It is well known that several definite integrals $I_n$ containing a parameter $n\in\mathbb N$ can be expressed recursively (e.g. doing integration by parts) in terms of $I_{n-1} $ or $I_{n-2} $, and thus written as some expression containing factorials, e.g. the gamma function itself $$\int_0^\infty t^{n}e^{-t}dt=n\int_0^\infty t^{n-1}e^{-t}dt=\cdots=n!$$

Or take the formula $I_n:=\int_0^\pi \sin^nx\;dx=\dfrac{n-1}nI_{n-2}$, allowing to obtain $$\int_0^\pi \sin^{2n}x\;dx=\dfrac{(2n)!}{2^{2n}n!^2}\pi\ \ \text{ and }\ \int_0^\pi \sin^{2n+1}x\;dx=\dfrac{2^{2n+1}n!^2}{(2n+1)!}$$ (which BTW easily yields the Wallis product).

As long as for such integrals the LHS is also defined for non-integer (say all positive real) $n$,

is it "automatically" guaranteed that replacing on the RHS $n!$ by $\Gamma(n+1)$ yields a valid formula?

We all know that the interpolation of the gamma function between the factorials to $\mathbb R$ is not unique but that the Bohr-Mollerup theorem assures a unique extension when adding the mild (?) condition of log-convexity. The problem: *the LHS expressions above don't "know" anything about log-convexity*...

We can ask a similar question (even though there is no simple recursion formula in this case) about the Riemann zeta function with, for $n>1$, $$\int_{0}^{\infty} \frac{t^n}{e^t - 1} \; \frac{dt}{t}=\zeta(n) \; \Gamma(n) $$ or, closely related, this interpolation of the Bernoulli numbers $$ 4n\int_{0}^{\infty} \frac{t^{2n}}{e^{2\pi t}-1} \frac{dt}{t}=4n\frac{2^{2n-1}}{2^{2n-1}-1}\int_{0}^{\infty} \frac{t^{2n}}{e^{2\pi t}+1} \frac{dt}{t}=(-1)^{n+1}B_{2n}$$ or this one of the Euler numbers $$ \int_{0}^{\infty} \frac{t^{2n}}{\cosh\frac{\pi t}2}\; dt =(-1)^{n}E_{2n}.$$ The two last ones are somewhat intriguing because for half-integers $n$, the integrals are obviously not $0$, unlike the odd Bernoulli and Euler numbers. For Euler numbers, this can of course be explained by the fact that the "entire" integral from $-\infty$ to $\infty$ does vanish in this case (odd function), but for the Bernoulli numbers the situation is different.

What is going on here?