Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:
$$\int\limits_0^\infty \frac{\sin x} x \,\mathrm  dx = \frac \pi 2$$
Well, can anyone prove this without using Residue theory?  I actually thought of using the series representation of $\sin x$:
$$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm  dt$$
but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
 A: Another iteration of this question came up, and I have an answer that isn't currently here. So I present yet another solution.
We want to show that $\int_{0} ^{\infty} \frac{\sin x }{x} \mathrm{d}x = \pi/2.$ 
First, let's show that it converges. We let $I_{ab} = \int_a^b \frac{\sin x}{x}$, and consider the limits $a \to 0, b \to \infty$. $a \to 0$ is easy, so we don't worry about it. $\frac{\sin x}{x}$ is continuous on this domain, so all we really want is for the upper limit to behave nicely.
Note that $I_{ab} = \int \frac{\sin x}{x} = \int \frac{1}{x} \frac{\mathrm{d} (1 - \cos x)}{\mathrm{d} x}$, and so we can use integration by parts. We then get
$$I_{ab} = \frac{1 - \cos b}{b} - \frac{1 - \cos a}{a} + \int_a^b \frac{1 - \cos x}{x^2}$$
This clearly converges. In fact, one can see that both $\cos$ terms disappear in the limit. It's more important to simply note that the integral converges.
Knowing that, we continue the trend of the other answers and show that $\displaystyle \int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2$
We show the following: $$1 + 2 \cos 2t + 2 \cos 4t + \ldots + 2 \cos 2nt = \frac{\sin(2n + 1)t}{\sin t}$$
We do this with $\sin a - \sin b = 2 \sin(\frac{a-b}{2}) \cos(\frac{a + b}{2})$, so that we also get $\sin(2k + 1)t - \sin(2k -1)t = 2\sin(t) \cos (2kt)$. Thus $1 + 2 \cos 2t + \ldots + 2 \cos 2nt = 1 + \frac{1}{\sin t} \left[ \sum \sin(2k+1)t - \sin(sk-1)t \right] $
$\phantom{1 + 2 \cos 2t + \ldots + 2 \cos 2nt} = 1 + \frac{1}{\sin t} [\sin(2n + 1)t - \sin t]$
$\phantom{1 + 2 \cos 2t + \ldots + 2 \cos 2nt} = \frac{\sin(2n + 1)t}{\sin t}$
We did this just so that we could then say that 
$$\int_0^{\pi/2} \frac{\sin (2n + 1)t}{\sin t} = \int_0 ^{\pi /2} (1 + 2 \cos 2t + 2 \cos 4t + \ldots + 2 \cos 2nt) = $$
$$\phantom{\frac{\sin (2n + 1)t}{\sin t}} = \frac{\pi}{2} + \left[ \sin 2t + \frac{\sin 4t}{2} + \ldots + \frac{\sin 2nt }{n}\right]_0^{\pi/2} = \frac{\pi}{2}$$
And thus we have it.
A: Here's another way of finishing off Derek's argument. He proves
$$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$
Let
$$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx=
\int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$
Let
$$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$
where
$$f(x)=\frac1{\sin x}-\frac1x.$$
We need the fact that if we define $f(0)=0$ then $f$ has a continuous
derivative on the interval $[0,\pi/2]$. Integration by parts yields
$$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$
Hence $I_n\to\pi/2$ and we conclude that
$$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$
A: In the book Advanced Calculus by Angus Taylor  it is shown that, if $a\gt 0$,
$$\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt=\arctan\dfrac{x}{a}.\tag{1}$$
If $x>0$, 
$$\displaystyle\int_0^{\infty}\dfrac{\sin xt}{t}dt=\dfrac{\pi}{2}\tag{2}$$
follows from $(1)$, observing that the integrand is $G(0)$ for
$$G(a)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt,\tag{3}$$
$G$ is uniformly convergent when $a\ge 0$, and $G(a)$ approaches $G(0)$ as $a$ tends to $0^+$.

Answer to Qiaochu: $(1)$ is proved as an application of the following theorem [Angus Taylor, Advanced Caluculus, p. 668] to $$F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt.$$ 
Let $$F(x)=\displaystyle\int_c^{\infty}f(t,x)dt$$
be convergent when $a\le x\le b$. Let $\dfrac{\partial f}{\partial x}$ be continuous in $t,x$ when $c\le t,a\le x\le b$, and let $\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$ converge uniformly on $[a,b]$. Then $$F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt.$$
A: Three (more or less) elementary proofs in one answer.
We may notice that:
$$ \int_{-\infty}^{+\infty}\frac{\sin x}{x}\,dx = \int_{-\pi/2}^{\pi/2}\sin(x)\left(\frac{1}{x}+\sum_{m\geq 1}\frac{(-1)^m 2x}{x^2-m^2\pi^2}\right)\,dx \tag{1}$$
since $\sin(x+\pi)=-\sin(x)$. We may study the singularities of
$$ \frac{1}{x}+\sum_{m\geq 1}\frac{(-1)^m 2x}{x^2-m^2\pi^2} = \sum_{m\in\mathbb{Z}}\frac{(-1)^m}{x-m\pi}\tag{2}$$
to deduce it is exactly $\frac{1}{\sin x}$, so the RHS of $(1)$ simply equals $\color{red}{\pi}$. Or we may notice that
$$ \forall \alpha>0,\qquad \int_{0}^{+\infty}\frac{\sin(\pi \alpha x)}{x}\,dx = C\tag{3}$$
and consider the Fourier series of a sawtooth-wave, divided by $x$:
$$ f(x) = \sum_{n\geq 1}\frac{2(-1)^{n+1}\sin(\pi nx)}{\pi nx}. \tag{4}$$
By $(3)$, $\int_{0}^{+\infty}f(x)\,dx$ equals $\frac{2C}{\pi}\log 2$. On the other hand $x\,f(x)$ is piecewise linear, hence:
$$\begin{eqnarray*}\int_{0}^{+\infty}f(x)\,dx &=& \int_{0}^{1}\frac{x}{x}\,dx+\int_{1}^{3}\frac{x-2}{x}\,dx+\int_{3}^{5}\frac{x-4}{x}\,dx+\ldots \\&=&1+\sum_{k\geq 1}\left(2-2k\log\frac{2k+1}{2k-1}\right)\tag{5}\end{eqnarray*}$$
and by summation by parts:
$$ \sum_{k=1}^{N}2k\log\frac{2k+1}{2k-1} = 2N\log(2N+1)-2\sum_{k=1}^{N-1}\log(2k+1)\\=2N\log(2N+1)-2\log((2N-1)!!)\tag{6} $$
so $\int_{0}^{+\infty}f(x)\,dx=\log 2$, then $\color{red}{C=\frac{\pi}{2}}$, follow from Stirling's approximation. In order to compute
$$ 2iC=\int_{0}^{+\infty}\frac{e^{ix}-e^{-ix}}{x}\,dx\tag{6}$$
we may also use the complex version of Frullani's theorem, leading to:
$$ 2iC = \lim_{\varepsilon\to 0^+}\text{Log}\left(\frac{i+\varepsilon}{-i+\varepsilon}\right)=\pi i.\tag{7} $$

It is interesting to point out that $(5)$ gives a nice by-product, i.e.
$$\begin{eqnarray*} \log(2) = 1+\sum_{k\geq 1}\left(2-2k\cdot 2\,\text{arctanh}\left(\frac{1}{2k}\right)\right) &=& 1-\sum_{k\geq 1}\sum_{n\geq 1}\frac{2}{(2n+1)(2k)^{2n}}\\&=&1-\sum_{n\geq 1}\frac{2\,\zeta(2n)}{(2n+1)4^n}\tag{8}\end{eqnarray*}$$
that can also be derived from:
$$ \pi x \cot(\pi x)=1-\sum_{n\geq 1}2x^{2n}\zeta(2n)\tag{9} $$
by integrating both sides over the interval $\left(0,\frac{1}{2}\right)$:
$$ \int_{0}^{1/2}\pi x\cot(\pi x)\,dx=\frac{1}{\pi}\int_{0}^{\pi/2}\frac{z\cos(z)}{\sin z}\,dz \stackrel{IBP}{=} -\frac{1}{\pi}\int_{0}^{\pi/2}\log(\sin z)\,dz\tag{10} $$
where the last integral can be computed through Riemann sums (!!!) since
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n\pi}\right)=\frac{2n}{2^n}.\tag{11}$$
A: Here's one more, just for the fun of it. For $\theta$ not an integer multiple of $2 \pi$, we have
$$\sum \frac{e^{i n \theta}}{n} = -\log(1-e^{i \theta}).$$
Taking imaginary parts, for $0 < \theta < \pi$, we have
$$\sum \frac{\sin (n \theta)}{n} = -\mathrm{arg}(1-e^{i \theta}) = \pi/2-\frac{\theta}{2}.$$
Draw the isosceles triangle with vertices at $0$, $1$ and $e^{i \theta}$ to see the second equality.
So $\displaystyle \sum \theta \cdot \frac{\sin (n \theta)}{n \theta} = \pi/2-\frac{\theta}{2}$. The right hand side is a right-hand Riemann sum for $\int \frac{\sin t}{t} dt$, with intervals of width $\theta$. So, taking the limit as $\theta \to 0$, we get $$\int\limits_0^\infty \frac{\sin t}{t} dt=\frac{\pi}{2}$$.
A: The answer is correct.
A related technique. Recalling the Laplace transform

$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$
  We can use the following relation
$$ \begin{align}
 \int_0^\infty F(u)g(u) \, du  & = \int_0^\infty f(u)G(u) \, du \\[6pt]
 L[f(t)] & = F(s) \\[6pt]
 L[g(t)] & = G(s)\end{align} $$

Let 

$$ G(u)=\frac{1}{u} \implies g(u)=1, $$

and 

$$ f(u)= \sin(u)  \implies F(u) = {\frac {1}{ \left( {u}^{2}+1 \right) }}. $$

Now, 

$$ \int_0^\infty \frac{\sin u}{u} \, dx = \int_0^\infty \frac{1}{\left( {u}^{2}+1 \right)} \, du = \frac{\pi}{2}$$

A: I think we should have Euler's original proof (from E675, translation available here). We start with
$$ \int_0^{\infty} x^{n-1} e^{-x} \, dx = \Gamma(n). $$
Changing variables gives
$$ \int_0^{\infty} x^{n-1} e^{-kx} \, dx = \frac{\Gamma(n)}{k^n}. $$
Euler now assumes that this still works if $k=p \pm iq$ is complex, provided $p>0$. (It does, but this needs an application of Cauchy's theorem.) We then have
$$ \int_0^{\infty} x^{n-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(n)}{(p \pm iq)^n}, $$
and if we write $p=f\cos{\theta}$, $q=f\sin{\theta}$, we can apply Euler's formula to obtain
$$ \int_0^{\infty} x^{n-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(n)}{f^n}(\cos{n\theta} \mp i\sin{n\theta}). $$
Adding and subtracting gives us the two integrals
$$ \int_0^{\infty} x^{n-1} e^{-px} \cos{qx} \, dx = \frac{\Gamma(n)}{f^n}\cos{n\theta} \\
\int_0^{\infty} x^{n-1} e^{-px} \sin{qx} \, dx = \frac{\Gamma(n)}{f^n}\sin{n\theta} $$
The second is the one we care about: taking the limit as $n \to 0$, the left-hand side makes sense, and (here we deviate from Euler's infinitesimal discussion of $\Gamma$ and sine to avoid needless controversy) $ \Gamma(n)\sin{n\theta} \sim \frac{1}{n}n\theta = \theta $, so, converting back to $p$ and $q$, we find
$$ \int_0^{\infty} \frac{1}{x} e^{-px} \sin{qx} \, dx = \theta = \arctan{\frac{q}{p}} $$
The result now follows almost as an afterthought, by putting $p=0$.
A: I would like to present yet another simple proof that goes through Fourier series. However, we will need the following theorem; we denote by $S_n(x;f)$ the $n$-th partial sum of the Fourier series of $f(\in L^1[-\pi,\pi]$ and $2\pi$-periodic) at $x$. Then:
Theorem. Riemann's principle of localization. If $f\in L^1[-\pi,\pi]$, then
$$
S_n(x,f)=\frac{1}{\pi}\int_{-\delta}^{\delta} f(x+t)\frac{\sin nt}{t}\, dt\, + o(1). \quad (\delta>0)
$$
Now, if we pick the function $f(x)\equiv 1$, then $S_n(x,f)\equiv 1$ for all $n,x$. Thus, by Riemann's principle of localization:
$$
1=\frac{1}{\pi}\int_{-\delta}^{\delta}\frac{\sin nt}{t}\, dt \, + o(1) = \frac{2}{\pi}\int_0^{n\delta} \frac{\sin t}{t}\, dt + o(1).
$$
Letting $n\to \infty$, we get
$$
1=\frac{2}{\pi}\int_0^{\infty}\frac{\sin t}{t}\, dt,
$$
which yields the desired result.
A: Another approach is to employ Laplace Transforms. 
$$I = \int_{0}^{\infty}\frac{\sin(x)}{x}\, \mathrm dx.$$
Let 
$$I(t) = \int_{0}^{\infty}\frac{\sin(xt)}{x} \,\mathrm dx$$
Take the Laplace Transform to yield 
\begin{align*}
\mathscr L[I(t)] &= \int_{0}^{\infty}\frac{\mathscr L[\sin(tx)]}{x}\,\mathrm dx\\ 
&= \int_{0}^{\infty}\frac{ 1}{x}\frac{x}{s^2 + x^2}\,\mathrm dx\\ 
&= \int_{0}^{\infty}\frac{1}{x^2 + s^2}\,\mathrm dx \\
&= \left[\frac{1}{s}\arctan\left(\frac{x}{s} \right) \right]_{0}^{\infty} \\
&= \frac{1}{s}\frac{\pi}{2}
\end{align*}
And so, to solve $I(t)$ we take the inverse Laplace transform:
\begin{align*}
I(t) &= \mathscr L^{-1}\left[\frac{1}{s}\frac{\pi}{2} \right] = \frac{\pi}{2}.1 = \frac{\pi}{2}
\end{align*}
Thus, 
$$\int_{0}^{\infty} \frac{\sin(x)}{x}\mathrm dx = I(1) = \frac{\pi}{2}$$
A: One easiest way to get this integral is to evaluate the following improper integral with parameter $a$:
$$ I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}dx, a\ge 0.$$
It is easy to see
$$I'(a)=-\int_0^\infty e^{-ax}\sin xdx=\frac{e^{-ax}}{a^2+1}(a\sin x+\cos x)\big|_0^\infty=-\frac{1}{a^2+1}.$$
Thus
$$I(\infty)-I(0)=-\int_0^\infty\frac{1}{a^2+1}da=-\frac{\pi}{2}.$$
Note $I(\infty)=0$ and hence $I(0)=\frac{\pi}{2}$.
A: To express the integral in terms of the series expansion coefficients can be done by invoking Glaisher's theorem, which is a special case of Ramanujan's master theorem. If $f(x)$ is an even function such that $\int_0^{\infty}f(x) dx$ exists, and the series expansion $f(x) = \sum_{0}^{\infty}(-1)^n c_n x^{2n}$ is valid in a neighborhood of $x = 0$, then we have:
$$\int_0^{\infty} f(x)dx = \frac{\pi}{2}c_{-\frac{1}{2}}\tag{1}$$
where $c_{-\frac{1}{2}}$ is to be interpreted as an analytic continuation of $c_n$ obtained by replacing factorials by gamma functions. Glaisher derived his theorem in a non-rigorous way, later Ramanujan formulated his master theorem, which was later rigorously proven by Hardy. That rigorous proof then does involve the residue theorem, but the arguments by Ramanujan and the earlier arguments by Glaisher don't involve any complex analysis.
Also while superficially the formula for the integral looks analogous to what you can get from applying the residue theorem, i.e. that a real integral is proportional to an expansion coefficient, unlike the reside theorem there is now no contour in the complex plane to consider over which the integral has to vanish. So, even if there exist no contour for evaluating the integral using the residue theorem, as long as you got analytic expressions for the expansion coefficients, you'll be able to write down the integral.
It's easy to justify the formula on heuristic grounds, an argument similar to the original argument by Glaisher works as follows. One introduces the operator $E$ that acts on the expansion coefficients as:
$$E c_n = c_{n+1}\tag{2}$$
This then allows one to write $f(x)$ formally as:
$$f(x) = \sum_{n=0}^{\infty}(-1)^n c_n x^{2n} =  \sum_{n=0}^{\infty}(-1)^n E^n c_0 x^{2n} = \frac{c_0}{1+E x^2}$$
Then if $E$ were a positive number, we would have:
$$\int_0^{\infty} \frac{c_0}{1+E x^2}dx = \frac{\pi}{2}E^{-\frac{1}{2}}c_0$$
One then assumes that this relation will still hold for $E$ the operator defined by Eq. (2), which implies the result given by Eq. (1).
For $f(x)=\dfrac{\sin(x)}{x}$, the $c_n$ are given by:
$$c_n = \frac{1}{(2n+1)!}$$
therefore $c_{-\frac{1}{2}} = 1$, the integral is therefore equal to $\dfrac{\pi}{2}$.
A: Here is a sketch of another elementary solution based on a proof in Bromwich's Theory of Infinite Series.
Using $\sin(2k+1)x-\sin(2k-1)x = 2\cos2kx\sin x$ and summing from k=1 to k=n we have
$$\sin(2n+1)x = \sin x \left( 1+ 2 \sum_{k=1}^n \cos 2kx \right),$$
and hence
$$ \int_0^{\pi/2} {\sin(2n+1)x \over \sin x} dx = \pi/2. \qquad (1)$$
Let $y=(2n+1)x$ and this becomes
$$ \int_0^{(2n+1)\pi/2} {\sin y \over (2n+1) \sin (y/(2n+1))} dy = \pi/2.$$
and since $\lim_{n \to \infty} (2n+1) \sin { y \over 2n+1} = y$ it suggests that there is a proof lurking in there somewhere.
So let's put
$$\begin{align}
I_n &=  \int_0^{n\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx
\ &= \sum_{k=0}^{n-1} \int_{k\pi/(2n+1)}^{(k+1)\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx.
\end{align}$$
Hence we have $I_n = u_0 – u_1 + u_2 \cdots + (-1)^{n-1}u_{n-1},$ where $u_k$ is a decreasing sequence of positive terms. We can see this from the shape of the curve
$y = \sin(2n+1)x / \sin x,$ which crosses the x-axis at $\pi/(2n+1), 2\pi/(2n+1),\ldots,n\pi/(2n+1).$ (I said that this is just a sketch, you have to check the details.)
Hence the sequence $I_n$ converges, and by (1) it converges to $\pi/2.$
Now if we make the substitution $y=(2n+1)x$ we see that
$$u_k = \int_{k\pi}^{(k+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy,$$
and since $I_n$ can be written as an alternating sequence of decreasing positive terms we can truncate the sequence wherever we like and the value of $I_n$ lies between two successive partial summations. Hence
$$ \int_{0}^{2m\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy < I_n <
 \int_{0}^{(2m+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy. \qquad (2)$$
for any m such that $2m+1 \le n.$ (Take $m=[\sqrt{n}],$ say, $n \ge 6.$)
Now $$\left| { \sin y \over y} - {\sin y \over (2n+1) \sin(y/(2n+1))} \right| 
< { \pi^2(2m+1)^2 \over 3(2n+1)^2}$$
and so this difference tends to zero uniformly in the interval $0 \le y \le (2m+1)\pi$
and so by taking the $\lim_{n \to \infty}$ in (2) we obtain
$$\int_0^{\infty} { \sin x \over x } dx = { \pi \over 2}.$$
A: Let
$$I=\int_0^\infty \frac{\sin x} x \, dx.$$
By the Schwinger parametrization we get
$$ I= \int_0^\infty \mathrm{d}t\int_0^\infty \sin{x}\exp(-t x)\, \mathrm{d}x.$$
The last integral can be evaluated by parts. Another simple way is using $\sin{x}=\Im\left[ \mathrm{e}^{-i x} \right]$:
$$\int_0^\infty \sin{x}\exp(-t x)\, \mathrm{d}x=\Im\int_0^\infty \mathrm{e}^{-(t-i)x}\, \mathrm{d}x=\frac{1}{1+t^2}.$$
Thereby,
$$I=\int_0^\infty \frac{\mathrm{d}t}{1+t^2}.$$
Here you can use again the Schwinger trick. However, 
$$\frac{\mathrm{d}\arctan(x)}{\mathrm{d}x}=\frac{1}{1+x^2},$$
Shuch that

$$\int_0^\infty \frac{\sin x} x \, dx=\arctan(\infty)-\arctan(0)=\frac{\pi}{2}$$

A: Here's a "proof" I came up with when asked to use this integral for a physics homework problem in undergrad (I think we were expected to cite wolfram alpha). Perhaps worth adding for the novelty, if not the rigor, I think:
Define 
$$I(a) = \int_0^\infty \frac{\sin(2\pi ax)}{x}dx =\frac{i}{2} \int_{-\infty}^\infty \frac{e^{-2\pi iax}}{x} dx$$
for $a \in \mathbb{R}$. Then 
$$I'(a)=\pi \int_{-\infty}^\infty e^{-2\pi iax} dx  =\pi \delta(a)$$
Where $\delta(x)$ is the Dirac delta "function". Integrating,
$$I(a)=\pi \theta(a)+C$$
Where $\theta$ is the Heaviside function, for some $C \in \mathbb{R}$. Since $I(-a)=-I(a)$ from its definition, we must have for $a>0$ that
$$\pi+C = I(a)=-I(-a)=-C $$
(since $\theta(a)=1$ and $\theta(-a)=0$). So $C=-\frac{\pi}{2}$, and our integral is
$$I \left( \frac{1}{2 \pi} \right)=\pi-\frac{\pi}{2}=\frac{\pi}{2}$$
Perhaps someone knows a way to interpret this rigorously in terms of distributional derivatives and what have you, but I'm not there yet.
A: Here is an unusual approach using what physicists have recently called Integration by differentiation. Full justification for the results quoted below can be found in the following paper.
Define $\partial_y := \frac{d}{dy}$. The following integration by differentiation method has been introduced (there are others, see paper in the link for details)
$$\int_0^\infty f(x) \, dx = \lim_{y \to 0^+} f(-\partial_y) \frac{1}{y}.\tag1$$
Something else we need is the following result. For a smooth function $f$
$$e^{a \partial_y} f(y) = f(y + a), \quad a \in \mathbb{C}. \tag2$$
Here $e^{a \partial_y}$ acts as a translation (shift) operator.
We are now in a position to proceed. Setting $f(x) = \sin x/x$ we have
\begin{align*}
\int_0^\infty \frac{\sin x}{x} dx &= \lim_{y \to 0^+} \frac{\sin (-\partial_y)}{-\partial_y} \frac{1}{y} \tag3\\
&= \lim_{y \to 0^+} \frac{\sin (\partial_y)}{\partial_y} \frac{1}{y}\\
&= \lim_{y \to 0^+} \frac{e^{i \partial_y} - e^{-i \partial_y}}{2i \partial_y} \frac{1}{y} \tag4\\
&= \frac{1}{2i} \lim_{y \to 0^+} \left (e^{i \partial_y} - e^{-i \partial_y} \right ) \log (y) \tag5\\
&= \frac{1}{2i} \lim_{y \to 0^+} \left (\log(y + i) - \log (y - i) \right ) \tag6\\
&= \frac{1}{2i} \lim_{y \to 0^+} \log \left (\frac{y + i}{y - i} \right ) \\
&= \lim_{y \to 0^+} \operatorname{arccot} (y) = \frac{\pi}{2},
\end{align*}
as expected.
Explanation
(3) Application of the integration by differentiation result given in (1).
(4) Expressed the sine function in terms of complex exponentials using: $\sin z = (e^{iz} - e^{-iz})/(2i)$,
(5) The antiderivative (the term $\frac{1}{\partial_y}$ is an antiderivative) of $\frac{1}{y}$ is $\log (y)$.
(6) Used the shift operator result given in (2).
A: By definition (Laplace Transform):
\begin{equation*}
            F(s)=L\left[\frac{\sin(t)}{t}\right]=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\arctan^{-1}\left(\frac{1}{s}\right)
\end{equation*}
Then, for $s=0$,
\begin{equation*}
            F(0)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt = \lim_{s\to0}\arctan^{-1}\left(\frac{1}{s}\right)=\lim_{u\to\infty}\arctan^{-1}(u)=\frac{\pi}{2}
\end{equation*}
A: Let's consider the integrals
$$I_1(t)=\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\mbox{ and }\qquad
I_2(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx,\qquad t\geq 0.$$
A direct calculation shows that $I_1(t)$ and $I_2(t)$ satisfy the ordinary differential equation
$$y''+y=\frac{1}{t},\qquad t>0.$$
Therefore, the difference $I(t)=I_1(t)-I_2(t)$ satisfy the homogeneous differential equation
$$y''+y=0,\qquad t>0,$$
hence it should be of the form 
$$I(t)=A\sin (t+B) $$
with some constants $A$, $B$. But  $I_1(t)$ and $I_2(t)$ both converge to $0$ as $t\to\infty$. This implies that $A=0$ and $I_1(t)=I_2(t)$ for all $t\geq 0$. Finally, we have that
$$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\lim_{n\to\infty}\left(\arctan(n)\right)-\arctan(0)=\frac{\pi}{2}.$$
A: I believe this can also be solved using double integrals.
It is possible (if I remember correctly) to justify switching the order of integration to give the equality:
$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$
This leads us to
$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.
$$\begin{align*}
I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\
&= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\
&= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.
\end{align*}$$
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$
A: Maybe I'm flogging a dead horse, but nobody has mentioned the standard suspiciously circular (see the comments) Fourier analytic proof yet:
Let $f(t)=1$ for $|t|<1$ and 0 otherwise. Then the Fourier transform is
$$ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt
= \int_{-1}^{1} e^{-i\omega t} dt
= \frac{e^{-i\omega} - e^{i\omega}}{-i\omega}
= \frac{2\sin\omega}{\omega}.$$
Fourier's inversion formula states that
$$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega $$
if $f$ is (say) differentiable at $t$.
In our case, we get in particular that
$$ 1 = f(0)
= \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega
= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2\sin\omega}{\omega} d\omega
= \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin\omega}{\omega} d\omega.
$$
(EDIT: Even if this is not really a proof, it's still a good thing to be aware of, since one can use similar ideas to integrate powers of $\sin\omega/\omega$, or integrals like these.)
A: \begin{align}
\int_{-\infty}^{\infty}{\sin(x) \over x}
\,{\rm d}x & =
\int_{-\infty}^{\infty}
\left({1 \over 2}\,\int_{-1}^{1}
{\rm e}^{{\rm i}kx}\,{\rm d}k\right)
\,{\rm d}x
\\[5mm] & =
\pi\int_{-1}^{1}\
\int_{-\infty}^{\infty}{\rm e}^{{\rm i}kx}\,{{\rm d}x \over 2\pi}\,\,{\rm d}k
\\[5mm] & = 
\pi\int_{-1}^{1}\delta(k)\,{\rm d}k = \pi
\end{align}
A: Another "proof" using distributions and Fourier transforms.
We treat the integral as a sine transform, rewrite it as a Fourier transform and use the fact that the Fourier transform of $\frac{1}{x}$ is $-i\pi \, \operatorname{sign}(\xi)$:
$$
\int_0^\infty \frac{\sin x}{x} dx 
= \frac{i}{2} \int_{-\infty}^{\infty} \frac{1}{x} e^{-ix} \, dx
= \frac{i}{2} \left. \int_{-\infty}^{\infty} \frac{1}{x} e^{-i\xi x} \, dx \right|_{\xi=1}
= \frac{i}{2} \left( -i\pi \, \operatorname{sign}(1) \right)
= \frac{\pi}{2}
.
$$
A: Define, for $\displaystyle x\geq 0,F(x)=\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\cos(x\cos \theta)\,d\theta$
Claims:
1)$\displaystyle F(x)=\frac{\pi}{2}-\int_0^x\frac{\sin \theta}{\theta}\,d\theta$
2)$\displaystyle \lim_{x\rightarrow\infty}F(x)=0$
Proof:

*

*Observe that,
\begin{align}
F(0)&=\frac{\pi}{2}\\
 F^\prime(x)&=-\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\sin(\theta+x\cos \theta)\,d\theta\\
&=\left[\frac{\text{e}^{-x\sin \theta}\sin(x\cos \theta)}{x}\right]_{\theta=0}^{\theta=\frac{\pi}{2}}\\
&=-\frac{\sin x}{x}
\end{align}
2)Observe that, for $\displaystyle 0\leq \theta\leq\frac{\pi}{2},0\leq \sin\theta<1$
For $x>0,0<a<1$,
\begin{align}\left|F(x)\right|&\leq \int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\,d\theta\\
\int_0^{\frac{\pi}{2}}\text{e}^{-x\sin \theta}\,d\theta&\overset{u=\sin\theta}=\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\\
&=\int_0^a \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du+\int_a^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\\
&\leq \int_0^a \frac{1}{\sqrt{1-u^2}}du+\text{e}^{-ax}\int_0^1 \frac{1}{\sqrt{1-u^2}}du\\
\end{align}
Therefore,
For all $0<a<1$,
$\displaystyle \lim_{x\rightarrow +\infty}\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du\leq \int_0^a \frac{1}{\sqrt{1-u^2}}du$ that is, $\displaystyle \lim_{x\rightarrow +\infty}\int_0^1 \frac{\text{e}^{-xu}}{\sqrt{1-u^2}}du=0$
(Thanks to Zephir for the help)
Thus,
$\displaystyle \lim_{x\rightarrow \infty}F(x)=0$ and,
$\boxed{\int_0^\infty\frac{\sin \theta}{\theta}\,d\theta=\frac{\pi}{2}}$
NB: I assume that $\displaystyle \int_0^\infty\frac{\sin \theta}{\theta}\,d\theta$ is converging (edit: no need to assume this)
Inspired by,
Uebungsbuch zum Studium der höheren Analysis
, a book by Oscar Schlömilch
A: I evaluated this integral in this answer where I started with
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(2kx)}{k}
&=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\
&=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\
&=\frac1{2i}\log(-e^{-i2x})\\[4pt]
&=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right)\tag{1}
\end{align}
$$
Setting $x=\frac a2$, we get
$$
\sum_{k\in\mathbb{Z}}\frac{\sin(ka)}{ka}=\frac\pi a\tag{2}
$$
where we set $\frac{\sin(ka)}{ka}=1$ when $k=0$. Multiplying $(2)$ by $a$ and setting $a=\frac1n$ yields
$$
\sum_{k\in\mathbb{Z}}\frac{\sin(k/n)}{k/n}\frac1n=\pi\tag{3}
$$
and $(3)$ is a Riemann Sum for the integral
$$
\int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{4}
$$
A: These proofs looked very intriguing the multiple ways to go about the same problem. I looked up toward the ceiling and then it dawned on me that there was another way to do this with this particular function as follows: 
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$The method of attack of use would be Laplace Transforms
$$f(t)=\dfrac{\sin(t)}{t}$$
$$ \lim_{t \to 0} ~ \dfrac{f(t)}{t} ~;~ \text{exist and is a finite number.}$$
$${\cal L} \left\{ \frac{\sin(t)}{t} \right\}=\int_0^\infty \! {\cal L}
\left\{ \sin(t) \right\} ~ \mathrm{d} \sigma=\int_0^\infty\! \frac{1}{\sigma^2+1} \mathrm{d} \sigma=\tan^{-1}(\sigma) ~ {\LARGE|_{\sigma=0}^{\sigma=\infty}}=\frac{\pi}{2}- \arctan(0)$$
So we see that we get the result of: $\dfrac{\pi}{2}~~~$ $\Big(\because~\arctan(0)=0 ~\Big)$.
A: Note: Laplace transforms, $$\int_{0}^{\infty}e^{-st}f(t)dt=L[f(t)]$$ $$L\left[\frac{f(t)}{t}\right]=\int_{s}^{\infty}L[f(t)]\ ds$$ & $$L[\sin t]=\frac{1}{1+s^2}$$
Now, we have $$\int_{0}^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}e^{-(0)x} \frac{\sin x}{x}\ dx$$$$=L\left[\frac{\sin x}{x}\right]_{s=0}$$ $$=\int_{s=0}^{\infty}L\left[\sin x\right]\ ds$$ $$=\int_{s=0}^{\infty}\frac{1}{1+s^2}\ ds$$ $$=[\tan^{-1}(s)]_{0}^{\infty}$$$$=\tan^{-1}(\infty)-\tan^{-1}(0)$$$$=\frac{\pi}{2}$$ 
A: This one I found in The American Mathematical Monthly from 1951 in the article 'A simple evaluation of an improper integral' written by Waclaw Kozakiewicz.
Theorem (Riemann). If $f(x)$ is Riemann integrable in the interval $a \leq x \leq b$, then:
$$\lim_{k \to +\infty} \int_a^b f(x) \sin kx \; dx = 0 \;.$$
Next, notice that:
$$\int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}\; dx = \frac{\pi}{2} \; ,n = 0,1,2,\ldots \quad (1)$$
and let:
$$\phi(x) = \begin{cases} 0 & , \;x = 0 \\ \frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} =\frac{2 \sin \frac{x}{2} - x}{2x \sin \frac{x}{2}} & ,\; 0 < x \leq \pi \; . \end{cases}$$
Then $\phi(x)$ is continuous and satisfies Riemann theorem, so choosing $k = n + \frac{1}{2}$ we write:
$$\lim_{n \to +\infty}\int_0^{\pi} \left(\frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} \right) \sin \left(n+\frac{1}{2}\right)x \; dx = 0 \;.$$
But taking $(1)$ into account we have:
$$\lim_{n \to +\infty} \int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{x} \; dx = \frac{\pi}{2}\;.$$
Using substitution $u = \left(n+\frac{1}{2}\right)x$ and knowing that $\int_0^{+\infty} \frac{\sin x}{x} \; dx$ converges we finally have:
$$\int_0^{+\infty} \frac{\sin x}{x} \; dx = \lim_{n \to +\infty} \int_0^{\left(n+\frac{1}{2}\right)\pi}\frac{\sin u}{u} \; du = \frac{\pi}{2}\;.$$
A: We can decompose interval $[0,+\infty)$ into intervals of length $\frac{\pi}{2}$. Then we'll have:
$$I = \int_0^{+\infty} \frac{\sin x}{x} \,dx = \sum_{n=0}^{+\infty} \int_{n\pi / 2}^{(n+1)\pi / 2} \frac{\sin x}{x} \,dx$$
Now consider the case when $n$ is even i.e. $n=2k$ and substitute $x = k\pi + t$:
$$\int_{2k\pi /2}^{(2k+1)\pi / 2} \frac{\sin x}{x} \,dx = (-1)^k \int_0^{\pi/ 2} \frac{\sin t}{k\pi + t} \, dt$$
and for odd $n$ we have $n=2k-1$ and we use substitution $x = k\pi-t$:
$$\int_{(2k-1)\pi /2}^{2k \pi / 2} \frac{\sin x}{x} \,dx = (-1)^{k-1} \int_0^{\pi/ 2} \frac{\sin t}{k\pi - t} \, dt$$
Hence we obtain:
$$I = \int_0^{\frac{\pi}{2}} \sin t \cdot \left[ \frac{1}{t} + \sum_{k = 1}^{+\infty} (-1)^k \left( \frac{1}{t+k\pi} + \frac{1}{t-k\pi} \right)  \right] \, dt$$
But in square bracket we have expansion of $\frac{1}{\sin x}$ into partial fractions, hence the result follows:
$$I = \int_0^{\frac{\pi}{2}} dt = \frac{\pi}{2}$$
A: See http://en.wikipedia.org/wiki/Dirichlet_integral for a proof using differentiation under the integral sign.
A: I'd add here the Feynman way, a very powerful, elegant and fast method to work out such things. You find here the example from $-\infty$ to $\infty$, but since the integrand is even, by dividing the result by 2 we get our required result.
http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
A: I got linked to this old question from a more recent one, and I hope that you don't mind me adding a somewhat bizarre way of doing calculating this integral, using Bessel functions. 
I'm aware of that this way is not the shortest way of obtaining the result, and the facts I give on Bessel functions are standard, and can be found in (probably) any book on Bessel functions. Therefore, some details will be left to be checked by the interested reader. 
I have never seen this way to calculate the integral $\int_0^{+\infty}\frac{\sin x}{x}\,dx$, but I claim no originality. If someone has seen it, please tell in a comment. Here it goes:
Let us define the $n$th Bessel function $J_n$ by the integral
$$
J_n(x)=\frac{1}{\pi}\int_0^\pi \cos(n\theta-x\sin \theta)\,d\theta.
$$
We will only work with the cases $n=0$ and $n=1$. The function $J_n$ solves the Bessel differential equation
$$
x^2y''(x)+xy'(x)+(x^2-n^2)y(x)=0,
$$
and moreover $D J_0(x)=-J_1(x)$ and $D(xJ_1(x))=xJ_0(x)$.
Our first statement is that
$$
\frac{\sin x}{x}=\int_0^1\frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy.\tag{1}
$$
Indeed, define $f$ as
$$
f(x)=\int_0^1\frac{xy J_0(yx)}{\sqrt{1-y^2}}\,dy
$$
Then
$$
f'(x)=\int_0^1 \frac{yJ_0(xy)-xy^2 J_1(xy)}{\sqrt{1-y^2}}\,dy
$$
and
$$
f''(x)=-\int_0^1 \frac{xy^3J_0(xy)+y^2J_1(xy)}{\sqrt{1-y^2}}\,dy,
$$
and so
$$
f''(x)+f(x)=\int_0^1 xy\sqrt{1-y^2}J_0(xy)-\frac{y^2}{\sqrt{1-y^2}}J_1(xy)\,dy=0.
$$
(Here we integrated by parts in the last step.) Moreover,
$$
f(0)=0,\quad\text{and}\quad f'(0)=\int_0^1 \frac{y}{\sqrt{1-y^2}}\,dy =1.
$$
Thus $f(x)=\sin x$ and the equality (1) follows. Thus, we can write
$$
\int_0^{+\infty} \frac{\sin x}{x}\, dx 
= \int_0^{+\infty}\int_0^1 \frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy\, dx.
$$
Next, we change the order of integration, and use the integral
$$
\int_0^{+\infty} J_0(u)\,du=1,\tag{2}
$$
to find that
$$
\int_0^{+\infty} \frac{\sin x}{x}\, dx = \int_0^1 \frac{1}{\sqrt{1-y^2}}\,dy=\arcsin 1-\arcsin 0=\frac{\pi}{2}.
$$
I remains to prove (2), which certainly follows by letting $x\to 0^+$ in
$$
\int_0^{+\infty} J_0(u)e^{-xu}\,du=\frac{1}{\sqrt{1+x^2}}.
$$
This is just the Laplace transform of $J_0$, and one can use the representation $J_0(u)=\frac{2}{\pi}\int_0^{\pi/2} \cos(u\cos\theta)\,d\theta$ to obtain it,
$$
\begin{aligned}
\int_0^{+\infty}J_0(u)e^{-xu}\,du &= \int_0^{+\infty} e^{-xu}\frac{2}{\pi}\int_0^{\pi/2}\cos(u\cos\theta)\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2} \int_0^{+\infty}e^{-xu}\cos(u\cos\theta)\,du\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2}\frac{x}{x^2+\cos^2\theta}\,d\theta\\
&=\frac{2}{\pi}\biggl[\frac{\arctan\Bigl(\frac{x\tan \theta}{\sqrt{1+x^2}}\Bigr)}{\sqrt{1+x^2}}\biggr]_0^{\pi/2}\\
&=\frac{1}{\sqrt{1+x^2}}.
\end{aligned}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

By "closing" a contour in the first quadrant ( a quarter circle of radius $\ds{R}$ ): 

\begin{align}
\int_{0}^{R\ >\ 0}{\sin\pars{x} \over x}\,\dd x & =
\Im\int_{0}^{R}{\expo{\ic x} - 1\over x}\,\dd x
\\[5mm] & =
-\,\Im\
\overbrace{\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} - 1 \over R\expo{\ic\theta}}\,R\expo{\ic\theta}\ic\,\dd\theta}
^{\ds{\mbox{along the arc}}}\ -\
\Im\
\overbrace{\int_{R}^{0}{\expo{-y} - 1 \over \ic y}\,\ic\,\dd y}
^{\ds{\mbox{along the}\ y\ \mbox{axis}}}
\\[5mm] & =
-\,\Re\int_{0}^{\pi/2}\bracks{\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}} - 1}\,\dd\theta
\\[5mm] & =
{\pi \over 2} -
\Re\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}}\,\dd\theta = \bbx{\pi \over 2}
\end{align}

Note that

\begin{align}
0 & < \verts{\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}}\,\dd\theta} <
\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta
\\[5mm] & <
\int_{0}^{\pi/2}\exp\pars{-\,{2R \over \pi}\,\theta}\,\dd\theta =
{\expo{-R} - 1 \over -2R/\pi}\,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\Large\to}\,\,\,{\large 0}
\end{align}
A: $$
\begin{aligned}
\text{Let }I(a)&=\int_{0}^{\infty} \frac{e^{-a x} \sin x}{x} d x \text {, where } a>0 \\
I^{\prime}(a)&=\int_{0}^{\infty} \frac{-x e^{-a x} \sin x}{x} d x \\
&=-\int_{0}^{\infty} e^{-a x} \sin x d x \\
&\stackrel{\text { IBP }}{=}-\frac{1}{a^{2}+1}
\end{aligned}
$$
Integrating w.r.t. $a$ yields
$$
\begin{aligned}
I(a) &=-\int \frac{d a}{a^{2}+1}+C=-\tan ^{-1} a+C
\end{aligned}
$$
Letting $a$ tends to infinity yields $C=\dfrac{\pi}{2}. $
$$
\therefore I(a)=-\tan ^{-1} a+\frac{\pi}{2}
$$$$
\text {Hence } \boxed{ \int_{0}^{\infty} \frac{\sin x}{x} d x=I(0)=\frac{\pi}{2}}
$$
