Integral of $\cos\left(\frac1x\right)\, dx$

Question

Is the following integral expression correct (neglecting the constant of integration)?

$$ \int\cos\left(\frac{1}{x}\right)dx = x^2\sin\left(2x\right) $$

When I take the derivative, it returns to the integral, so it's ok, right? For some reason I can't find this integral over the internet, and Wolfram gives a strange answer.

Answer 1

Let us check by differentiating your claimed antiderivative.

Call $F(x) = x^2 \sin(2x)$. Then $F'(x) = 2x\sin(2x) + 2x^2\cos(2x)$. Now we ask, is $F'(x)$ the same as $\cos(1/x)$?

Although there are many ways to understand it, they are not the same. For instance, $\cos(1/x)$ is bounded by $1$ always, while $F'(\pi)$ is much larger.

So they are not the same. In fact, $\displaystyle \int \cos(1/x) dx$ has no elementary antiderivative.

Answer 2

$$\int \cos(1/x)dx = \frac{2x \cos(1/x) - i \cdot \Gamma(0, \frac{i}{x}) + i \cdot \Gamma(0, -\frac{i}{x})}{2}+C$$

$$\frac{d}{dx} x^2\sin(2x)=2x\sin(2x)+2x^2\cos(2x)$$