Try to evaluate the integral: $$ \int e^{-x^2}\dfrac{\Gamma(-1/2+ix)\Gamma(-1/2-ix)}{\Gamma(ix)\Gamma(-ix)}\Gamma(-1/4+ix)\Gamma(-1/4-ix)K_{2ix}(z)x, $$ either over $\mathbb{R}$ or over $\mathbb{R}^+$ (there is some symmetry here).
Trying to use the fact that $$ \dfrac{\Gamma(-1/2+ix)\Gamma(-1/2-ix)}{\Gamma(ix)\Gamma(-ix)} = \dfrac{x\tanh(\pi x)}{1+4x^2} $$ and $$ \int_0^\infty\dfrac{x\tanh(\pi x)}{\Gamma(3/4+ix/2)\Gamma(3/4-ix/2)}K_{ix}(a)K_{ix}(b)dx= \dfrac{1}{2}\sqrt{\dfrac{\pi ab}{a^2+b^2}}\exp\{-\sqrt{a^2+b^2}\} $$ from Gradshteyn and Ryzhik. No luck so far. Would appreciate any help.