Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

Question

A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$$

Well, can anyone prove this without using Residue theory. I actually thought of doing this: $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \lim_{t \to \infty} \int_{0}^{t} \frac{1}{t} \Bigl( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \Bigr) \ dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.

Answer 1

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 $$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$

Answer 2

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 $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$

Answer 3

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 $$\int_0^{\infty} { \sin x \over x } dx = { \pi \over 2}.$$

Answer 4

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 $$\int\limits_0^\infty \frac{\sin t}{t} dt=\frac{\pi}{2}$$.

Answer 5

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.)

Answer 6

See http://en.wikipedia.org/wiki/Dirichlet_integral for a proof using differentiation under the integral sign.

Answer 7

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 $$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\arctan(\infty)-\arctan(0)=\frac{\pi}{2}.$$

Answer 8

In my Advanced Calculus book (Angus Taylor) it is shown that, if $a\gt 0$,

$\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt=\arctan\dfrac{x}{a}\qquad (1)$

If $x>0$,

$\displaystyle\int_0^{\infty}\dfrac{\sin xt}{t}dt=\dfrac{\pi}{2}\qquad (2)$

follows from $(1)$, observing that the integrand is $G(0)$ for

$G(a)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt\qquad (3)$,

$G$ is uniformly convergent when $a\ge 0$, and $G(a)$ approaches $G(0)$ as $a$ tends to $0^+$.

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Answer to Qiaochu: $(1)$ is proved as an application of the following theorem 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 $t,x$ for $c\lt t$,$a\lt x\lt b$, and let $\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$ converge uniformly on $[a,b]$. Then $F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$".

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PS. It was a surprise to me to see the evaluation of $F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt$:

$F(x)=\dfrac{\pi}{2}\qquad$ if $x>0$,

$F(x)=-\dfrac{\pi}{2}\qquad$ if $x<0$,

$F(x)=0\qquad\qquad$ if $x=0$.

Answer 9

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

$$f(t)=\dfrac{\sin(t)}{t}$$

$$ \lim_{t \to 0} ~ \dfrac{f(t)}{t} ~;~ \text{exist and is a finite number.}$$

$~~~~~\cal{L} \left\{ \dfrac{\sin(t)}{t} \right\}=\displaystyle\int_0^{\infty} \! \cal{L} \left\{ \sin(t) \right\} ~ \mathrm{d} \sigma=\int_0^{\infty}\! \dfrac{1}{\sigma^{2}+1} \mathrm{d} \sigma=\tan^{-1}(\sigma) ~ {\LARGE|_{\sigma=0}^{\sigma=\infty}}=\frac{\pi}{2}-$ arctan($0$)

So we see that we get the result of: $\dfrac{\pi}{2}~~~$ $\Big(\because~$arctan($0)=0 ~\Big)$.

Answer 10

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: $$\int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)\pi}{2 \sin \frac{x}{2}}\; dx = \frac{\pi}{2} \; ,n = 0,1,2,\ldots \quad (1)$$ 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:

$$\int_0^{+\infty} \frac{\sin x}{x} \; dx = \lim_{n \to +\infty} \int_0^{\left(n+\frac{1}{2}\right)\pi}\frac{\sin u}{u} \; du = \frac{\pi}{2}\;.$$

Answer 11

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.

Answer 12

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

Answer 13

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: $$I = \int_0^{\frac{\pi}{2}} dt = \frac{\pi}{2}$$

Answer 14

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 $I(0)=\frac{\pi}{2}$.