What have i done? $I$ is improper at $0$ and $+\infty$. The function inside is continuous on $(0 ;+\infty)$ and then integrable on any closed interval contained in $(0 ;+\infty).$ $$ I=\int_0^1 \frac{1}{\sqrt{t}} \sin \left(t+\frac{1}{t}\right) \cdot d t+\int_1^{+\infty} \frac{1}{\sqrt{t}} \sin \left(t+\frac{1}{t}\right) \, dt $$ $$ \begin{array}{l} \text { As }\left|\frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \right| \leqslant \frac{1}{\sqrt{t}} \text { and }\left(\int_0^1 \frac{1}{\sqrt{t}} \cdot d t\right) \text { converges } , \\ \text { the comparison test states that } \int_0^1 \frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \,dt \text { is } \\ \text { absolutely convergent. Thus } I_1 = \int_0^1 \frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \, d t \text { converges. } \end{array} $$ $$ \begin{array}{l} I_2=\int_1^{+\infty} \frac{1}{\sqrt{t}} \cdot \sin \left(t+\frac{1}{t} \right) \, d t \\ I_2=\int_1^{+\infty}\left(\frac{1}{\sqrt{t}} \sin (t) \cos \left(\frac{1}{t}\right)+\frac{1}{\sqrt{t}} \sin \left(\frac{1}{t}\right) \cos (t)\right) \,dt \\ \text { The taylor expansion of } \cos \left(\frac{1}{t}\right) \text { and } \sin \left(\frac{1}{t}\right) \\ \text { at first order about } +\infty \text { gives: } \\ \cos \left(\frac{1}{t}\right)=1-\frac{1}{2 \cdot t^{2}} \\ \sin \left(\frac{1}{t}\right)=\frac{1}{t} \\ \text { Thus, } \end{array} $$ $$ I_2=\int_1^{+\infty}\left(\frac{1}{\sqrt{t}} \cdot \sin (t) \cdot\left(1-\frac{1}{2 t^{2}}\right)+\frac{1}{\sqrt{t}} \cos (t) \cdot \frac{1}{t}\right) \cdot d t $$ $$ I_2=\int_1^{+\infty}\left(\frac{\sin (t)}{\sqrt{t}}-\frac{\sin (t)}{2 t^{5 / 2}}+\frac{\cos (t)}{t^{3 / 2}}\right) \cdot d t $$ For the same previous reasons (when studying $ I_1$ convergence. The convergence of $ \int_{1}^{+\infty} \frac{\sin (t)}{t^{1 / 2}} \cdot d t $ can be proved using Abel criterion) ), $ I_2$ converges, as sum of convergent integrals. My questions: Is my method correct? Is there any other nice way to study the convergence of $I$? I stopped the taylor expansion of $\cos(1/x)$ and $\sin(1/x)$ at the first order; but will it always work (stopping it at the first order)?can i have a counterexample? Thanks in advance!
2 Answers
We can go ahead and actually compute the value of this integral by recognizing
$$I = \int_0^\infty\frac{2\,dt}{2\sqrt{t}}\sin\left((\sqrt{t})^2+\frac{1}{(\sqrt{t})^2}\right)$$
which suggests using the substitution $s = \sqrt{t}$
$$I = \int_0^\infty 2\sin\left(s^2+ \frac{1}{s^2}\right) = \int_{-\infty}^\infty \sin\left(\left[s-\frac{1}{s}\right]^2+2\right)$$
Then we can use the following theorem
$$\operatorname{p.v.}\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx = \operatorname{p.v.}\int_{-\infty}^\infty f(x)\,dx$$
to compute the previous integral
$$I = \operatorname{p.v.}\int_{-\infty}^\infty \sin(x^2+2)\:dx = \sin(2)\int_{-\infty}^\infty \cos(x^2)\:dx + \cos(2)\int_{-\infty}^\infty \sin(x^2)\:dx$$
$$I = \sqrt{\frac{\pi}{2}}\left[\sin(2)+\cos(2)\right]$$
using the known value of $\int_{\Bbb{R}}\sin(ax^2)\:dx = \int_{\Bbb{R}}\cos(ax^2)\:dx = \sqrt{\frac{\pi}{2a}}$
We can write your integral $I_2$ as
$$I_2 =\int_1^{\infty} \frac{1}{\sqrt{t}}\left[ \sin \left(t+\frac{1}{t}\right) - \sin t \right]\, dt + \int_1^{\infty} \frac{\sin t}{\sqrt{t}}\,dt.$$
Now $|\sin y-\sin x| \le |y-x|$ by the MVT. Thus the absolute value of the first integrand is $\le t^{-3/2}.$ Hence the first integral converges absolutely. The second integral converges by Dirichlet's theorem. (You wrote "the Abel criterion"; I think we mean the same thing.) It follows that $I_2$ converges.
