I highly suspect that ∫∞0cos(coshx)cosh(αx)dx
converges for 0≤α<1
(If true, it obviously also converges for −1<a<0.)
I can show that the integral converges for α=0:
∫∞0cos(coshx)dx=∫∞1cos(u)√u2−1du
which converges by Dirichlet's test
I can also show that the integral doesn't converge for α=1:
∫∞0cos(coshx)cosh(x)dx=∫∞1ucos(u)√u2−1du
which doesn't converge since ucos(u)√u2−1∼cos(u) for large values of u
For other values of α between 0 and 1, I'm not sure what to do. I don't know how to express cosh(αx) in terms of cosh(x).
Answer
For α∈[0,1), let t=ex. Then
∫R0cos(coshx)cosh(αx)dx=∫eR1cos(t+t−12)tα+t−α2tdt.
Noticing that
cos(t+t−12)=cos(t2)+O(12t)as t→∞,
we find that the integral converges as R→∞ by Dirichlet's test.
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