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: $$\int_0^\infty \frac{\sin x} x \, dx = \lim_{t \to \infty} \int_0^t \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \, dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.

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    $\begingroup$ note that from $\int\limits_0^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$, we can get $\int_0^\infty\frac{\sin(x^n)}{x}dx=\frac{\pi}{2n}$ by a simple change of variables $\endgroup$ – user49084 Nov 12 '12 at 2:04
  • $\begingroup$ @Chappers are you sure $\int_0^t t \mathrm{d}t$ is a correct integral... $\endgroup$ – Hexacoordinate-C Aug 20 '15 at 15:24
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    $\begingroup$ Since no one has mentioned them yet, G.H. Hardy wrote two articles about approximately 12 different ways of doing this integral: in 1909 and Math. Gaz. 8 (July 1916) pp. 301–303., although the latter is not very easy to find online. Both are available in his Collected Works and The G. H. Hardy Reader. $\endgroup$ – Chappers Feb 15 '17 at 23:56

28 Answers 28


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


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

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    $\begingroup$ @Americo: I heard of this one a long time back from one of my teachers. I thought this will be well known, but I guess I could be mistaken about that. $\endgroup$ – Aryabhata Sep 23 '10 at 16:26
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    $\begingroup$ This is also the technique used in R. Durrett (2005), Probability theory and examples, 3rd ed., Duxbury, p. 470. It is Exercise 6.6 on that page. The justification of exchanging the order of integration actually comes from considering the strip $(0,a) \times (0,\infty)$ and observing that $\int_0^a \int_0^\infty |e^{-xy} \sin x| \,\mathrm{d} y\,\mathrm{d} x \leq a$, from whence Fubini's theorem can be applied. To get the result, we take $a \to \infty$. $\endgroup$ – cardinal Sep 18 '11 at 12:09
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    $\begingroup$ this methods is elegant,and I computer the integral $\int e^{-xy}\sin x \text{dx}$: $$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big) \end{align*} $$ $\endgroup$ – Laura Feb 4 '13 at 8:30
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    $\begingroup$ @Ale: As regards your comment, yes, Fubini is applied correctly as stated in my comment. For each fixed $a$, the associated integral is bounded. Hence, Fubini can be used to compute the integral in two ways. Rearranging gives you a bound for $|\int_0^a \frac{\sin x}{x} - \frac{\pi}{2}|$ as a function of $a$. Then, take $a \to \infty$ to get the result. (Note that some argument like this is necessary since $\frac{\sin x}{x}$ is not integrable on $(0,\infty)$ in the Lebesgue sense.) Hope this helps. Cheers. :-) $\endgroup$ – cardinal Feb 17 '15 at 2:07
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    $\begingroup$ This method is basically the same as the usage of an inverse and a regular Laplace transform. The required substitution can be obtained systematically through contour integrals, but usually the point of using it to avoid complex integration... $\endgroup$ – 1010011010 Dec 23 '15 at 20:53

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

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    $\begingroup$ Sorry for digging out a 4 years old post, but how does one justify that the limit of the sum is actually the integral we are interested in? I only know that this kind of Riemann sums works for bounded intervals and I'm not convinced this works for improper integrals. Is there a general result which justifies this? $\endgroup$ – Wojowu Mar 26 '16 at 20:14

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

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    $\begingroup$ +1 This is esentially the Feynman method mentioned in "Chris's sis" answer; in the linked pdf there a justification for deriving under the integral is provided. $\endgroup$ – leonbloy Aug 10 '13 at 0:48
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    $\begingroup$ Why does $I(\infty) = 0$? Instead of $I(\infty) - I(0) = ...$, shouldn't it be $I(\infty) - \lim_{ a \rightarrow 0^{+} }I(a) = ...$ because you haven't shown $I(0)$ converges? So it still remains to show $\lim_{ a \rightarrow 0^{+} }I(a) = I(0)$? $\endgroup$ – LucasSilva May 20 '15 at 5:53
  • $\begingroup$ @LucasSilva, I omitted the details. $\endgroup$ – xpaul May 20 '15 at 15:22
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    $\begingroup$ @xpaul : The issue with $I(0)$ is still unresolved. The issue also exists on the wikipedia page: en.wikipedia.org/wiki/… $\endgroup$ – LucasSilva May 20 '15 at 15:37
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    $\begingroup$ The issue also exists in the link referred to by @leonbloy : ocw.mit.edu/courses/mathematics/… $\endgroup$ – LucasSilva May 20 '15 at 16:28

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


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

  • $\begingroup$ Where does $I_2$ come from? $\endgroup$ – J. M. is a poor mathematician Oct 30 '10 at 14:09
  • $\begingroup$ @ J.M.: I think one could arrive at the idea by inspecting the integral $I$ which appears in Moron's solution. $\endgroup$ – Andrey Rekalo Oct 30 '10 at 14:30
  • $\begingroup$ Why does $I_1(t)$ go to $0$ as $t$ goes to $\infty$? $\endgroup$ – LucasSilva May 20 '15 at 5:44

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.)

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    $\begingroup$ Often proofs of the Fourier inversion theorem use the value of the sine integral. Certainly the one I learned as an undergraduate did. To me this is reminiscent of the argument that $\lim_{x\to0}(\sin x)/x=1$ by L'Hopital's rule :-( $\endgroup$ – Robin Chapman Oct 13 '10 at 10:14
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    $\begingroup$ @Robin Chapman: Hmm, that's true. Very good point. Maybe that's why nobody gave this answer! PS. You need to get rid of the reflex to hit the Return key before you're done writing your comments. :) $\endgroup$ – Hans Lundmark Oct 13 '10 at 10:33
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    $\begingroup$ +1, because posts like these make me want to properly learn fourier analysis as soon as possible. Ps. the proof of the inversion formula in Rudin's book doesn't get anywhere near of making use of the value in this integral, as far as I can remeber. $\endgroup$ – Sam Apr 22 '11 at 14:00
  • $\begingroup$ @RobinChapman Can you point me to a reference to a proof the Fourier inversion theorem that uses the value of the sine integral? $\endgroup$ – LucasSilva May 20 '15 at 17:26
  • $\begingroup$ @LucasSilva: Robin hasn't been active on this site for several years (unfortunately). $\endgroup$ – Hans Lundmark May 20 '15 at 20:05

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 $$ \int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{4} $$


$$ \int_{-\infty}^{\infty}{\sin(x) \over x}\,{\rm d}x = \int_{-\infty}^{\infty}\left\lbrack{1 \over 2}\,\int_{-1}^{1}{\rm e}^{{\rm i}kx}\,{\rm d}k\right\rbrack \,{\rm d}x = \pi\int_{-1}^{1}{\rm d}k \int_{-\infty}^{\infty}{{\rm d}x \over 2\pi}\,{\rm e}^{{\rm i}kx} = \pi\int_{-1}^{1}{\rm d}k\,\delta(k) = \pi $$


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 $$\int_{0}^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}e^{-(0)x} \frac{\sin x}{x}\ dx$$$$=L\left[\frac{\sin x}{x}\right]_{s=0}$$ $$=\int_{s=0}^{\infty}L\left[\sin x\right]\ ds$$ $$=\int_{s=0}^{\infty}\frac{1}{1+s^2}\ ds$$ $$=[\tan^{-1}(s)]_{0}^{\infty}$$$$=\tan^{-1}(\infty)-\tan^{-1}(0)$$$$=\frac{\pi}{2}$$


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


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

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

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

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    $\begingroup$ How is $\cal{L}\{\frac{\sin(t)}{t}\}=\int^{\infty}_0 \cal{L}\{\sin(t)\}d\sigma$? And also, how did you get a $\sigma$ there? $\endgroup$ – Aditya Agarwal Dec 30 '15 at 9:34

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:

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

  • $\begingroup$ Is it then $\sin((n+1/2) x)$ in (1) instead of the $\pi$ in it ? $\endgroup$ – user42761 Jun 25 '13 at 13:32

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


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.



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


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, $$ \begin{aligned} \int_0^{+\infty}J_0(u)e^{-xu}\,du &= \int_0^{+\infty} e^{-xu}\frac{2}{\pi}\int_0^{\pi/2}\cos(u\cos\theta)\,d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2} \int_0^{+\infty}e^{-xu}\cos(u\cos\theta)\,du\,d\theta\\ &=\frac{2}{\pi}\int_0^{\pi/2}\frac{x}{x^2+\cos^2\theta}\,d\theta\\ &=\frac{2}{\pi}\biggl[\frac{\arctan\Bigl(\frac{x\tan \theta}{\sqrt{1+x^2}}\Bigr)}{\sqrt{1+x^2}}\biggr]_0^{\pi/2}\\ &=\frac{1}{\sqrt{1+x^2}}. \end{aligned} $$

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    $\begingroup$ Just the other day I stumbled upon the fact that $\frac{\sin x}{x}=\int_0^1\frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy $ when I was playing around with the Abel transform, and I immediately wondered if anyone had ever thought of using it to evaluate the Dirichlet integral. (+1) $\endgroup$ – Random Variable May 13 '17 at 2:15

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.

  • 1
    $\begingroup$ You really helped me with this. I'm a math freshman and currently have only studied basic academic math courses. This really helped me proof this for my homework. :) $\endgroup$ – Ory Band May 5 '12 at 16:02
  • 1
    $\begingroup$ Aren't there a few $dx$s and $dt$s missing? $\endgroup$ – JMCF125 Apr 5 '14 at 12:26

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 $$F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt.$$

  • $\begingroup$ Yes, but how is 1) proven? $\endgroup$ – Qiaochu Yuan Sep 22 '10 at 20:15
  • 1
    $\begingroup$ @Qiaochu Yuan: Agree, but it is contained in en.wikipedia.org/wiki/Dirichlet_integral too. :) $\endgroup$ – AD. Sep 22 '10 at 20:34
  • $\begingroup$ @Qiaochu Yuan: I deleted my original comment due to some errors and incorporated it in my answer. $\endgroup$ – Américo Tavares Sep 22 '10 at 20:42
  • $\begingroup$ yes, that is the same method Rasmus describes. $\endgroup$ – Qiaochu Yuan Sep 22 '10 at 20:52

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 $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n\pi}\right)=\frac{2n}{2^n}.\tag{11}$$

  • $\begingroup$ proving the partial fraction decomposition of $\frac{1}{\sin(z)}$ is supposed to be elementary :) ? $\endgroup$ – reuns Jul 3 '16 at 14:01
  • $\begingroup$ @user1952009: well, it is not terribly difficult to show that $\frac{1}{\sin z}$ has simple poles at $\pi\mathbb{Z}$ and the residues match the residues of $\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{z-\pi n}$ at the same points. $\endgroup$ – Jack D'Aurizio Jul 3 '16 at 14:06
  • $\begingroup$ @user1952009: anyway, it can also be derived from the Weierstrass product of the sine function through logarithmic differentiation. $\endgroup$ – Jack D'Aurizio Jul 3 '16 at 14:07
  • $\begingroup$ yes, the only way I know for all this is proving $\frac{1}{\sin(x)}$ or $\frac{1}{e^x-1}$ or $\tan(x)$ minus their poles are bounded entire functions, so by Liouville it reduces to a constant, and for $1/\sin(x)$ the constant should be easy to find $\endgroup$ – reuns Jul 3 '16 at 14:18
  • 1
    $\begingroup$ oh ok got it $\sum_{n=-N}^N \frac{(-1)^n}{2n-1} = 1+ \frac{(-1)^n}{-2N-1}$, tks didn't notice $\endgroup$ – reuns Jul 3 '16 at 14:30

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

$$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$


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


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


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

  • 4
    $\begingroup$ @user1729: I don't think there is anything wrong with adding a 17th answer, so long as it contributes something new. On the surface, it appears that it does. Nevertheless, if one ponders where the LT relation above comes from, one will see that this method is precisely that of Aryahbata below, but packaged a little differently. If this method referenced Aryahbata's solution and pointed out how it could be formalized, and under what conditions, then this would be genuinely new. But as it doesn't, it appears to be a rehash. $\endgroup$ – Ron Gordon Aug 10 '13 at 0:30
  • $\begingroup$ @MhenniBenghorbal: correctness $\ne$ usefulness. The votes are based on usefulness. But you know this already. BTW the downvote is not mine. But if it were, how would that knowledge help you? $\endgroup$ – Ron Gordon Aug 10 '13 at 0:38
  • $\begingroup$ @RonGordon: I do not know why you people assume it is not useful? It is an approach to solve the problem and I used it before to solve other problems! It is totally correct and useful answer and simplifies the integral in a nice way as it was commented by one of the users beneath my related answer in the provided link. $\endgroup$ – Mhenni Benghorbal Aug 10 '13 at 4:24
  • $\begingroup$ @RonGordon I seems to me that this is just a fleshing out of Night Owl's answer (certainly, they have the same global idea). Perhaps I am mistaken? Also, for what it is worth, the downvote it not mine. $\endgroup$ – user1729 Aug 11 '13 at 15:09

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 $p=0$.


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.



$$I=\int_0^\infty \frac{\sin x} x \, dx.$$

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 $\sin{x}=\Im\left[ \mathrm{e}^{-i x} \right]$:

$$\int_0^\infty \sin{x}\exp(-t x)\, \mathrm{d}x=\Im\int_0^\infty \mathrm{e}^{-(t-i)x}\, \mathrm{d}x=\frac{1}{1+t^2}.$$


$$I=\int_0^\infty \frac{\mathrm{d}t}{1+t^2}.$$

Here you can use again the Schwinger trick. However,


Shuch that

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


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}


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 $-i\pi \, \operatorname{sign}(\xi)$: $$ \int_0^\infty \frac{\sin x}{x} dx = \frac{i}{2} \int_{-\infty}^{\infty} \frac{1}{x} e^{-ix} \, dx = \frac{i}{2} \left. \int_{-\infty}^{\infty} \frac{1}{x} e^{-i\xi x} \, dx \right|_{\xi=1} = \frac{i}{2} \left( -i\pi \, \operatorname{sign}(1) \right) = \frac{\pi}{2} . $$


Another approach is to employ Laplace Transforms.

$$I = \int_{0}^{\infty}\frac{\sin(x)}{x}\, \mathrm dx.$$


$$I(t) = \int_{0}^{\infty}\frac{\sin(xt)}{x} \,\mathrm dx$$

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*}


$$\int_{0}^{\infty} \frac{\sin(x)}{x}\mathrm dx = I(1) = \frac{\pi}{2}$$


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 $\dfrac{\pi}{2}$.


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.


protected by Mathmo123 Jul 27 '16 at 12:37

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