An integral formula involving parabolic cylinder functions (Fourier Transform of parabolic cylinder functions and an exponential function)

Question

Table of Integrals, Series, and Products (8th edition)(I.S.Gradshteyn,I.M.Ryzhik,Daniel Zwillinger, and Victor Moll) contains the following formula.(Page 852)

7.741 5.

$$\int_{0}^{\infty} e^{-\frac{1}{2}x^2}\cos(bx)\left(D_{2\nu-\frac{1}{2}}(x)+D_{2\nu-\frac{1}{2}}(-x)\right)\mathrm dx=\frac{ 2^{\frac{1}{4}-2\nu}\pi^{\frac{1}{2}}b^{2\nu-\frac{1}{2}}e^{-\frac{1}{4}b^2}}{\csc((\nu+\frac{1}{4})\pi)}$$ $$\operatorname{Re} \nu>\frac{1}{4}, \quad b>0$$

As far as I've checked, this formula is wrong.

Does anyone know the correct formula?

Answer 1

$$ \int_0^\infty e^{-x^2/2}\cos(bx) [D_{2\nu-1/2}(x)+D_{2\nu-1/2}(-x)]dx $$ $$ = \int_0^\infty e^{-x^2/2}\cos(bx) \frac{2^{\nu+3/4}\sqrt{\pi}}{\Gamma(3/4-\nu)}e^{-x^2/4}{}_1F_1(-\nu+1/4;1/2;x^2/2)dx $$ $$ = \frac{\sqrt{2}\pi}{\Gamma(3/4-\nu)} (2/3)^{\nu+1/4} e^{-b^2/3} {}_1F_1(-\nu+1/4;1/2;-\frac{2b^2}{3}) $$ The derivation of that formula is in vixra:2207.0148.