I am interested in this integral (for $a,b\in\mathbb{R}$): $$ F(a,b;y) := \int\limits_0^y \text{d}x \, \frac{\sin(ax+b/x)}{x} $$ Usually for integrals of this type it is useful to transform $\tilde{x} = 1/x$, but in my case the integral boundaries will get changed as well: $$ F(a,b;y) := \int\limits^\infty_{1/y} \text{d}\tilde{x} \, \frac{\sin(b\tilde{x}+a/\tilde{x})}{\tilde{x}} $$ So I conclude that this transformation is only useful for the integral over the entire real line, and not in my case.
In the case of $b=0$ one has of course $F(a,0;y) = \text{Si}(a\,y)$. Is there a special function for the case $b\not=0$? Any ideas?
I would enjoy to have a special function for that.
The only thing I have in mind is to simplify the problem using $x=\frac y a$ and $c=ab$ to make $$\int\frac{\sin \left(a x+\frac{b}{x}\right)}{x}\,dx=\int\frac{\sin \left(y+\frac{c }{y}\right)}{y}\,dy$$ and then expand the integrand as a series built at $c=0$ to get $$\frac{\sin \left(y+\frac{c }{y}\right)}{y}=\sum_{n=0}^\infty \frac{ \sin \left(y+n\frac{\pi }{2}\right)}{y^{n+1}\,n!}c^n$$ and then use $$\int \frac{\sin(y)}{y^k}\,dy=-\frac{1}{2} y^{-k} \left((-i y)^k \Gamma (1-k,-i y)+(i y)^k \Gamma (1-k,i y)\right)$$ $$\int \frac{\cos(y)}{y^k}\,dy=-\frac{1}{2} i y^{-k} \left((-i y)^k \Gamma (1-k,-i y)-(i y)^k \Gamma (1-k,i y)\right)$$