Improper integral: $\int_1^\infty\frac{\sin(\sqrt{x})}{\sqrt{x}}dx $.

Question

mathematica is reporting that the improper integral $\int_1^\infty\frac{\sin(\sqrt{x})}{\sqrt{x}}dx $ coverges to $2\cos(1)$. However, when I try to confirm this by actually integrating it using u-substitution, I end up with $-2\lim\limits_{n=1}^\infty\left(\cos n - \cos 1\right)$. I am thinking we cannot determine the first limit here.(oscillation). Any help would be appreciated.

Answer 1

OP, you are correct, $$\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=-2\cos\sqrt{n}+2\cos 1$$ Hence the improper integral is $$\int_1^\infty\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}-2\cos\sqrt{n}+2\cos 1$$ And the latter limit does not exist. Your computer algebra system (mathematica) is giving an incorrect answer. Alpha gives the same incorrect answer, probably because it's got mathematica under the hood.