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
$$int_0^infty dx frac{sin x}{x} = int_0^{pi/2} dtheta frac{sin theta}{sin theta} = pi$$
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 $pi$.