value of the integral at all for any value of s in the critical strip!

Can anyone provide a mathematical explanation for why and how this is the case?

Answer 1

Compare with the following:

For $\Re(s) > 1$, $$\frac1{s-1}=\int_0^1 x^{s-2}dx$$ For $\Re(s) < 1$, $$\frac1{s-1}=-\int_1^\infty x^{s-2}dx$$

This is also how we prove your formula, as for $\Re(s) > 1$, $$\Gamma(s)\zeta(s)-\frac1{s-1}=\int_0^1 (\frac1{e^x-1}-\frac1x) x^{s-1}dx+\int_1^\infty \frac1{e^x-1} x^{s-1}dx$$ and the RHS stays analytic for $\Re(s)> 0$, thus providing the analytic continuation of the LHS.

Answer 2

$$\int_0^1 x^{s-2}dx=\left.\frac{x^{s-1}}{s-1}\right|_0^1=\frac1{s-1}$$

$$\int_1^\infty x^{s-2}dx=\left.\frac{x^{s-1}}{s-1}\right|_1^\infty=-\frac1{s-1}$$

since if $\;s=a+bi\;,\;\;a,b\in\Bbb R\;,\;0<a<1\;$, then

$$\left|x^{s-1}\right|=\left|e^{\left(a-1+bi\right)\log x}\right|=e^{(a-1)\log x}\xrightarrow[x\to\infty]{}0\,,\,\,\text{because}\;\;a-1<0\,,\,\,\log x\xrightarrow[x\to\infty]{}\infty$$

Thus adding $\;-\frac1x\;$ to the parentheses in the integrand is just like adding zero to the whole integral