Proof that $\int_0^\infty\ln^{s-1}(t+1)dt=\int_0^\infty\frac{\ln^{s-1}(t+1)}{t}dx$

Question

I proved that$$\int_0^\infty\ln^{s-1}(t+1)dt=\int_0^\infty\frac{\ln^{s-1}(t+1)}{t}dt.$$ When $s$ is a zero of the zeta function $\zeta(s)$. Is my proof correct?:$$0=\int_0^\infty\frac{t^{s-1}}{e^t-1}dt\\ =\int_0^\infty\frac{(t-1)\ln^{s-1}(t+1)}{t}dt \text{ (by substitution of $u=e^t-1$ then change variables)}\\ =\int_0^\infty \ln^{s-1}(t+1)dt-\int_0^\infty\frac{\ln^{s-1}(t+1)}{t}dt\\ \implies\int_0^\infty\ln^{s-1}(t+1)dt=\int_0^\infty\frac{\ln^{s-1}(t+1)}{t}dt$$I am not sure if I was allowed to separate the integrals, so I am feeling skeptical.

Answer 1

$$u=e^t-1$$ $$t=\ln(u+1)$$ $$dt=\frac{du}{u+1}$$ $$\int_0^\infty\frac{t^{s-1}}{e^t-1}dt=\int_0^\infty\frac{\ln^{s-1}(u+1)}{u(u+1)}du$$ Perhaps you might have made a mistake in your substitution. Also, as the comments mentioned, the domain for $s$ doesn't include the roots of the zeta function. For instance, using a substitution we find the following integral diverges for all real $s$ $$\int_0^\infty\ln^{s-1}(t+1)dt=\int_0^\infty u^{s-1}(e^u-1)du=\Gamma(s)-\infty$$