Of course, can be explained by \[\int_a^b x\frac{du}{dx} \;dx = -c + [xu]_a^b\]
Where \(c\) is the integral from \(a\) to \(b\) of \(u\). That way if the \(xu\) term vanishes, the identity holds...
Thus the operation \(d/d\mathbb{I}\) was mapping the integral \[\frac{d}{d\mathbb{I}}\int_a^b f(x) dx \to \int_a^b xf'(x) dx\]
and the answer \[\frac{d}{d\mathbb{I}}I \to -I + [xf(x)]_a^b\]
Where for the class of functions \(x^ne^{-x}\) , \([xf(x)]_{-\infty}^\infty\) will always vanish.
This way any analogous scenario for the derivative of a sum, should follw the summation by parts, or Abel summation formula.
We can keep applying the same concept such that \[\int_a^b f(x) \;dx= C \\ \int_a^b xf'(x) \;dx = -C + [xf(x)]_a^b \\ \int_a^b xf'(x)+x^2f''(x) \;dx = C - [xf(x)]_a^b + [x^2f'(x)]_a^b \\ \int_a^b x^2f''(x)\;dx= 2C - 2[xf(x)]_a^b + [x^2f'(x)]_a^b \\ \int_a^b 2x^2f''(x) + x^3f'''(x) \;dx = -2C +2[xf(x)]_a^b -[x^2f'(x)]_a^b +[x^3f''(x)]_a^b \\ \int_a^b x^3f'''(x) \;dx = -6C +6[xf(x)]_a^b -3[x^2f'(x)]_a^b +[x^3f''(x)]_a^b\]
This progression continiues and one can find that \[\int_a^b x^nf^{(n)} \;dx = \sum_{k=0}^n \frac{(-1)^nn!}{(-1)^kk!}B_n\]
Where \(B_0=C\) and \(B_i=[x^if^{(i-1)}]_a^b\), for \(i=1\dots n\).
So for example we have \[\int_0^\infty x^n\frac{d^n}{dx^n}\bigg(\frac{x^{s-1}}{e^x-1}\bigg)\;dx=(-1)^nn!\zeta(s)\Gamma(s)+\sum_{k=1}^n\frac{(-1)^{k+n}n!}{k!}\bigg[x^k\frac{d^{k-1}}{dx^{k-1}}\bigg(\frac{x^{s-1}}{e^x-1}\bigg)\bigg]_0^\infty\]
Or rearranging \[\zeta(s)=\frac{(-1)^n}{n!\Gamma(s)}\Bigg[\int_0^\infty x^n\frac{d^n}{dx^n}\bigg(\frac{x^{s-1}}{e^x-1}\bigg)\;dx-\sum_{k=1}^n\frac{(-1)^{k+n}n!}{k!}\bigg[x^k\frac{d^{k-1}}{dx^{k-1}}\bigg(\frac{x^{s-1}}{e^x-1}\bigg)\bigg]_0^\infty\Bigg]\]
Which for \(n=0\) reclaims the equation \[\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1} \;dx\]