# Convergence of $I=\int_{0}^{+\infty} \frac{1}{\sqrt{t}} \cdot \sin \left(t+\frac{1}{t}\right) \, dt$

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!

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.