Is integral convergent? $\int_1^\infty \frac{\sqrt{x}}{1+x} \sin(2x)dx$ I have a problem with following integral:
$\int_1^\infty \frac{\sqrt{x}}{1+x} \sin(2x)dx$
I was trying to prove convergence (or divergence) of this integral, however without any success.
My best guess at this time is that this integral is not convergent, but I have no proof of that.
 A: Hint
We have
$$\frac{\sqrt x}{1+x}\sin(2x)\sim_\infty \frac{\sin(2x)}{\sqrt x}$$
Now we integrate by parts we get
$$\int_1^\infty \frac{\sin(2x)}{\sqrt x}dx=-\frac12\frac{\cos(2x)}{\sqrt x}\Bigg|_1^\infty-\frac14\int_1^\infty\frac{\cos(2x)}{x^{3/2}}dx$$
and clearly that the last integral is convergent so the given integral is convergent.
A: Hint: Try applying Dirichlet's test. I will leave it to you to break the integrand into two pieces.
A: Hint: With the change of variable $y=\sqrt x$, your integral becomes:
$$\int_{1}^{\infty}\frac{2y^2}{1+y^2}\sin(2y^2)dy=2\int_{1}^{\infty}\sin(2y^2)dy-2\int_{1}^{\infty}\frac{1}{1+y^2}\sin(2y^2)dy\ .$$
The first one is a Fresnel integral (http://en.wikipedia.org/wiki/Fresnel_integral).
A: Hint: 


*

*Split the integral up in small intervals with $X_0 = [1,\pi/2]$, $X_j= [\pi j/2, \pi(j+1)/2]$, $j=1,...,N$. We have $I= \sum_{j=0}^N I_j$ with $I_j= \int_{X_j} dx \sqrt{x} \sin(2 x)/(1+x)$ and $N\to \infty$.

*Find a bound of the absolute value of $I_j$ (this should be easy be observing that $\sqrt{x}/(1+x)$ is monotonously decreasing on $x\in [1,\infty]$. You should find a bound which goes to 0 for $j\to \infty$.

*Observe that the sum $\sum_j I_j$ is alternating and apply Dirichlets criterion.
A: In the same spirit as Sami Ben Romdhane's answer, you could notice that,for large value of $x$, $$\frac{\sqrt{x}}{1+x}=\sum_{n=0}^\infty \frac{(-1)^n}{x^{n+\frac{1}{2}}}$$ and arrive to Sami Ben Romdhane's conclusions.
