# 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)

$$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}\frac{1}{2}\\ 0 \end{matrix}&\frac{x^2}{y^2+a^2}\end{array}\right) \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}0\\ 0 \end{matrix}&\frac{x^2}{y^2}\end{array}\right)\, G^{1,0}_{0,2}\left(\begin{array}{c|c}\begin{matrix}-\\\frac{1}{2},\, 0\end{matrix}&\frac{x^2 \,b^2}{4}\end{array}\right) \, \mathrm{d}x$$

• The integral seems to be expressible as $\int_1^\infty \frac{e^{-b t y} \sin(a b \sqrt{t^2-1})}{b(t^2-1)} \; dt$ – Benedict W. J. Irwin May 25 '18 at 15:16

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.