integral of $x^{-\frac{1}{x}}\cos x$ from $1$ to $\infty$ does not converge:
Solution 1:
By integration by parts, one has $$\int_1^{+\infty}x^{-\frac{1}{x}}\cos xdx=x^{-\frac{1}{x}}\sin x\bigg|_1^{+\infty}-\int_1^{+\infty}x^{-\frac{1}{x}}\bigg(-\frac{1}{x^2}+\frac{\ln x}{x^2}\bigg)\sin xdx.$$ Noting that the integral above converges and the first term diverges, one can conclude that the original integral diverges.