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

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.

• $u=\sqrt{x}, du = \frac{1}{2\sqrt{x}}dx$
– JEM
Feb 11, 2016 at 3:01
• the OP tried this already. Feb 11, 2016 at 3:01
• The first limit doesn't exist Feb 11, 2016 at 3:02
• This bug (nice catch!) might be worth posting to mathematica.stackexchange.com . At least one SE user here and there, @daniel-lichtblau is a Wolfram employee - there are probably others - and might pass it on to the developers. Feb 11, 2016 at 3:16
• The Mathematica Question is here if you are interested: mathematica.stackexchange.com/questions/106162/… Feb 11, 2016 at 13:32

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.
• Perhaps, mathematica is choosing something like $(2\pi n +\pi/2)^2$ as sequence tending to infinity? (sorry for my previous - incorrect - comment) Feb 11, 2016 at 3:18