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  • A definite integral

    From Robert Gragg@21:1/5 to All on Tue Feb 7 07:32:01 2023
    Can Maple provide a result for the integral
    \int_{-\pi/2}^{\pi/2} e^{-i a cos\phi} \cos^2\phi d\phi
    with a>0?
    Or maybe an asymptotic result for a>>1?

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  • From William Unruh@21:1/5 to Robert Gragg on Wed Feb 8 06:53:11 2023
    On 2023-02-07, Robert Gragg <robertfgragg@gmail.com> wrote:
    Can Maple provide a result for the integral
    \int_{-\pi/2}^{\pi/2} e^{-i a cos\phi} \cos^2\phi d\phi
    with a>0?
    Or maybe an asymptotic result for a>>1?

    Have you tried?

    This is of course just
    2 int_1/sqrt(2)^1 e^{-iax} x^2/sqrt{1-x^2} dx

    =2 -int _1/sqrt(2)^1 (e^-iax)(x-ia}sqrt(1-x^2}dx -e^{-ia}

    or something like that. (x=cos(phi))

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  • From Wasell@21:1/5 to Robert Gragg on Thu Feb 9 17:32:11 2023
    On Tue, 7 Feb 2023 07:32:01 -0800 (PST), in article <ac6dded8-1440-4858-9ef8- 3fc8bb634f15n@googlegroups.com>, Robert Gragg wrote:

    Can Maple provide a result for the integral
    \int_{-\pi/2}^{\pi/2} e^{-i a cos\phi} \cos^2\phi d\phi
    with a>0?
    Or maybe an asymptotic result for a>>1?


    Per your question, I gave Maple this:

    f := a -> int(exp(-I*a*cos(phi))*cos(phi)^2, phi = -1/2*Pi .. 1/2*Pi)

    Maple claimed that

    f(a) = Pi*(-a*StruveH(0, a)*I + a*BesselJ(0, a) +
    StruveH(1, a)*I - BesselJ(1, a))/a

    and futhermore that

    limit(f(a), a = infinity) = 0

    The last equality is heuristically supported by:

    seq(abs(evalf(f(10^n))), n = 1 .. 5) =
    0.7917525236,
    0.2506651220,
    0.07926654025,
    0.02506628274,
    0.007926654597

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