# How to show int (sinx /x) dx (x-0..infinity) is convergent

• September 6th 2012, 08:44 PM
kyb03316
How to show int (sinx /x) dx (x-0..infinity) is convergent
I'm 16 years old in south korea. In this semester, I learn advanced calculus 2 ( technique of integration ~infinite series.. in stewart calculus)

Today i got the pop calculus test. In test, final problem is

"Determine whether int (sinx /x) dx (x-0..infinity) is convergent or not

if convergent, what is value of it?"

Because of it, i gonna crazy. By wolfram alpha, its value is pi/2 ( of course , convergent).

How to prove it is convergent?! (DO NOT USE COMPLEX, MULTIVARIABLE INTEGRATION)
• September 6th 2012, 08:45 PM
kyb03316
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
x=0..infinity
• September 6th 2012, 09:09 PM
Prove It
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Quote:

Originally Posted by kyb03316
I'm 16 years old in south korea. In this semester, I learn advanced calculus 2 ( technique of integration ~infinite series.. in stewart calculus)

Today i got the pop calculus test. In test, final problem is

"Determine whether int (sinx /x) dx (x-0..infinity) is convergent or not

if convergent, what is value of it?"

Because of it, i gonna crazy. By wolfram alpha, its value is pi/2 ( of course , convergent).

How to prove it is convergent?! (DO NOT USE COMPLEX, MULTIVARIABLE INTEGRATION)

Sine integral - Calculus
• September 6th 2012, 11:08 PM
johnsomeone
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Some quick thoughts:
1. That's hard for a pop calculus test, so don't feel bad.
2. It's improper at x=0 and x->infinity, but actually the function is well behaved at x=0, so it's only infinity you need to worry about.
3. The graph will oscillate between the graphs of y=1/x and y=-1/x, since sin(x) oscillates between -1 and 1.
4. If it converges, it's because the positive and negative parts are roughly canceling out over every 2pi "period" (it's not really periodic of course). If the function were just 1/x, it would diverge, so the simple test of using |sin(x)/x|<=1/x fails - it blows up as x->infinity, as the integral is the natural log.
5. When the chunks are allowed to cancel, over each "period", it starts to "look like" 1/x^2 instead of 1/x, and thus the integral over R should converge.
That's because INT{sin(x)/x over [2npi, 2(n+1)pi]} = INT{sin(x)/x over [(2n)pi, (2n+1)pi]} + INT{sin(x)/x over [(2n+1)pi, (2n+2)pi]}
Now in 2nd integral, substitute x=y+pi (dy=dx), and get
= INT{sin(x)/x over [2npi, (2n+1)pi]} + INT{sin(y+pi)/(y+pi) over [(2n)pi, (2n+1)pi]}
= INT{ ( sin(x)/x + sin(x+pi)/(x+pi) ) over [(2n)pi, (2n+1)pi]} (replace the dummy y with the dummy x and combine the integrals)
= (use sin(x+pi)=-sin(x)) INT{ ( sin(x)/x - sin(x)/(x+pi) ) over [(2n)pi, (2n+1)pi]} = INT{ ( sin(x) * ( pi / (x^2+pi x) ) over [(2n)pi, (2n+1)pi]}
= pi * INT{ sin(x)/( (x+pi/2)^2 - (pi/2)^2 ) over [(2n)pi, (2n+1)pi]} = (via t=x+pi/2) pi * INT{ sin(t-pi/2)/( t^2 - (pi/2)^2 ) over [(2n+1/2)pi, (2n+3/2)pi]}
= pi * INT{ -cos(t)/( t^2 - (pi/2)^2 ) over [(2n+1/2)pi, (2n+3/2)pi]} = (-pi) * INT{ cos(t)/( t^2 - (pi/2)^2 ) over [(2n+1/2)pi, (2n+3/2)pi]}
So you can see that, for t large, it's roughly cos(t)/t^2, whose absolute value is bounded by 1/t^2, which is integrable over all of R (away from 0). Moreover, it's not even as big as that, since removed roughly half the domain when did the integral&algebraic manipulations.
We don't have an integral 0 to infinity anymore, but rather a sum from n = 1 to infinity of those individual integrals. The case n=0 needs to be handled separately, as it produces an improper integral (that actually isn't, but still, need to take care). However, that doesn't impact the question of convergence.
There's one other worry: this is choosing to look at the integral at very special endpoints as the interval goes to infinity, endpoints choosen carefully to maximize the cancellation so that the total is small. That can create a false impression of convergence. [[ Consider a step function that's +1 between on [n,n+1] for n an even pos integer, and -1 on [n,n+1] for n an odd pos integer. The integral on [0,2k] is 0 for k a pos integer, making it look like it converges "just let k go to infinity", but it doesn't. ]] However, that's not the case here, as |sin(x)/x| < 1/x, so the areas of the wobbles are, at their largest, getting smaller and smaller, approaching zero.
This isn't very precise - it's off the top of my head (sorry if I made an algebra mistake anywhere) and I'd have to think about formulating it exactly. But it's a rough argument to justify saying that the integral converges.
• September 7th 2012, 12:09 AM
Vlasev
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
This is known as Dirichlet's Integral and if you look at the page, there are plenty of interesting ways to prove it! As a challenge, prove it in your own way!
• September 7th 2012, 08:58 PM
SworD
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Here's a sketch, fill in depending on how much rigor you want.

Let $a_n = \int_{\pi n}^{\pi(n+1)} \frac{\sin(x)}{x} dx$

Then $\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \sum_{0}^{\infty} a_n$

$\sum_{0}^{\infty} a_n$ converges because it is an alternating series, and $\lim_{n \to \infty} a_n = 0$. So the integral converges too.
• September 7th 2012, 09:48 PM
Vlasev
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Quote:

Originally Posted by SworD
$\sum_{0}^{\infty} a_n$ converges because it is an alternating series, and $\lim_{n \to \infty} a_n = 0$. So the integral converges too.

You need to be somewhat careful. The terms $|a_n|$ need to converge to 0 monotonically.
• September 7th 2012, 10:09 PM
SworD
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
You are right, but in this case indeed they do. I did say it was only a sketch :P

But thanks for that reminder, I didn't immediately realize that this was necessary.
• September 8th 2012, 02:20 AM
johnsomeone
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Quote:

Originally Posted by SworD
Here's a sketch, fill in depending on how much rigor you want.

Let $a_n = \int_{\pi n}^{\pi(n+1)} \frac{\sin(x)}{x} dx$

Then $\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \sum_{0}^{\infty} a_n$

$\sum_{0}^{\infty} a_n$ converges because it is an alternating series, and $\lim_{n \to \infty} a_n = 0$. So the integral converges too.

There's one other subtle point besides monotinicity that needs to be considered (but that also can be handled).

Technically, this equation $\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \sum_{0}^{\infty} a_n$

isn't justified. That's despite it being very intuitive, and, it turns out, being actually correct it seems (though that won't always be the case).

That's because $\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \lim_{R \to \infty} \int_{0}^{R} \frac{\sin(x)}{x} dx$

That limit has to include all positive reals, not stoping just on the "special" ones that are integer multiples of pi. So you need to break the integral to a "generic R" into a partial sum of your series, PLUS an integral from the last integer multiple of pi less than R, to R.

$\int_{0}^{R} \frac{\sin(x)}{x} dx = \sum_{0}^{N_R-1} a_n + \int_{\pi N_R}^{R} \frac{\sin(x)}{x} dx$, where ${N_R}$ is the greatest integer less than $R/{\pi}$.

However, since the absolute value of that additional integral will be small (in fact, less than the next series term in absolute value - and so definitely going to 0), this correction won't change the result, and you can still take the limit, and get that the integral converges because the series does. In other words, this is just something to account for, but it can be accounted for, and so the gist of your argument (a good one!) remains.

As an example of why this matters, consider a similar problem where your approach, without proper care, would lead you to a false conclusion. So here's an INCORRECT argument:

Let $b_n = \int_{2n\pi}^{2n\pi + 2\pi} {sin(x)} dx$

Notice that $b_n = 0$ for all integers n.

Then $\int_{0}^{\infty} sin(x) dx = \sum_{0}^{\infty} b_n$

But $\sum_{0}^{\infty} b_n = \sum_{0}^{\infty} (0)$ which converges (trivially, to 0). So the integral converges too (and in fact, by above has value 0).

But of course, the integral actually doesn't converge.
• September 29th 2012, 05:50 PM
Krizalid
Re: How to show int (sinx /x) dx (x-0..infinity) is convergent
Just integrate by parts to prove the convergence.