Another Hot Integral

The substitution $y = x^{-1}$ gives $I_1 \; = \; \int_0^\infty e^{-x^2-x^{-2}}\,dx \; = \; \int_0^\infty e^{-(x-x^{-1})^2-2}\,dx \; = \; \int_0^\infty e^{-(y-y^{-1})^2-2}y^{-2}\,dy$ and hence $I_1 \; =\; \tfrac12\int_0^\infty e^{-(x-x^{-1})^2-2}\,(1 + x^{-2})\,dx \; =\; \tfrac12\int_{\mathbb{R}} e^{-z^2-2}\,dz \; = \; \tfrac12\sqrt{\pi}e^{-2}$ Similary, since $\begin{aligned} \int_{X^{-1}}^X \cos(x^2 + x^{-2})\,dx & = \; \int_{X^{-1}}^X \cos\big[(x-x^{-1})^2+2\big]\,dx \; = \; \int_{X^{-1}}^X \cos\big[(y-y^{-1})^2+2\big] y^{-2}\,dy \\ & = \; \tfrac12\int_{X^{-1}}^X \cos\big[(x-x^{-1})^2+2\big]\,(1 + x^{-2})\,dx \; = \; \tfrac12\int_{-(X-X^{-1})}^{X - X^{-1}}\cos(u^2+2)\,du \end{aligned}$ we see, letting $X \to \infty$ , that $I_2 \; = \; \int_0^\infty \cos(x^2+x^{-2})\,dx \; = \; \tfrac12\int_{\mathbb{R}}\cos(u^2+2)\,du \; = \; \tfrac{\sqrt{\pi}}{2\sqrt{2}}\big(\cos 2 - \sin 2\big)$ Finally, integrating the analytic function $e^{i(z^2-z^{-2})}$ about (almost) the octant wedge defined by the $x$ -axis, the line $y=x$ , and circles radius $R$ and $\varepsilon$ about the origin, we can show, letting $R \to \infty$ and $\varepsilon \to 0$ that $I_3 \; = \; \int_0^\infty e^{i(x^2 - x^{-2})}\,dx \; = \; \tfrac{1}{\sqrt{2}}(1+i)\int_0^\infty e^{-(x^2+x^{-2})}\,dx \; = \; \tfrac{1}{\sqrt{2}}(1+i)I_1 \; =\; \tfrac{1}{2\sqrt{2}}\sqrt{\pi}(1+i)e^{-2}$ and hence $I_4 \; = \; \int_0^\infty \cos(x^2 - x^{-2})\,dx \; = \; \mathfrak{Re}(I_3) \; = \; \tfrac{1}{2\sqrt{2}}\sqrt{\pi}e^{-2}$ Thus $\int_0^\infty \sin(x^2)\sin(x^{-2})\,dx \; =\; \tfrac12\int_0^\infty \big[\cos(x^2-x^{-2}) - \cos(x^2 + x^{-2})\big]\,dx \; = \; \tfrac12(I_4 - I_2) \; = \; \tfrac{1}{4\sqrt{2}}\sqrt{\pi}\big[\sin2 - \cos2 + e^{-2}\big]$ so that $A + B + C = 32 + 2 - 2 = \boxed{32}$ .