I need to determine about convergence of integral below.I found out that it doesn't converge absolutely but i have problem with determining about conditional convergence.I am not sure how to use Dirichlet test in this case.I would be grateful for any advice.Please don't use taylor expansion or similar techniques.
$$ \int_{0}^{\infty} \frac{\sin{\left(x+\frac{1}{x}\right)}}{x^{a}}\ dx $$
By splitting $\mathbb{R}^+$ as $(0,1]\cup(1,+\infty)$ and performing the substitution $x\mapsto\frac{1}{x}$ on the second "half" we have that
$$ \int_{0}^{+\infty}\frac{\sin\left(x+\frac{1}{x}\right)}{x^a}\,dx=\int_{0}^{1}\left(\frac{\sin\left(x+\frac{1}{x}\right)}{x^{a-1}}+\frac{\sin\left(x+\frac{1}{x}\right)}{x^{1-a}}\right)\,\frac{dx}{x} $$ and the RHS can be written (via $x+\frac{1}{x}\mapsto 2z$) as $$ \int_{1}^{+\infty}\frac{\sin(2z)}{\sqrt{z^2-1}}\left[\left(z+\sqrt{z^2-1}\right)^{a-1}+\left(z+\sqrt{z^2-1}\right)^{1-a}\right]\,dz. \tag{A}$$ For any $a\in(0,2)$ the function $$ \frac{1}{\sqrt{z^2-1}}\left[\left(z+\sqrt{z^2-1}\right)^{a-1}+\left(z+\sqrt{z^2-1}\right)^{1-a}\right] $$ is decreasing to zero from some point on, hence the integral is convergent by Dirichlet's test ($\sin(2z)$ has a bounded primitive). If $a<0$ or $a>2$ the $(A)$-integral is blatantly divergent, so you only have to study the case $a=0$ (which is equivalent to the case $a=2$). It leads to the outcome $-\pi Y_1(2)$, where $Y$ is a Bessel function of the second kind. The substitution $z=\cosh\theta$ immediately allows to recognize the relation between your integral and the Laplace transform of $\sin(2\cosh\theta)$.