$\lim_{y \to 0^{+}} \int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \frac{\pi}{2}$ I have shown using the application of Dominated Convergence Theorem (i.e., interchange of differentiation and integral) that
$$
\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \frac{\pi}{2}-\tan^{-1}y
$$
for $(x,y) \in [0,\infty) \times (0,\infty)$.
Now I want to conclude that
$$
\int_{0}^{\infty} \frac{\sin x}{x} dx =\int_{0}^{\infty} \lim_{y \to 0+}e^{-yx} \frac{\sin x}{x} dx =\lim_{y \to 0+}\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \lim_{y \to 0+} \frac{\pi}{2}-\tan^{-1}y =  \frac{\pi}{2}
$$
(The hint which is given with this part is to use DCT.)
To interchange the limit and the integral sign, using DCT, I need to find a non-negative function $g$ which dominates $ |(e^{-yx}\sin x) /x|$ and is integrable, i.e., $g \in L^{1}([0,\infty))$.
The only function I could think of is (obviously) $|\sin x/x|$. But I don't know whether $|\sin x/x|$ is integrable.
 A: The interchange of the limit and integral is in question here. Rather than the dominated convergence theorem, we can apply the following theorem and the Dirichlet test.

If $f:[a,\infty)\times [0,\infty) \to \mathbf{R}$ is continuous and
convergence of the improper integral $F(y) =\int_a^\infty f(x,y) \,
 dx$ is uniform, then $F$ is continuous on $[0,\infty)$ and $$\lim_{y
 \to 0+}F(y) = \int_a^\infty \lim_{y \to 0+}f(x,y) \, dx  =
 \int_a^\infty f(x,0) \, dx$$

With $f(x,y) =  e^{-yx} \frac{\sin x}{x}$ for $x > 0$ we can extend to a continuous function by defining $f(0,y) = 1$. Note that $\int_0^c \sin x \, dx$ is uniformly bounded for all $c > 0$ and $y\geqslant 0$. Furthermore $\left|\frac{e^{-yx}}{x}\right| \leqslant \frac{1}{x}$, and, hence, $\frac{e^{-yx}}{x} \searrow 0$ monotonically and uniformly for all $y \geqslant 0$. By the Dirichlet test the improper integral of $x \mapsto f(x,y)$ over $[,\infty)$ converges  uniformly for all $y \geqslant 0$.
Thus,
$$\lim_{y \to 0+}\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x}\, dx =\int_{0}^{\infty}  \lim_{y \to 0+}e^{-yx} \frac{\sin x}{x}\, dx = \int_{0}^{\infty}  \frac{\sin x}{x}\, dx $$
A: $$I(y)=\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx =\Im \int_{0}^{\infty} \frac{e^{-(y-i)x}}{x} dx$$
$$J(y)=\Im \int_{0}^{\infty} e^{-(y-i)x} dx=\Im \frac{1}{y-i}=\frac{1}{1+y^2}$$
$$I(y)=-\int J(y) dy+C=-\tan^{-1} y+C$$
As $I(0)=\pi/2 \implies C=\pi/2$
$$\implies I(y)=\pi/2-\tan^{-1}y \implies \lim_{y\to 0} I(y)=\pi/2$$
