Generalize $ \int_0^\infty \frac{t^3 \, dt}{e^t + 1} = \frac{7 \cdot 31}{8} \zeta(4)$ to the number field $\mathbb{Q}(i)$

Question

I found in a physics textbook an integral formula, that mentions both an improper integral and a zeta function:

$$ \int_0^\infty \frac{t^3 \, dt}{e^t + 1} = \frac{7 \cdot 31}{8} \zeta(4) = \frac{7 \, \pi^4}{120} $$

Is it possible to express the Dedekind zeta function over the number field $K = \mathbb{Q}(i)$ as an integral transform? The ring of integers $\mathcal{O}_K = \mathbb{Z}[i]$:

$$ \zeta_K(4) = \sum_{I \subseteq \mathcal{O}_K} \frac{1}{N_{K/Q}(I)^4} = \sum_{ (a+bi)\subseteq \mathbb{Z}[i]} \frac{1}{(a^2 + b^2)^2} $$

There is some redundancy as the ideals $(a+bi)$ and $(a-bi)$ are the same, i.e. $(a+bi)\mathbb{Z}[i] = (a-bi)\mathbb{Z}[i]$.

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This seems to be the Laplace transform or the Mellin transform (e.g. integrate over $(\mathbb{R}^\times, \frac{dx}{x})$.

• Integral Representation of the Zeta Function: $\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$

• Prove $\int_0^\infty \frac{\ln^2x\ln(1+x)}{x(1+x)} dx=7\zeta(4)$

Indeed there seems to be something mechanical about this entire set of calculations.