A known integral takes the form of
$$\zeta(3)=\frac{1}{2}\int_{0}^{\infty} \frac{t^2}{e^t-1}dt$$
Through Wolfram part of the integral converges to
$$\int_{0}^{\infty} \frac{t}{e^t-1}dt = \frac{\pi^2}{6}$$
which, I assume is due to the Bernoulli numbers. Since the above part converges, can we break the integral into
$$\zeta(3)=\frac{1}{2}\int_{0}^{\infty} \frac{t}{e^t-1}dt\int_{0}^{\infty} {t}\:dt=\frac{\pi^2}{12}\int_{0}^{\infty} {t}\:dt$$
Is my logic incorrect? Please let me know if I have made a mistake.
Your logic is incorrect because $$\int_a^b f(x)\cdot g(x)dx\not\equiv\int_a^b f(x)dx\cdot\int_a^b g(x)dx$$