I want to solve this integral(1): $\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\cos(bx)\,dx. \tag{1}$ From reference I got, $\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\sin(bx)\,dx=\frac{1}{b}\exp\left[-(2ab)^{1/2}\right]\cos\left[(2ab)^{1/2}\right], \tag{2}$ $\int_{0}^{\infty}\operatorname{erf}\left(\sqrt{\dfrac ax}\right)\cos(bx)\,dx=\frac{1}{b}\exp\left[-(2ab)^{1/2}\right]\sin\left[(2ab)^{1/2}\right].\tag{3}$
Now, my question is: In eq(1) putting $\operatorname{erfc} = 1-\operatorname{erf}$, and getting $\int_{0}^{\infty}\cos(bx)$ which is not converging.
Similarly If in Eq(2) I put $\operatorname{erfc} = 1-\operatorname{erf}$, L.H.S. should not converge as $\int_{0}^{\infty}\sin(bx)$ term will come, then how integral (2) converges and integral(1) does not converge?
Also if solution or approximation for integral(1) is possible, then please help. I really appreciate for any help from you. Thanks in advance.
Reference: Ng, E.W., Geller, M., (1968) A table of integrals of the error functions, journal of research of the national bureau of standards, 73B, 1.