Solving the Integral $\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$ Is there any possibily to solve the following integral
$$\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$$
with $a>0$, $y>0$ and $-\pi/2<\mathrm{arg}(b)<0$.
I assume, the result is connected to Bessel and Struve functions. Thank you.
Edit:
Using integration by parts with
$$\int\mathrm{cos}(b \, x)\,\mathrm{d}x=\frac{\mathrm{sin}(b \, x)}{b}$$
$$\frac{\mathrm{d}}{\mathrm{d}x}( \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right))= - \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}}$$
and the limits
$$ \lim_{x\to0}  \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0$$
$$ \lim_{x\to\infty}  \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0 $$
Gives the integral
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x$$
if that makes anything simpler...
Edit2:
Mathematica tells me that the last integrand can be presented as the product of three G-functions. Inhere it is said that the integral of the product of three G-functions can be computed under certain restrictions. Sadly it is not mentioned which restrictions. Does anybody know anything about this?
It would be:
$$\frac{1}{x^2+y^2+a^2} = \frac{1}{\sqrt{\pi} \, \sqrt{y^2+a^2}} \,\mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right]$$
$$\frac{1}{x^2+y^2} = \frac{1}{y^2} \,\mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right]$$
$$\mathrm{sin}(b\,x)= \sqrt{\pi} \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{ 0\} \right\},\tfrac{x^2 \, b^2}{4}\right]$$
which finally results in
$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, \mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right] \, \mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right] \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{0 \} \right\},\tfrac{x^2 \, b^2}{4}\right] \, \mathrm{d}x$
The $\mathrm{MeijerG}$ are defined according to Mathematica syntax. Or (I hope I converted this correctly)
$$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}\frac{1}{2}\\ 0 \end{matrix}&\frac{x^2}{y^2+a^2}\end{array}\right) \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}0\\ 0 \end{matrix}&\frac{x^2}{y^2}\end{array}\right)\, G^{1,0}_{0,2}\left(\begin{array}{c|c}\begin{matrix}-\\\frac{1}{2},\, 0\end{matrix}&\frac{x^2 \,b^2}{4}\end{array}\right) \, \mathrm{d}x$$
 A: First, we simplify the parameters:
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x= \frac{a}{by} \int_0^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
$$c^2=1+\frac{a^2}{y^2}, \qquad r=b y$$
I assume $b$ is real for this calculation (which may not be the case for the OP).
So we want to find:
$$f(r,c)=\int_0^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
For $c=1$ (which means $a=0$), we have a closed form as a modified Bessel function:
$$f(r,1)=r K_0(r)$$
This gives the following idea. Write:
$$\sqrt{c^2+t^2}=\sqrt{1+\frac{a^2}{y^2}+t^2}=\sqrt{1+t^2} \sqrt{1+\frac{a^2}{y^2(1+t^2)}}$$
And expand the latter as a binomial series. Then we have:
$$f=\sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2} \left( \frac{a}{2y} \right)^{2k} \int_0^\infty \frac{t \sin rt ~dt}{(1+t^2)^{3/2+k} }$$
Amazingly enough all the integral have a closed form:
$$\int_0^\infty \frac{t \sin rt ~dt}{(1+t^2)^{3/2+k} }=\frac{r^{k+1}}{(2k+1)!!} K_k (r)$$
Here $!!$ means double factorial: $(2k+1)!!=1 \cdot 3 \cdot 5 \cdots (2k+1)$.
Finally we have for the original integral:

$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x= a \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2 (2k+1)!!} \left( \frac{a  }{2}\sqrt{\frac{b }{y}} \right)^{2k} K_k (b y)$$


We could also attempt residues on the integral:
$$\int_{-\infty}^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
but it has a square root and so will involve the branch cuts, I'm not sure how to deal with it.
