From Gradshteyn&Ryzhik 3.692.6 we know that ∫∞−∞(cos(√x2−1)−cos(√x2+1)) dx=π(J1(1)+I1(1))
I'd like you to help me and all kinds of methods are welcomed. Thank you.
Answer
Here's a sketch of a proof. I intend to use Ramanujan's Master Theorem (RMT), which states that for a function F(x)=∑∞k=0ϕ(k)(−x)k/k!, with ϕ(0)≠0, then
∫∞0xn−1F(x)dx=Γ(n)ϕ(−n).
This is true for where the integral converges.
A proof for non-integer n appears in L. Bougoffa, ArXiv 1902.01539v1, 5 Feb 2019. By splitting the integral at x=0 and scaling, we'll show the equivalent
(1)∫∞0(cos(√2u−a2 )−cos(√2u+a2 ))du√u=π a√2(J1(a)+I1(a)).
The formula of the OP is the special case of a→1. Naturally we will let n=1/2 in RMT so we need the Taylor expansion of the function in the big parentheses. I worked out the first 40 terms (with a symbolic computer program) and discovered a pattern:
(2)cos(√2u−a2 )−cos(√2u+a2 )=√πa/2∞∑k=0(−Jk−1/2(a)+Ik−1/2(a))(−u/a)k/k!
which means
ϕ(k)=√πa2a−k(−Jk−1/2(a)+Ik−1/2(a)).
Putting in k=−1/2, Γ(1/2)=√π, and −J−1(a)=J1(a) gives the answer (1).
Of course I haven't proved (2), but it would surprise me if this expansion is not known. I have checked (1) numerically for many $0, using the PrincipalValue-> True argument in the numerical integration, with the point about which the principal value is taken at u=a. Thus I think the suitable generalization is
$$
(3)\quad \int_{-\infty}^\infty \Big( \cos{(\sqrt{x^2-a^2}\ )} - \cos{(\sqrt{x^2+a^2}\ )} \Big) dx =
\pi \ a\Big( J_1(a) + I_1(a) \Big), \quad 0
as long as the integral is interpreted as a principal value.
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