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Integrate (sinx)^2/x

oooppp2

New member
Oct 21, 2013
19
I have tried the substitutions I know og, but I just can't solve it. It would be great if someone helped me on this one.


sin.png
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Re: integrate (sinx)^2/x

I have tried the substitutions I know og, but I just can't solve it. It would be great if someone helped me on this one.


View attachment 1535
Welcome to MHB, oooppp2! :)

So which means do you have available?
To give you a heads-up: the integral diverges.
Can you prove it?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I have tried the substitutions I know og, but I just can't solve it. It would be great if someone helped me on this one.


View attachment 1535
A substitution won't help you. I'd suggest writing [tex]\displaystyle \begin{align*} \sin^2{(x)} = \frac{1}{2} - \frac{1}{2}\cos{(2x)} \end{align*}[/tex], and then applying the MacLaurin Series for Cosine

[tex]\displaystyle \begin{align*} \cos{(t)} &= 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \\ \cos{(2x)} &= 1 - \frac{ (2x)^2}{2} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \dots \\ &= 1 - 2x + \frac{2}{3}x^4 - \frac{4}{45}x^6 + \dots \end{align*}[/tex]

Use this series to get a series for [tex]\displaystyle \begin{align*} \sin^2{(x)} \end{align*}[/tex], divide everything through by x and see what you get when you integrate...
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Interestingly

\(\displaystyle \int^{\infty}_0 \frac{\sin^n(x)}{x}\, dx\)

only converges for odd powers of $n$ , I cannot prove it though.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Interestingly

\(\displaystyle \int^{\infty}_0 \frac{\sin^n(x)}{x}\, dx\)

only converges for odd powers of $n$ , I cannot prove it though.
I expect you need to do a complex contour integral...
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have tried the substitutions I know og, but I just can't solve it. It would be great if someone helped me on this one.


View attachment 1535
In any interval $\displaystyle n\ \pi \le x \le (n+1)\ \pi$ is...

$\displaystyle \frac{\sin^{2} x}{x} \ge \frac{\sin^{2} x}{(n+1)\ \pi}\ (1)$

... and because is...

$\displaystyle \int_{n\ \pi}^{(n+1)\ \pi} \sin^{2} x\ d x = \frac{\pi}{2}\ (2)$

... we have...

$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x}\ d x > \frac{1}{2}\ \sum_{n=0}^{\infty} \frac{1}{n+1}\ (3)$

But the series in (3) diverges so that also the integral in the OP diverges...

Kind regards


$\chi$ $\sigma$
 

oooppp2

New member
Oct 21, 2013
19
In any interval $\displaystyle n\ \pi \le x \le (n+1)\ \pi$ is...

$\displaystyle \frac{\sin^{2} x}{x} \ge \frac{\sin^{2} x}{(n+1)\ \pi}\ (1)$

... and because is...

$\displaystyle \int_{n\ \pi}^{(n+1)\ \pi} \sin^{2} x\ d x = \frac{\pi}{2}\ (2)$

... we have...

$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x}\ d x > \frac{1}{2}\ \sum_{n=0}^{\infty} \frac{1}{n+1}\ (3)$

But the series in (3) diverges so that also the integral in the OP diverges...

Kind regards


$\chi$ $\sigma$
Thanks. But I cannot follow your solution, can you provide some more steps?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Thanks. But I cannot follow your solution, can you provide some more steps?
I'm afraid to have been misundestood... the problem You proposed has no solutions because the integral...

$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x}\ dx$

... diverges...

Kind regards

$\chi$ $\sigma$
 

oooppp2

New member
Oct 21, 2013
19
I want to understand this, but now, the solution is too short for me to understand. (First semester of calculus.)

Did I not study enough, or does it come later, that

$\displaystyle \int_{n\ \pi}^{(n+1)\ \pi} \sin^{2} x\ d x = \frac{\pi}{2}\ (2)$

This was unknown to me.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I want to understand this, but now, the solution is too short for me to understand. (First semester of calculus.)

Did I not study enough, or does it come later, that

$\displaystyle \int_{n\ \pi}^{(n+1)\ \pi} \sin^{2} x\ d x = \frac{\pi}{2}\ (2)$

This was unknown to me.
Use the identity:

\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
In any interval $\displaystyle n\ \pi \le x \le (n+1)\ \pi$ is...

$\displaystyle \frac{\sin^{2} x}{x} \ge \frac{\sin^{2} x}{(n+1)\ \pi}\ (1)$

... and because is...

$\displaystyle \int_{n\ \pi}^{(n+1)\ \pi} \sin^{2} x\ d x = \frac{\pi}{2}\ (2)$

... we have...

$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x}\ d x > \frac{1}{2}\ \sum_{n=0}^{\infty} \frac{1}{n+1}\ (3)$

But the series in (3) diverges so that also the integral in the OP diverges...

Kind regards


$\chi$ $\sigma$
Very nice and classic. I reckon this is a way of transforming the interval [a,infinity) to a bounded interval using the periodicity of the function.

- - - Updated - - -

I expect you need to do a complex contour integral...
The result for \(\displaystyle n=1\) is quite known and it can be done in a number of ways , but haven't tried it for higher powers. I wonder if there is a general formula .
 

chisigma

Well-known member
Feb 13, 2012
1,704
The result for \(\displaystyle n=1\) is quite known and it can be done in a number of ways , but haven't tried it for higher powers. I wonder if there is a general formula .
Let's try to solve...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{3} x}{x}\ d x\ (1)$

We start from the trigonometric identity...

$\displaystyle \sin^{3} x = \frac{3}{4}\ \sin x - \frac{1}{4}\ \sin 3\ x\ (2)$

... and (2) permits us to write...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{3} x}{x}\ d x = \frac{3}{4}\ \int_{0}^{\infty} \frac{\sin x}{x}\ dx - \frac{1}{4}\ \int_{0}^{\infty} \frac{\sin 3 x}{x}\ dx = \frac{\pi}{4}\ (3)$

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Let's try to solve...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{3} x}{x}\ d x\ (1)$

We start from the trigonometric identity...

$\displaystyle \sin^{3} x = \frac{3}{4}\ \sin x - \frac{1}{4}\ \sin 3\ x\ (2)$

... and (2) permits us to write...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{3} x}{x}\ d x = \frac{3}{4}\ \int_{0}^{\infty} \frac{\sin x}{x}\ dx - \frac{1}{4}\ \int_{0}^{\infty} \frac{\sin 3 x}{x}\ dx = \frac{\pi}{4}\ (3)$
The integral...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{5} x}{x}\ dx\ (1)$

... is solved in similar way using the identity...

$\displaystyle \sin^{5} x = \frac{10\ \sin x - 5\ \sin 3 x + \sin 5 x}{16}\ (2)$

... so that is...

$\displaystyle \int_{0}^{\infty} \frac{\sin ^{5} x}{x}\ dx = \frac{3}{16}\ \pi\ (3)$

Clearly it is not difficult to find a general formula for n odd...

Kind regards

$\chi$ $\sigma$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
If $f(x)$ is continuous and $\pi$-periodic on $\mathbb{R}$, then $ \displaystyle \int_{-\infty}^{\infty} f(x) \frac{\sin x}{x} \ dx = \int_{0}^{\pi} f(x) \ dx $.


$ \displaystyle \int_{-\infty}^{\infty} f(x) \frac{\sin x}{x} \ dx = \sum_{k=-\infty}^{\infty} \int^{(k+1) \pi}_{k \pi} f(x) \frac{\sin x}{x} \ dx $

$ \displaystyle = \sum_{k=-\infty}^{\infty} \int^{\pi}_{0} f(u + k \pi) \frac{\sin (u + k \pi)}{u + k \pi} \ du = \sum_{k = -\infty}^{\infty} \int_{0}^{\pi} f(u) (-1)^{k} \frac{\sin u}{u + k \pi} \ du $

$ \displaystyle = \int_{0}^{\pi} f(u) \sin u \sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{u + k \pi} \ du = \int_{0}^{\pi} f(u) \sin u \csc u \ du $

$ \displaystyle = \int_{0}^{\pi} f(u) \ du $



$ \displaystyle \int_{0}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \sin^{2n} (x) \frac{\sin x}{x} \ dx$

$ \displaystyle = \frac{1}{2} \int_{0}^{\pi} \sin^{2n} (x) \ dx = \int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \ dx = \int_{0}^{\frac{\pi}{2}} \sin^{2(n+\frac{1}{2})-1} (x) \cos^{2(\frac{1}{2}) -1} (x) \ dx $

$ \displaystyle = \frac{1}{2} B \Big( n+\frac{1}{2},\frac{1}{2} \Big) = \frac{1}{2} \frac{\Gamma(n+ \frac{1}{2}) \sqrt{\pi}}{n!} = \frac{\pi}{2^{2n}} \frac{(2n-1)!}{n! (n-1)!} \frac{2n}{2n}$

$ \displaystyle = \frac{\pi}{2^{2n+1}} \binom{2n}{n} $
 
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oooppp2

New member
Oct 21, 2013
19
What about this argument:

we know how the graph of sin^2(x) looks, since we can write it in terms of cos(2x).

and sin^2(x) is divided by x, so we can actually understand how the graph looks for any x except zero (put it in you grapher and see).

for the given function, we have f(x)=-f(-x) for all x except 0.

so lets take the limit of f(x) as x -> 0 from both positive and negative direction. We'll se that f approaches. thus, even for x close to 0 the function is defined.

now, since the area from x = - infty to x=0 is equal to the area from x = infty to x=0 we see on the graph that they cancel out, and the answer should be: zero.

Is this reasoning wrong? If so, why?
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
That's called the Cauchy principal value of the integral. It's a common way to assign a value to some divergent integrals by taking a limit in a symmetrical fashion. If an integral converges, it is equal to its Cauchy principal value. This fact can sometimes be very useful.
 

oooppp2

New member
Oct 21, 2013
19
That's called the Cauchy principal value of the integral. It's a common way to assign a value to some divergent integrals by taking a limit in a symmetrical fashion. If an integral converges, it is equal to its Cauchy principal value. This fact can sometimes be very useful.
So, I'm correct?
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
$\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{0} \frac{\sin^{2} x}{x} \ dx + \lim_{N \to \infty} \int_{0}^{N} \frac{\sin^{2} x}{x} \ dx$ does not converge.

But $ \displaystyle \text{PV} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{N} \frac{\sin^{2}x}{x} \ dx = 0 $.


As shown in this thread, the issue is not the behavior of the function near $x=0$. It's the behavior of the function as $x \to \pm \infty$.
 

oooppp2

New member
Oct 21, 2013
19
$\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{0} \frac{\sin^{2} x}{x} \ dx + \lim_{N \to \infty} \int_{0}^{N} \frac{\sin^{2} x}{x} \ dx$ does not converge.

But $ \displaystyle \text{PV} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{N} \frac{\sin^{2}x}{x} \ dx = 0 $.


As shown in this thread, the issue is not the behavior of the function near $x=0$. It's the behavior of the function as $x \to \pm \infty$.
Wouldn't it be very obvious how it behaves when x approaches infty – since for very large x, the denominator would be very large, but the numerator is periodic. So the fraction approaches 0. Correct or wrong?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
The fact that $ \displaystyle \lim_{x \to \infty} \frac{\sin^{2} x}{x} = 0$ does not tell you anything about the convergence of the integral.
 
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oooppp2

New member
Oct 21, 2013
19
So the view that integral from a to b for f(x) is the area under that curve, is not an exact view in this case?
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
Only if $b= \infty$ (or if $a=-\infty$). Otherwise the signed area (a term that some snotty mathematicians despise) is finite.

I said something in my last post (which I deleted) that wasn't true for this integral. I hope you didn't read it and get more confused.
 

Prove It

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MHB Math Helper
Jan 26, 2012
1,403
Only if $b= \infty$ (or if $a=-\infty$). Otherwise the signed area (a term that some snotty mathematicians despise) is finite.

I said something in my last post (which I deleted) that wasn't true for this integral. I hope you didn't read it and get more confused.
I must be a snotty mathematician :p
 

oooppp2

New member
Oct 21, 2013
19
Only if $b= \infty$ (or if $a=-\infty$). Otherwise the signed area (a term that some snotty mathematicians despise) is finite.

I said something in my last post (which I deleted) that wasn't true for this integral. I hope you didn't read it and get more confused.
And in this case, a is indeed $-\infty$ and b is indeed $\infty$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
If the limit of a function near infinity exits that doesn't tell us anything about the convergence of the integral , take the following example

\(\displaystyle \lim_{n \to \infty} \int_1^{n} \frac{1}{x}\, dx \)

I think you can prove the integral diverges .