Approximation of integral of error function? My research requires to approximate an integral that involves an error function. The integral is,
$$
\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\cos(bx)\,dx. \tag{1}
$$
I found two formulas that are very similar to above integral (Ng and Geller, 1968),
$$
\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\sin(bx)\,dx=\frac{1}{b}\exp\left[-(2ab)^{1/2}\right]\cos\left[(2ab)^{1/2}\right], \tag{2}
$$
$$
\int_{0}^{\infty}\operatorname{erf}\left(\sqrt{\dfrac ax}\right)\cos(bx)\,dx=\frac{1}{b}\exp\left[-(2ab)^{1/2}\right]\sin\left[(2ab)^{1/2}\right].\tag{3}
$$
I am trying to represent equation (1) as a product of exponential and trigonometric functions, which is very important for my following work, but I didn't figure it out how the equations (2) and (3) are obtained. 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.
 A: Not really a full answer, but too long for a comment. If I try to numerically integrate your (1) in Mathematica with
a = 0.7; b = 3.1; NIntegrate[Erfc[Sqrt[a/x]] Cos[b x], {x, 0, Infinity}];
(-1/b) Exp[-Sqrt[2 a b]] Sin[Sqrt[2 a b]]

and compare with the 'exact' result, I get agreement (the value is $-0.0350084...$) between the two.
BUT
Mathematica gives me a flag: 
NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral 
estimate, but the integral might be divergent

Such flag is not given when computing the same integral with $\mathrm{erf}$, instead of $\mathrm{erfc}$.
Now, the more I look at it, the more indeed your integral (1) looks divergent to me. There are a number of red flags:


*

*If you plot the integrand, the oscillations do not 'die out' but persists all the way to infinity. [This does not happen with the $\mathrm{erf}$ version]

*Knowing that $\mathrm{erfc}(z)=1-\mathrm{erf}(z)$, your integral (1) can be effectively written as 
$$
\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\cos(bx)\,dx=
\int_{0}^{\infty}\cos(bx)dx-\int_{0}^{\infty}\operatorname{erf}\left(\sqrt{\dfrac ax}\right)\cos(bx)dx\ ,
$$
i.e. the sum of a divergent integral minus a (probably) convergent one.

*The paper you quote displays a rather suspicious 'asymmetry' (eqs. 4.5.16 and 4.5.17), as you have certainly noted too. If both the integrals with $\mathrm{erf}$ and $\mathrm{erfc}$ were convergent, then I see no reason to omit the other two 'obvious' integrals (where you swap sine with cosine) in the list!


None of this is a smoking gun per se, but my feeling is that (1) is divergent (as well as its sine version), and the paper you quote has a number of typos (except perhaps not the typo I thought it had in a comment I left earlier underneath your question!) - while similar integrals with $\mathrm{erf}$ are convergent (both in their sine and cosine version).
In terms of proving any assertion about such integrals, maybe an approach worth attempting is to write your (1) (or its $\mathrm{erf}$ version) as
$$
\mathrm{Re}\int_{0}^{\infty}\operatorname{erfc}\left(\sqrt{\dfrac ax}\right)\exp(\mathrm{i}bx)\,dx\ ,
$$
and then try integration by parts (the $\mathrm{erfc}$ has a 'nice' derivative, while $\exp(\mathrm{i}bx)$ has a nice antiderivative). This should conclusively prove whether my hunch is correct (no time to do it myself now, though...). 
A: For $a>0$, $\operatorname{erf} \sqrt{a/x}$ is $O(1)$ at positive zero and $O(x^{-1/2})$ at infinity. So the integrals of $\operatorname{erf} \sqrt{a/x} \sin b x$ and $\operatorname{erf} \sqrt{a/x} \cos b x$ converge, provided that $b$ is real and nonzero, and the integrals with $\operatorname{erfc}$ diverge.
The values for $b>0$ are
$$\int_0^\infty \operatorname{erf} \sqrt{\frac a x} \sin b x \,dx =
 \frac 1 b -\frac 1 b e^{-\sqrt{2 a b}} \cos \sqrt{2 a b} ,\\
\int_0^\infty \operatorname{erf} \sqrt{\frac a x} \cos b x \,dx =
 \frac 1 b e^{-\sqrt{2 a b}} \sin \sqrt{2 a b}.$$
These can be computed by differentiating the integrand wrt $a$, making the indefinite integral doable in terms of $\operatorname{erf}$, after which the limit  is expressible in terms of elementary functions. Then one has to integrate the result and recover the term independent of $a$. The latter can be done by substituting $a=0$.
