how to evaluate $\int_0^1\sin(\frac{1}{x})dx$?

Question

How can I evaluate the integral of $$\int_0^1\sin\left(\frac{1}{x}\right)dx.$$ Maybe it needs the cosine integral to evaluate it, but I cannot understand it very well.

Thanks a lot.

Answer 1

This integral cannot be done in terms of elementary functions, in general. That being said, we can try to get somewhere:
Let $\frac{1}{x} = t$. Then $-\frac{1}{x^2} dx = dt$, which means that $dx = -\frac{1}{t^2} dt$. So $$\int_0^1 \sin\!\left(\frac{1}{x}\!\right)\, dx = \int_\infty^1 -\frac{\sin(t)}{t^2}\, dt = \int_1^\infty \frac{\sin(t)}{t^2}\,dt = \sin(1) + \int_1^\infty \frac{\cos(t)}{t}\, dt$$ by integration by parts. Use the substitution:
$dv = \frac{1}{t^2} dt$, $u = \sin(t)$,
$v = -\frac{1}{t}$, $du = \cos(t) dt$

Using the definition of the cosine integral $$\mathrm{ci}(x) = -\int_x^\infty \frac{\cos(t)}{t}\, dt$$ we get that $$\int_0^1 \sin\!\left(\frac{1}{x}\!\right)\, dx = \sin(1) - \mathrm{ci}(1)$$

And I don't believe this is number going to be elementarily expressible, but it can be approximated. (The approximate value of the integral is around 0.504)

(Thank you to J.M. for his correction of a piece of misused terminology.)