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- Thread starter oooppp2
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- Mar 5, 2012

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Welcome to MHB, oooppp2!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

So which means do you have available?

To give you a heads-up: the integral diverges.

Can you prove it?

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 CosineI 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

[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...

- Jan 17, 2013

- 1,667

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

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

I expect you need to do a complex contour integral...

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

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

- Feb 13, 2012

- 1,704

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

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

... 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

$\chi$ $\sigma$

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- #7

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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

I'm afraid to have been misundestood... the problem You proposed hasThanks. But I cannot follow your solution, can you provide some more steps?

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

...

Kind regards

$\chi$ $\sigma$

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- #9

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.

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- #10

Use the identity:

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.

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

- Jan 17, 2013

- 1,667

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.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

$\chi$ $\sigma$

- - - Updated - - -

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 .I expect you need to do a complex contour integral...

- Feb 13, 2012

- 1,704

Let's try to solve...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 .

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

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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

The integral...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...

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

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

... 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

$\chi$ $\sigma$

- 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} $

$ \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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- Thread starter
- #15

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?

- Jan 31, 2012

- 253

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- #17

So, I'm correct?

- Jan 31, 2012

- 253

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 $x \to \pm \infty$.

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- #19

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?

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 $x \to \pm \infty$.

- 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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- #21

- Jan 31, 2012

- 253

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

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.

- Thread starter
- #24

And in this case, a is indeed $-\infty$ and b is indeed $\infty$.

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.

- Jan 17, 2013

- 1,667

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

I think you can prove the integral diverges .