Regularization of integral

Question

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!

Answer 1

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.

Answer 2

Your second example is wrong, I do not know your source, but the correct answer is $-1/5$. This can be seen using two different methods, one using Laplace transform and the other reducing the problem to Dirichlet regularization of a series.

Here is the code in Mathematica and results of regularizing your examples and another function using the both methods:

Regularize := 
 Inactivate[reg[Integrate[f[x], {x, 0, Infinity}]], Integrate] == 
   Limit[s Sum[f[s x], {x, 1, Infinity}, 
       Regularization -> "Dirichlet"] // FullSimplify, s -> 0] // 
  TraditionalForm
Regularize2 := 
 Inactivate[reg[Integrate[f[x], {x, 0, Infinity}]], Integrate] == 
   Evaluate[
    Evaluate[LaplaceTransform[ f[t], t, x]] /. x -> 0 // 
     FullSimplify] // TraditionalForm

For the function $\cosh^4(x)$ one of the methods fails but the other works.