I need to check conditional/absolute convergence of the integral:
$$f(x) = \int_{1}^{\infty}\cos(x^{2})\,\mathrm dx$$
I tried for a long time and I can't understand what I should do.
I know that $\int_{1}^{\infty}\cos(x^{2})\,\mathrm dx$ converges but I can't find if it's conditional convergence or absolute convergence.
On
An idea:
$$t:=x^2\implies dx=\frac{dt}{2\sqrt t}\implies \int\limits_0^\infty\cos x^2dx=\int\limits_0^{\sqrt\frac{\pi}2}\cos x^2dx+\int\limits_{\sqrt\frac{\pi}2}^\infty\cos x^2dx=$$
$$=\int\limits_0^{\sqrt\frac{\pi}2}\cos x^2dx+\frac12\int\limits_{\pi/2}^\infty \frac{\cos t}{\sqrt t}dt$$
the first integral above poses no problem, and for the second we do the following:
$$\int_{\pi/2}^\infty\frac{\cos t}{\sqrt t}du=\sum_{n=1}^\infty\int\limits_{(2n-1)\pi/2}^{\frac{(2n+1)\pi}2}\frac{\cos t}{\sqrt t}dt=:\sum_{n=1}^\infty a_{n}$$
Please do note that we got an alternating series (why?) , and now we check what happens with the monotony of the absolute value of the general term sequence integrating, again by substitution:
$$a_{n+1}=\int\limits_{\frac{(2n+1)\pi}2}^{\frac{(2n+3)\pi}2}\frac{\cos t}{\sqrt t}dt\;\;,\;\;u=t-\pi\;,\;\;du=dt\implies$$
$$a_{n+1}=\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos(u+\pi)}{\sqrt{u+\pi}}du=-\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt{u+\pi}}du$$
and thus we get that
$$|a_{n+1}|=\left|\;-\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt{u+\pi}}du\;\right|\le\left|\;\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt u}du\;\right|=|a_n|\;\;\;\;\;\text{(why?)}$$
This shows the above is a Leibniz series and thus it converges...
The integral does not converge absolutely. Use the substitution $u = x^2$ to show that it's enough to prove that the following integral diverges to $\infty$: $$ \int_1^\infty \left|\frac{\cos(u)}{2\sqrt{u}}\right| \,du $$
Now follow the same steps as in this answer and the fact that $\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$ diverges.