test for conditional and absolute convergence find value of a parameter $\alpha$ at which integral converges absolutely and at which conditional
$$\int\limits_0^\infty \frac{x + 1}{x ^ {\alpha}}\sin(x)\,dx$$
We can consider 2 cases: area of $0$ and area of $\infty$.
1) in area of $0$ $\frac{x + 1}{x ^ {\alpha}}\sin(x) > 0$ so there is no difference beetween absolute and conditional convergence. $\frac{x + 1}{x ^ {\alpha}}\sin(x)  \sim \frac{1}{x^{\alpha - 1}} $ => $\alpha > 2$. What can i do in area of $\infty$?
 A: To find the values of $\alpha$ for which the integral converges conditionally, we can use Dirichlet's Test.  
Note that the integral of the sine function is bounded over all intervals and $\frac{x+1}{x^\alpha}$ monotonically decreases for $\alpha>1$.  Therefore, from Dirichlet's Test, the integral
$$\int_1^\infty \frac{(x+1)\sin(x)}{x^\alpha}\,dx$$
converges for $\alpha >1$.
To show when the convergence is conditional, and not absolute, we write the integral of absolute value as
$$\begin{align}
\int_\pi^{N\pi} \frac{|\sin(x)|(1+x)}{x^\alpha}\,dx&=\sum_{k=1}^{N-1}\int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)|(1+x)}{x^\alpha}\,dx\\\\
&\ge 2\sum_{k=1}^{N-1} \frac{1+k\pi}{\left((k+1)\pi\right)^\alpha}
\end{align}$$
which by the comparison test diverges for $\alpha \le 2$.  It is straightforward to see that the convergence is absolute for $\alpha >2$.
Therefore, the integral $\int_1^\infty \frac{\sin(x)(1+x)}{x^\alpha}\,dx$ converges absolutely for $\alpha >2$, and conditionally for $1<\alpha\le2$.
Putting this together with the fact that the integral $\int_0^1 \frac{\sin(x)(1+x)}{x^\alpha}\,dx$ converges for $\alpha <2$, we find that the integral $\int_0^\infty \frac{\sin(x)(1+x)}{x^\alpha}\,dx$ converges conditionally for $1<\alpha <2$ and diverges otherwise.
A: Focusing only on the part of the interval away from $x=0$, say $(1,\infty)$:
For large $x$, the integrand is approximately $\sin(x)/x^{\alpha - 1}$. We have the bound
$$\left|\frac{\sin(x)}{x^{\alpha - 1}}\right| \leq \frac{1}{x^{\alpha - 1}}$$
so if $\alpha > 2$ then the tail of the integral converges absolutely.
For $\alpha = 2$ and large $x$, the integrand is approximately $\sin(x)/x$, which is conditionally integrable.
More generally, for any $1 < \alpha \leq 2$ and large $x$, the integrand is approximately
$$\sin(x)\cdot\frac{1}{x^{\alpha - 1}}$$
The factor $1/x^{\alpha - 1}$ is monotonically decreasing, and the factor $\sin(x)$ oscillates periodically in sign, so the integral behaves similarly to an alternating series and converges conditionally. You can fill in the details by breaking the integral into pieces taken over intervals of the form $[n\pi, (n+1)\pi]$.
A: *

*Step 1. If $\alpha < 1$, the integral is trivially divergent;

*Step 2. If $\alpha=1$ the integral is divergent, too, but that is less trivial;

*Step 3. If $\alpha > 1$, since $\frac{x+1}{x^\alpha}$ is decreasing to zero while $\sin(x)$ has a bounded primitive, the integral over $(1,+\infty)$ is conditionally convergent due to Dirichlet's test;

*Step 4. In a right neighbourhood of zero the integrand function behaves like $x^{1-\alpha}$, so we must have $\alpha<2$ to ensure integrability over such a neighbourhood;

*Step 5. So we have that the integral is conditionally convergent for $\color{red}{1<\alpha<2}$;

*Step 6. The absolute convergence still requires $\alpha<2$, but with such a constraint the integral over $(1,+\infty)$ is not absolutely convergent.

