Evaluation of Painful Integral:

Allow me to put it like this: $~\displaystyle\int_{-1}^\infty\frac{\sin x}x\cdot\frac{dx}{\sqrt{1+x}} ~=~ \pi~\bigg[C\bigg(\sqrt{\frac2\pi}\bigg)+S\bigg(\sqrt{\frac2\pi}\bigg)\bigg],~$ which

can easily be generalized to $~\displaystyle\int_{-\tfrac1a}^\infty\frac{\sin x}x\cdot\frac{dx}{\sqrt{1+ax}} ~=~ \pi~\bigg[C\bigg(\sqrt{\frac2{a~\pi}}\bigg)+S\bigg(\sqrt{\frac2{a~\pi}}\bigg)\bigg],~$

where C and S are the two Fresnel integrals. Everything else is doomed to have $($ generalized $)$

hypergeometric functions in its expression : $$\int_{-\tfrac1a}^0\frac{\sin x}x\cdot\frac{dx}{\sqrt{1+ax}} ~=~ \frac 2a~_2F_3\bigg(\bigg\{\frac12~,~1\bigg\}~,~\bigg\{\frac34~,~\frac54~,~\frac32\bigg\}~,~-\frac1{4a^2}\bigg),$$ for instance.

We have: $$\mathcal{L}(\sin x)=\frac{1}{1+t^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{x\sqrt{1+ax}}\right)=\operatorname{Erf}\left(\sqrt{\frac{t}{a}}\right)$$ hence for $y=+\infty$ it follows that: $$ \lim_{y\to +\infty}I(y) = \int_{0}^{+\infty}\operatorname{Erf}\left(\sqrt{\frac{t}{a}}\right)\frac{dt}{1+t^2}=\int_{0}^{+\infty}\operatorname{Erf}(u)\frac{2a^2 u}{1+a^2 u^4}\,du$$ and by the residue theorem:

$$ \lim_{y\to +\infty}I(y) = \pi a\left( C\left(\sqrt{\frac{2\pi}{a}}\right)-C\left(\sqrt{\frac{2\pi}{a}}\right)^2+S\left(\sqrt{\frac{2\pi}{a}}\right)-S\left(\sqrt{\frac{2\pi}{a}}\right)^2\right)$$

where $C$ and $S$ are given by the Fresnel integrals: $$ C(z)=\int_{0}^{z}\cos\left(\frac{\pi t^2}{2}\right)\,dz,\qquad S(z)=\int_{0}^{z}\sin\left(\frac{\pi t^2}{2}\right)\,dz.$$

For different values of $y$ we can perform almost the same steps, i.e. to write $\frac{1}{x\sqrt{1+ax}}$ as the Laplace transform of an error function, switch the order of integration and compute the last integral through the residue theorem. Obviously we'll have to deal with an incomplete integral and with an "incomplete Laplace transform", too, i.e. with: $$ \int_{0}^{v}\sin x e^{-sx}\,dx = \frac{1}{1+s^2}-\frac{e^{-sv}}{1+s^2}\left(\cos v-s\sin v\right).$$

We try to solve the special case of $y=\infty$ , assuming that $a>0$.

First step: Relabeling $a =1/b$ ,substitute $b(q-1)=x$ $$ I(\infty,b)=\int_{1}^{\infty}\frac{\sin\left[b(q-1)\right]}{q-1}\frac{1}{\sqrt{q}}dq $$

differentiate w.r.t to b yields $$ \partial_b I(\infty,b)=\int_{1}^{\infty}\frac{\cos\left[b(q-1)\right]}{\sqrt{q}}=\int_{1}^{\infty}\frac{\sin (b) \sin (b q)+\cos (b) \cos (b q)}{\sqrt{q}}dq $$ Setting $q=p^2$ brings this integrals to a nice form:

$$ \partial_b I(\infty,b)=2\int_{1}^{\infty}\sin (b) \sin (b p^2)+\cos (b) \cos (b p^2)dp $$

By definition of the Fresnel integrals $C(x), S(x)$ this yields $$ \partial_b I(\infty,b)=\sqrt{\frac{\pi }{2 b}} \left(-2 C\left(\sqrt{\frac{2 b}{\pi }}\right) \cos (b)-2 S\left( \sqrt{\frac{2b}{\pi }}\right) \sin (b)+\sin (b)+\cos (b)\right) $$

Integrating back to w.r.t to $b$
(integration by parts will work because $\partial_x C^2(x)= C(x)\cos(\frac{\pi x^2}{2})$ and the same for $S(x)$)

yields

$$ I(\infty,b)=\pi \left(-C\left( \sqrt{\frac{2b}{\pi }}\right)^2+C\left( \sqrt{\frac{2b}{\pi }}\right)-S\left(\sqrt{\frac{2b}{\pi }}\right)^2+S\left( \sqrt{\frac{2b}{\pi }}\right)\right) $$ where the constant of integration is fixed by the condition $I(\infty, 0)=0$