Regularization of integral How does one regularize a divergent integral of the form,
$$
I = \int_0^\infty dx \, \cosh^4 x \hspace{0.5 in} ?
$$
Regularizing via simple subtraction of divergences (as we commonly do in physics) is not a viable option, since expanding $\cosh^4x$ in a power series reveals an infinite number of divergent contributions, with no finite piece to remain. However, I do know that there must be some regularization technique out there enabling us to evaluate this, since we have a similar result
$$
\int_0^\infty dx \, \cosh^4 x \sinh x = -{8 \pi \over 15}
$$
in common use. Any tips on how to obtain this sort of thing  (or even reference to a source where this is calculated) would be very helpful!
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.
The answer by Simply Beautiful Art is totally wrong because of his last step.

$$I=\int_0^\infty\frac38~\mathrm dx=\frac38 \int_0^\infty 1 dx=\frac38\left(\int_0^{1/2}1dx+\sum_{k=1}^\infty 1\right)=\frac38(\frac12-\frac12)=0$$

The last integral can only be regularized to zero.