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        riddles >> putnam exam (pure math) >> Definite Integral
        
(Message started by: Barukh on Jul 24^th, 2004, 10:46pm)

Title: Definite Integral
Post by Barukh on Jul 24^th, 2004, 10:46pm

Evaluate the following integral:

[int]₀^[infty] sin(x)/x dx

Title: Re: Definite Integral
Post by Jack Huizenga on Jul 25^th, 2004, 12:37am

[hide]
The answer: \pi/2. The following solution requires knowledge of basic Fourier series. If you own a LaTeX distribution, you can paste the following solution into a .tex document and compile it; this problem involves way too many symbols for a solution to be easily read otherwise.

Let $0<\delta < \pi$, and define $f_\delta:R\to R$ by
$f_\delta(x)=1$ if $|x|\leq \delta$, $f_\delta(x)=0$ if
$\delta<|x|\leq \pi$, and $f_\delta(x+2\pi)=f_\delta(x)$ for every
$x$. Then $f_\delta$ is periodic with period $2\pi$, and is the
pointwise limit of its Fourier series everywhere $f$ is
continuous, so for $x\neq \pm \delta$ we have
$$f_\delta(x)=\frac{a_0}{2}+\sum_{n=1}^\infty \left(a_n \cos nx+b_n\sin
nx\right),$$ where $$a_k =\frac{1}{\pi}\int_{-\pi}^{\pi}
f_\delta(t) \cos kt \, dt$$ $$b_k=\frac{1}{\pi}\int_{-\pi}^\pi
f_\delta(t)\sin kt \, dt.$$ Since the form of $f_\delta$ is
particularly nice, these integrals are very easy to calculate: for
$k>0$, we have
$$a_k=\frac{1}{\pi}\int_{-\pi}^\pi f_\delta(t) \cos kt \,
dt=\frac{1}{\pi}\int_{-\delta}^\delta \cos kt \,
dt=\left.\frac{\sin{kt]{k\pi}\right|_{-\delta}^\delta=\frac{2\sin{k\delta]{k\pi}$$and$$
b_k= \frac{1}{\pi} \int_{-\pi}^\pi f_\delta(t) \sin kt \,
dt=\frac{1}{\pi}\int_{-\delta}^\delta \sin kt \,dt=0.$$ Also,
$$a_0=\frac{1}{\pi}\int_{-\pi}^\pi f_\delta(t)\, dt=\frac{2\delta}{\pi}.$$ Therefore
$$f_\delta(x)=\frac{\delta}{\pi}+\sum_{n=1}^\infty \frac{2\sin n\delta}{n\pi}
\cos(nx).$$ In particular,
$$1=f_\delta(0)=\frac{\delta}{\pi}+\sum_{n=1}^\infty \frac{2\sin
n\delta}{n\pi},$$ so $$\sum_{n=1}^\infty \frac{\sin
n\delta}{n}=\frac{\pi-\delta}{2}.$$

With this identity, we are ready to calculate the integral at
hand. We note briefly that $\int_0^\infty \frac{\sin x}{x} \, dx$
clearly converges since the integrals over each interval $[n,n+1]$
form an alternating sequence whose terms successively decrease in
absolute value and converge to zero. Then since $\frac{\sin
x}{x}$ is continuous on every interval $[0,N]$, we have that
$$\int_0^\infty \frac{\sin x}{x} \, dx = \lim_{\delta\to 0}
\sum_{n=1}^\infty \frac{\sin n\delta}{n\delta}\cdot
\delta=\lim_{\delta\to 0}\sum_{n=1}^\infty \frac{\sin
n\delta}{n}=\lim_{\delta\to 0}
\frac{\pi-\delta}{2}=\frac{\pi}{2}.$$[/hide]

Title: Re: Definite Integral
Post by TenaliRaman on Jul 25^th, 2004, 12:52am

this is probably an easier way,
L(sinx) = 1/(s²+1)
L(sinx/x) = \int_{s}^{oo} f(s) ds = pi/2 - tan^-1(s)
Use definition and let s tend to 0 and we are done!

Title: Re: Definite Integral
Post by Jack Huizenga on Jul 25^th, 2004, 2:11am

What is L in your solution?

Title: Re: Definite Integral
Post by Barukh on Jul 25^th, 2004, 3:45am

Jack, I converted your solution to PostScript and attached it here, so everybody interested may easily read it. Also: if you register, you will be able to do the same thing.

on 07/25/04 at 02:11:24, Jack Huizenga wrote:

What is L in your solution?

It's Laplace transform (IMHO). And yes, I had this particular solution in mind. Bravo, TenaliRaman! :D

Title: Re: Definite Integral
Post by Sir Col on Jul 25^th, 2004, 8:41am

on 07/25/04 at 00:37:14, Jack Huizenga wrote:

If you own a LaTeX distribution...

And if I don't?! :'(

Title: Re: Definite Integral
Post by Barukh on Jul 25^th, 2004, 8:55am

on 07/25/04 at 08:41:09, Sir Col wrote:

And if I don't?! :'(

...then you can see my previous post. :D

Title: Re: Definite Integral
Post by Sir Col on Jul 25^th, 2004, 3:14pm

I can't view postscript either! :(

Title: Re: Definite Integral
Post by towr on Jul 25^th, 2004, 3:29pm

Surely you can download a viewer somewhere.. (gsview32.exe Should also be linked via the miktex website somewhere, and otherwise if you download miktex you can compile and view it yourself :P )

Title: Re: Definite Integral
Post by Eigenray on Dec 8^th, 2004, 9:56pm

Here's a nice proof just using Fubini's Theorem:
Integrate
f(x,y) = e^-xysin x
over the region 0<x<a, 0<y:
[int]₀^a sin(x)/x dx = [pi]/2 - cos a [int]₀^[infty] e^-ay/(1+y²)dy - sin a [int]₀^[infty] ye^-ay/(1+y²)dy.
It follows that for a[ge]1,
|[int]₀^a sin(x)/x dx - [pi]/2| [le] 2/a.

(Found while skimming the appendix of Durrett's "Probability: Theory and Examples".)