Integral involving the error function

Question

Is there a closed form solution to the integrals \begin{align} I_{c} &= \int_{0}^{\infty} \cos(a x) \, \operatorname{erf}(b x) \, dx \\ I_{s} &= \int_{0}^{\infty} \sin(a x) \, \operatorname{erf}(b x) \, dx \hspace{5mm} ? \end{align}

Answer 1

$$\int \cos(a x) \operatorname{erf}(b x) \, dx = \frac{\sin (a x) \text{erf}(b x)}{a}+\frac{e^{-\frac{a^2}{4 b^2}} \left(\text{erfi}\left(\frac{a}{2 b}+i b x\right)-i \text{erf}\left(\frac{2 b^2 x+i a}{2 b}\right)\right)}{2 a} + C.$$

$$\int \sin(a x) \operatorname{erf}(b x) \, dx = \, \, -\frac{\cos (a x) \text{erf}(b x)}{a}+\frac{e^{-\frac{a^2}{4 b^2}} \left(\text{erf}\left(\frac{2 b^2 x-i a}{2 b}\right)+\text{erf}\left(\frac{2 b^2 x+i a}{2 b}\right)\right)}{2 a} + C.$$

Because of this $I_c$ and $I_s$ does not converges on $[0,\infty)$.

If you are interested in other cases:

$$\int_0^1 \cos(a x) \operatorname{erf}(b x) \, dx = \frac{2 \sin (a) \text{erf}(b)+e^{-\frac{a^2}{4 b^2}} \left(\text{erfi}\left(\frac{a}{2 b}-i b\right)+\text{erfi}\left(\frac{a}{2 b}+i b\right)-2 \text{erfi}\left(\frac{a}{2 b}\right)\right)}{2 a}.$$

For $a=b=1$ it is

$$\int_0^1 \cos(x) \operatorname{erf}(x) \, dx = \text{erf}(1) \sin (1)+\frac{-i \text{erf}\left(1+\frac{i}{2}\right)-2 \text{erfi}\left(\frac{1}{2}\right)+\text{erfi}\left(\frac{1}{2}+i\right)}{2 \sqrt[4]{e}}.$$

And the other one

$$\int_0^1 \sin(a x) \operatorname{erf}(b x) \, dx = -\frac{2 \cos (a) \text{erf}(b)+e^{-\frac{a^2}{4 b^2}} \left(\text{erfc}\left(b+\frac{i a}{2 b}\right)+i \text{erfi}\left(\frac{a}{2 b}+i b\right)-1\right)}{2 a}.$$

For $a=b=1$ it is

$$\int_0^1 \sin(x) \operatorname{erf}(x) \, dx = \text{erf}(1) (-\cos (1))+\frac{\text{erf}\left(1+\frac{i}{2}\right)-i \text{erfi}\left(\frac{1}{2}+i\right)}{2 \sqrt[4]{e}}$$