The value of $$\int_0^\infty {\frac{1}{{1 + {{\text{x}}^2}}}} {\text{dx}} + \int_0^\infty {\frac{{\sin {\text{x}}}}{{\text{x}}}} {\text{dx}}$$ is $\pi$.
To see this, we can use the residue theorem. Consider the following contour integral:
$$\oint_C dz \frac{1}{z(1+z^2)}$$
where $C$ is a keyhole contour in the complex plane, as shown below.
[asy]
unitsize(1 cm);
draw((0,-1.2)–(0,1.2));
draw((-1.2,0)–(1.2,0));
real g(real x) {
return sqrt(1+x^2);
}
draw(graph(g,-1.2,1.2),red);
draw((0,0)–(1.2,0.6));
draw((0,0)–(-1.2,0.6));
draw((1.2,0.6)–(1.2,-0.6),dashed);
draw((-1.2,0.6)–(-1.2,-0.6),dashed);
label(“$Re(z)$”, (1.2,0.2), E);
label(“$Im(z)$”, (0.2,1.2), N);
label(“$z=0$”, (0,0), S);
label(“$z=i$”, (0,1), E);
label(“$z=-i$”, (0,-1), W);
[/asy]
The contour integral is equal to the sum of the integrals over the four line segments. The integrals over the vertical line segments go to zero as the radius of the semicircles goes to infinity. Therefore, the contour integral is equal to
$$\int_0^\infty dx \frac{1}{1+x^2} + i \int_0^\infty dx \frac{x}{1+x^2} + \int_{-\infty}^0 dx \frac{1}{1+x^2} – i \int_{-\infty}^0 dx \frac{x}{1+x^2}$$
The real part of the contour integral is therefore
$$\int_0^\infty dx \frac{1}{1+x^2} – \int_{-\infty}^0 dx \frac{1}{1+x^2}$$
Using the substitution $x \mapsto -x$, we can write this as
$$2 \int_0^\infty dx \frac{1}{1+x^2}$$
The imaginary part of the contour integral is zero.
Therefore, the residue theorem tells us that
$$2 \pi i \cdot 2 \pi i \cdot 0 = \pi \cdot \frac{1}{2} + \pi \cdot \frac{-1}{2}$$
or
$$\int_0^\infty dx \frac{1}{1+x^2} = \pi$$
Now, we can use the substitution $x \mapsto \sin \theta$ to write
$$\int_0^\infty dx \frac{\sin x}{x} = \int_0^{\pi/2} d\theta \frac{\sin \theta}{\sin \theta} = \pi$$
Therefore, the value of $$\int_0^\infty {\frac{1}{{1 + {{\text{x}}^2}}}} {\text{dx}} + \int_0^\infty {\frac{{\sin {\text{x}}}}{{\text{x}}}} {\text{dx}}$$ is $\pi$.