# Integrate (sinx)^2/x

#### oooppp2

##### New member
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
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 \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...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
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
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
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
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

Staff member
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
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
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
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
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

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

##### Well-known member
MHB Math Helper
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
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

##### Well-known member
MHB Math Helper
$\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{0} \frac{\sin^{2} x}{x} \ dx + \lim_{N \to \infty} \int_{0}^{N} \frac{\sin^{2} x}{x} \ dx$ 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
$\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x} \ dx = \lim_{N \to \infty} \int_{-N}^{0} \frac{\sin^{2} x}{x} \ dx + \lim_{N \to \infty} \int_{0}^{N} \frac{\sin^{2} x}{x} \ dx$ 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
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
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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

#### oooppp2

##### New member
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
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 .