I don't remember how to solve this integral ;_;
I tried $u=\frac{1}{x}$ to format it into a frullani integral, but I ended up having this:
$$\int_{0}^{\infty}\frac{1}{u^2}\left ( \cos(u)+\frac{\sin(u)}{u}-2 \right )du$$
Is there a trig-identity I'm not aware of or is there a better way?
Best Answer
Rewrite the original integrand as
$$I = \int_0^\infty\frac{1}{2}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}-1\right)+\frac{3}{2}\left(\cos\frac{1}{x}-1\right)\:dx \equiv I_1 + I_2$$
The first one has a direct antiderivative
$$I_1 = \frac{1}{2}\left[x^2\sin\frac{1}{x}-x\right]_0^\infty = 0$$
and the second one can be done with your substitution and the half angle identity
$$I_2 = \frac{3}{2}\int_0^\infty \frac{\cos u - 1}{u^2}du = -\frac{3}{2}\int_0^\infty \frac{2\sin^2 \frac{u}{2}}{u^2}du = \boxed{-\frac{3\pi}{4}}$$
which happens to be the final answer.