Non-elementary functions with elementary proofs!

If we define $g(x) \; = \; \int_0^\infty \frac{t e^{-xt}}{1 + t^2}\,dt \hspace{2cm} x > 0$ then (differentiating within the integral sign) $g''(x) + g(x) \; = \; \int_0^\infty te^{-xt}\,dt \; =\; x^{-2} \hspace{2cm} x > 0$ Looking for a solution of the form $g(x) \,=\, e^{ix}a(x) + e^{-ix}b(x)$ , subject to the condition $e^{ix} a'(x) + e^{-ix}b'(x) \; = \; 0$ we deduce that we require $ie^{ix} a'(x) - ie^{-ix}b'(x) \; = \; x^{-2}$ and hence that $a'(x) \; =\; \frac{e^{-ix}}{2ix^2} \hspace{2cm} b'(x) \; = \; -\frac{e^{ix}}{2ix^2}$ and so $a(x) \; = \; \alpha - \frac{1}{2i}\int_x^\infty \frac{e^{-it}}{t^2}\,dt \hspace{2cm} b(x) \; = \; \beta + \frac{1}{2i}\int_x^\infty \frac{e^{it}}{t^2}\,dt$ so that $g(x) \; = \; \alpha e^{ix} + \beta e^{ix} + \int_x^\infty \frac{\sin(t-x)}{t^2}\,dt$ for some constants $\alpha,\beta$ . Since $g(x) \to 0$ as $x \to \infty$ , we deduce that $\alpha = \beta = 0$ , and so $g(x) \; = \; \int_x^\infty \frac{\sin(t-x)}{t^2}\,dt$ Integrating by parts, we see that $g(x) \; =\; \Big[-\frac{\sin(t-x)}{t}\Big]_x^\infty + \int_x^\infty \frac{\cos(t-x)}{t}\,dt \; = \; \int_x^\infty \frac{\cos(t-x)}{t}\,dt \; = \; -\sin x \,\mathrm{Si}(x) - \cos x \,\mathrm{Ci}(x)$ (I am using the notation of this question. It is more normal to write $\mathrm{si}(x)$ here). Thus $\int_0^\infty \big(\sin x \,\mathrm{Si}(x) + \cos x \,\mathrm{Ci}(x)\big)\,dx \; =\; -\int_0^\infty g(x)\,dx \; = \; -\int_0^\infty \frac{dt}{1+t^2} \; = \; -\tfrac12\pi$ making the answer $1 + 1 + 2 = \boxed{4}$ .

Redefining the zone of integration:

$\displaystyle -\int_0^\infty\int_x^\infty\sin(x)\frac{\sin(t)}{t}dtdx=-\int_0^\infty\int_0^t\sin(x)\frac{\sin(t)}{t}dxdt = \int_0^\infty\frac{\cos(t)\sin(t)}{t}-\frac{\sin(t)}{t}dt$

Falling a similar procedure: $\displaystyle\int_0^\infty\cos(x)\text{Ci}(x)dx=-\int_0^\infty\frac{\sin(t)\cos(t)}{t}dt$

Now using the Dirichlet integral $\displaystyle\int_0^\infty\frac{\sin(x)}{x}dx=\frac{\pi}{2}$ The answer becomes $\bigg|\frac{\pi}{4}-\frac{\pi}{2}-\frac{\pi}{4}\bigg|=\frac{\pi}{2}$

define two integrals: $(I_1,I_2)=\left(\int_0^\infty \sin(x) SI(x) dx,\int_0^\infty \cos(x) Ci(x) dx \right)$ consider the first integral: $I_1= \int_0^\infty \sin(x) SI(x) dx=\int_0^\infty \sin(x) \int_\infty^x \dfrac{\sin(t)}{t}dt dx$ by using integration by parts $\int a'b = ab-\int ab'\\ a'= \sin(x) \to a = -\cos(x), b = \int_\infty^x \dfrac{\sin(t)}{t}dt \to b' = \dfrac{\sin(x)}{x}\\ I_1 = -\cos(x) \int_\infty^x \dfrac{\sin(t)}{t}dt |_0^\infty - \int_0^\infty \dfrac{-\cos(x)\sin(x)}{x} dx\\ I_1 = \dfrac{\pi}{2} + \dfrac{1}{2} \int_0^\infty \dfrac{\sin(2x)}{x} dx= \dfrac{3\pi}{4}$

similarly, $I_2 = \sin(x) \int_\infty^x \dfrac{\cos(t)}{t}dt |_0^\infty- \int_0^\infty \dfrac{\cos(x)\sin(x)}{x} dx$ using l'hopital rule on the first part $\lim_{x\to 0} \sin(x) \int_\infty^x \dfrac{\cos(t)}{t}dt =\lim_{x\to 0} \dfrac{\int_\infty^x \dfrac{\cos(t)}{t}dt}{\csc(x)}\\ =\lim_{x\to 0} \dfrac{\cos(x)/x}{-\cos(x)/\sin^2(x)} =-\lim_{x\to 0} \dfrac{\sin^2 (x)}{x} =0$ $I_2 = 0 - \dfrac{\pi}{2}=- \dfrac{\pi}{2}$ $I_1+I_2 = \boxed{\dfrac{1\pi^1}{2}}$