Regularization of integral

One may note that:

$$I=\int_0^\infty\cosh^4(x)~\mathrm dx=\frac1{16}\int_0^\infty e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x}~\mathrm dx$$

Using $\displaystyle\int_0^\infty e^{-ax}~\mathrm dx=\frac1a$, we may regularize our integral to

$$I=\int_0^\infty\frac38~\mathrm dx$$

Which may be regularized to $\frac38\zeta(0)=-\frac3{16}$.

Same method for the second integral.