$\int_{0}^{\infty} \frac{(\sin(x) )}{ x} \,\mathrm dx$ and
$\int_{0}^{\infty} \frac{(\sin(x) \arctan(x))}{ x} \,\mathrm dx$
These are convergent. How to prove that??
I using the comparison test.
Case 1. x is in [0,1], then $\ \frac{(\sin(x) )}{ x} \,$ is smaller than 1.
$\int_{0}^{1} \frac{(\sin(x) )}{ x} \,\mathrm dx$ is convergent.
Case 2. x is in [1,∞). $\int_{1}^{b} \frac{(\sin(x) )}{ x} \,\mathrm dx$ = -$\ \frac{(\cos(x) )}{ x} \,$ + cos(1) - $\int_{1}^{b} \frac{(\cos(x) )}{ x^2} \,\mathrm dx$
Since,(if b→∞)..................lim$\ \frac{(\cos(b) )}{ b} \,$ = 0,
$\int_{1}^{\infty} \frac{(\sin(x) )}{ x} \,\mathrm dx$ = cos(1) - $\int_{1}^{\infty} \frac{(\cos(x) )}{ x^2} \,\mathrm dx$
If showing existence f(x) on [0,1] then f(x) is convergent. This sentence is true? And there's another way to prove the second question? someone said to me it is easy by using the Trigonometric Series Test.. This is ok???
You can apply the similar idea for the first one, so I'll try to show the second.
Notice that $f(x)=\frac{\sin(x) \arctan(x)}{x}$ is continuous the origin, which shows that the following integral exists $$\int_{0}^{1}\frac{\sin(x)\arctan(x)}{x}dx$$
For the existence on the interval $[1,\infty)$, an integration by parts yields $$\int_{1}^{\infty}\frac{\sin(x)\arctan(x)}{x}dx=\lim_{R\rightarrow \infty}-\frac{\cos(x)\arctan(x)}{x}|_{1}^{R} +\int_{1}^{R}\frac{\cos(x)}{x(x^{2}+1)}dx-\int_{1}^{R}\frac{\cos(x)\arctan(x)}{x^{2}}dx$$
It is now relatively easy to see the convergence, since the integrands decay fast enough.