How would you integrate $\sin(1/x)$?

Question

I have found out that if i use u-substitution here's what i get :

$$\int ( \sin(u) ) \, dx$$

\begin{align*} u &= 1/x = x^{-1}, \\ du / dx &= -x^{-2}, \\ dx / du &= -x^2, \end{align*}

$$\int ( -\sin (u)x^2) \, du$$

$$= \text{?}$$

as you can see this cannot be equal to $\cos u$ because there is the term in $x$ as well. i know that we should generally divide by the derivative of the function but here the derivative is not a constant. my question is : what to do when the derivative is not a constant?

Is there a reason you're trying to calculate $\int \sin(1/x), dx$, e.g., you were specifically asked to, this particular integral arises while doing something else, you made up a problem for practice...? I ask because this does not have a closed formula. — Andrew D. Hwang, Jun 04 '21 at 23:52
@AndrewD.Hwang oh i was watching a youtube video called 100 integrals — sweet lovely, Jun 04 '21 at 23:54
You can get rid of the $x$ using $x = 1/u$. That said, not all integrals have simple expressions in terms of basic functions. Even very simple seeming integrals may require special functions, and even those may not be enough. — eyeballfrog, Jun 05 '21 at 00:38
@sweetlovely I don't think that video exists from blackpenredpen, so can you give us the link of the video? — Toby Mak, Jun 05 '21 at 01:12
To answer your title - try maclaurin expansion of sin x, then integrate — Gaurang, Jun 05 '21 at 04:58
There actually is a way to integrate like so in closed form. It uses a special function. — Тyma Gaidash, Aug 05 '21 at 13:57

score 1 · Answer 1 · answered Jun 05 '21 at 00:40

1

You can briefly continue by making the substitution $x^2=\frac{1}{u^2}$:

$$\int -\frac{\sin u}{u^2} du = -\int \frac{\sin u}{u^2} du$$

The only problem is that you won't be able to express $\int \frac{\sin u}{u^2} du$ using elementary functions.

answered Jun 05 '21 at 00:40

Kman3

2,479

Тyma Gaidash · Answer 2 · 2021-08-06T18:09:41.850

Here are a few basic ways to integrate. I will use sine and cosine properties like their expansions. The first step uses integration by parts and a series expansion switching it and the integral by integrating term by term:

$$\mathrm{\int sin\left(\frac1x\right)dx=x\, sin\left(\frac1x\right)+\int\frac{cos\left(\frac1x\right)}{x}dx= x\,sin\left(\frac1x\right)+\int\frac1x\sum_{n=0}^\infty \frac{(-1)^n\left(\frac1x\right)^{2n}}{(2n)!}= x\,sin\left(\frac1x\right)+\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\int x^{-2n-1}dx= C+x\,sin\left(\frac1x\right)+\sum_{n=0}^\infty \frac{(-1)^n\left(\frac1x\right)^{2n}}{2n(2n)!}}$$

This matches the main definition given for the Cosine Integral Ci(y). This means we have the special function closed form of:

$$\mathrm{\int sin\left(\frac1x\right)dx=C+ x\,sin\left(\frac1x\right)-Ci\left(\frac1x\right)}$$

There is also a way to use a sine series right away without integration by parts:

$$\mathrm{\int sin\left(\frac1x\right)dx=C+ x\,sin\left(\frac1x\right)-Ci\left(\frac1x\right)=\int\sum_{n=0}^\infty \frac{(-1)^n\left(\frac1x\right)^{2n+1}}{(2n+1)!}dx=C-\sum_{n=0}^\infty \frac{(-1)^n}{2n(2n+1)!x^{2n}}}$$ Here is a beautiful complex plot of the function. Please correct me and give me feedback!

How would you integrate $\sin(1/x)$?

2 Answers2