Evaluate $\int_{0}^{\pi} x\sin\big(\frac{1}{x}\big)\cos x \,dx$ I wonder if an integral of the form $$\int_{0}^{\pi} x\sin\Bigl(\frac{1}{x}\Bigr)\cos x \,dx$$ which can be further simplified to
$$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,\cos x\, dx=\cos x\biggl(\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,dx\biggr)\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\sin x \bigg(\int \frac{\sin (x^{-1})}{(x^{-1})}dx\bigg)dx$$ 
will have an explicit form. One simplifies the integral
$$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}dx=\int_{0}^{\pi} \int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx\,dt$$
where we know that $\int_{0}^{\infty}e^{-xt}dt=\frac{1}{x}$.
We also know that $$\int_{0}^{\infty}e^{-(xt)}\sin (x)\,dx=\frac{1}{t^2+1}.$$
Any idea on how to compute 
$$\int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx?$$

With regard to the last definite integral, I do not think that we can compute it because of the singularities.
However, what is "amazing" is that the antiderivative does exist (given by a CAS)
$$\int e^{-\frac{t}{x}} \sin \left(\frac{1}{x}\right)\,dx=-\frac{1}{2} i e^{-\frac{t+i}{x}} \left(e^{\frac{t+i}{x}} \left((t-i)
   \text{Ei}\left(-\frac{t-i}{x}\right)-(t+i)
   \text{Ei}\left(-\frac{t+i}{x}\right)\right)+\left(-1+e^{\frac{2 i}{x}}\right)
   x\right)$$
