I would like to know if there is a way of expressing the integral $$G(a) = \int_{-\infty}^{\infty} \sin(ax)\frac{\sqrt{x^2 + 1}}{x} \; dx$$ in terms of known functions.
Numerically I have observed that $$G(a) \rightarrow \frac{4}{a}$$ as $|a|\rightarrow 0$ and $$G(a) \rightarrow \pm \pi $$ as $a \rightarrow \pm\infty$.
Thanks.
The integral is not convergent, due to bad infrared (large $x$) behaviour, and even its Cauchy principal value $$ \lim_{M\to\infty}\int_{|x|<M}\frac{\sin ax}{x}\sqrt{1+x^2}dx $$ is not well defined, since the integrand is even.
If one really wants to give a meaning to such an integral, it is possible to follow this procedure: consider the integral of the function $$ f(z) = \frac{e^{iaz}}{z^2}\sqrt{1+z^2},\text{ for }a>0, $$ along the contour below:
We can split the integration on the various pieces of the contour and, for $R\to\infty$, the integral on the big arcs tends to zero by Jordan's lemma, due to $a>0$. For finite $\varepsilon$, which keeps us safe from the pole at the origin, we let the contour approach the branch cut, yielding: \begin{align} \int_{|x|>\varepsilon}\frac{e^{iax}}{x^2}\sqrt{1+x^2}dx =& \left(e^{-i\pi/2}-e^{i\pi/2}\right)\int_1^{+\infty}\frac{e^{-ay}}{y^2}\sqrt{y^2-1}\,idy\\ &+\frac{i}{\varepsilon}\int_0^{\pi} e^{-i\varphi} d\varphi - a \int_0^{\pi}d\varphi + \mathcal O(\varepsilon)\\ =&2\int_1^{+\infty}\frac{e^{-ay}}{y^2}\sqrt{y^2-1}\, dy+\frac{2}{\varepsilon}-a\pi+\mathcal O(\varepsilon). \end{align} Differentiating with respect to $a$ eliminates the $a$-independent pole in $\varepsilon$, so that in the limit $\varepsilon\to0$: $$ PV\int_{-\infty}^{+\infty}\frac{e^{iax}}{x}\sqrt{1+x^2}dx = 2i \int_1^{+\infty}\frac{e^{-ay}}{y}\sqrt{y^2-1}\, dy +i \pi; $$ the real part is, of course, $$ PV\int_{-\infty}^{+\infty}\frac{\cos ax}{x}\sqrt{x^2+1}\, dx =0, $$ whereas the imaginary part, dropping the principal value, yields $$ \int_{-\infty}^{+\infty}\frac{\sin ax}{x}\sqrt{x^2+1}\, dx = 2 \int_1^{+\infty}\frac{e^{-ay}}{y}\sqrt{y^2-1}\,dy+\pi\equiv I(a)+\pi, $$ as a consequence of the above regularisation.
Now, letting $y=\cosh\psi$, $$ I(a) = \int_{-\infty}^{+\infty} \frac{e^{-a\cosh\psi}}{\cosh\psi}\sinh^2\psi d\psi $$ \begin{align} I'(a) &= -\int_{-\infty}^{+\infty}e^{-a\cosh\psi}\sinh^2\psi d\psi = \frac{e^{-a\cosh\psi}}{a}\sinh\psi\Big|_{-\infty}^{+\infty}-\frac{1}{a}\int_{-\infty}^{+\infty}e^{-a\cosh\psi}\cosh\psi d\psi\\ &= 0 + \frac{1}{a}\frac{d}{da}\int_{-\infty}^{+\infty}e^{-a\cosh\psi}d\psi=\frac{2}{a}K_0'(a), \end{align} where we have recognised the integral representation of Bessel's $K_0$ function. $K'_0(z)=-K_1(z)$, so, up to a constant $c$ $$ I(a) = 2\int_{a}^{+\infty}\frac{K_1(s)}{s}ds + c, $$ which is convergent for $a>0$ since $K_1(z)$ decreases as $e^{-z}z^{-1/2}$ for large $z$. The constant is zero by Lebesgue's theorem: $$ \lim_{a\to\infty}I(a) = \int_{1}^{+\infty}\frac{1}{y^2}\lim_{a\to\infty}e^{-ay}\sqrt{y^2-1}\,dy=0 $$ since $e^{-ay}<e^{-y}$ for large $a$.
So $${ \int_{-\infty}^{+\infty}\frac{\sin ax}{x}\sqrt{x^2+1}\, dx = 2\int_{a}^{+\infty}\frac{K_1(s)}{s}ds+\pi,\text{ for }a>0.} $$
A similar regularisation using a contour in the lower-half plane gives $${ \int_{-\infty}^{+\infty}\frac{\sin ax}{x}\sqrt{x^2+1}\, dx = -2\int_{|a|}^{+\infty}\frac{K_1(s)}{s}ds-\pi,\text{ for }a<0.} $$
To sum up: $$\boxed{ \int_{-\infty}^{+\infty}\frac{\sin ax}{x}\sqrt{x^2+1}\, dx = \text{sign}(a)\left(2\int_{|a|}^{+\infty}\frac{K_1(s)}{s}ds+\pi\right),\text{ for }a\neq0.} $$ This reproduces the limit $\pm\pi$ as $a\to\pm\infty$, and also the divergent behaviour $2/a$ as $a\to0,$ given by $K_1(z)\approx 1/z$ for small $z$.
Distributional meaning: consider the tempered distribution $$ \tau(x)\equiv \frac{\sqrt{1+x^2}}{x}\in\mathscr S'(\mathbb R), $$ where we leave the $PV$ implicit; as such, $\tau$ has a well-defined Fourier transform given by the following action on functions $\psi\in\mathscr S(\mathbb R)$ of fast decrease $ \langle \widehat\tau(\xi),\psi(\xi)\rangle=\langle \tau(x),\widehat\psi(x)\rangle $. Now, since $1\in C^\infty$, \begin{align} \langle \tau(x), \widehat \psi(x)\rangle =& \langle x\cdot x^{-1} \tau(x), \widehat \psi(x)\rangle\\ =&\langle x^{-1} \tau(x), x \widehat \psi(x)\rangle\\ =&-i\langle x^{-1}\tau(x), \widehat{\psi’} (x) \rangle\\ =&-i\langle \widehat{x^{-1}\tau(x)}(\xi), \psi’(\xi) \rangle\\ =&i\langle \widehat{\left(x^{-1}\tau(x)\right)}’(\xi), \psi(\xi) \rangle; \end{align} finally $$ \widehat {\tau} (\xi) = i\partial_\xi \left[\widehat{x^{-1}\tau(x)}(\xi)\right] $$ in the sense of tempered distributions.