Help With Solving Integral of (sinx)^2/x

[vilhelm]

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

 

[I like Serena]

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.https://www.physicsforums.com/attachments/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]

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.https://www.physicsforums.com/attachments/1535

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

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

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

 

[alyafey22]

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]

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]

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.https://www.physicsforums.com/attachments/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$\chi$ $\sigma$

 

[vilhelm]

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]

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

$\chi$ $\sigma$

 

[vilhelm]

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]

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:

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

 

[alyafey22]

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]

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]

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

$\chi$ $\sigma$

 

[polygamma]

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} $

 

[vilhelm]

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 let's 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?

 

[polygamma]

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.

 

[vilhelm]

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?

 

[polygamma]

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

 

[vilhelm]

$\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{0} \frac{\sin^{2} x}{x} \ dx + \lim_{N \to \infty} \int_{0}^{N} \frac{\sin^{2} x}{x} \ dx$ 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?

 

[polygamma]

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.

 

[vilhelm]

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?

 

[polygamma]

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]

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

 

[vilhelm]

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 $\infty$.

 

[alyafey22]

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 .

 

[DreamWeaver]

Interestingly

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

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

This paper here might clear things up -->

http://129.81.170.14/~vhm/papers_html/final16.pdf

 

[alyafey22]

This paper here might clear things up -->

http://129.81.170.14/~vhm/papers_html/final16.pdf

It is a very interesting paper , I did take a look at the first pages in the past but didn't finish it. That is a remainder to read it all , thanks !

 

[DreamWeaver]

No worries, friend... :D