Does the integral \int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt converge?

[alyafey22]

$$\int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt$$

Can we say the following :

$$\int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt=\int^{\epsilon}_{0} \frac{\log(t) \sin(t) }{t} \, dt+\int^{\infty}_{\epsilon} \frac{\log(t) \sin(t) }{t} \, dt$$

1-$$\int^{\epsilon}_{0} \frac{\log(t) \sin(t) }{t} \, dt \leq \int^{\epsilon}_{0} \log(t) dt <\infty$$

2-$$\int^{\infty}_{\epsilon} \frac{\log(t) \sin(t) }{t} \, dt$$

If that is correct , how to check near infinity ?

 

[alyafey22]

If we use integration by parts we get the following

$$\int^{\infty}_{\epsilon} \frac{\log(t)\sin(t) }{t}\,dt=\int^{\infty}_{\epsilon} \frac{\cos(t) }{t^2}\,dt-\int^{\infty}_{\epsilon} \frac{\log(t)\cos(t) }{t^2}\,dt$$

$$\int^{\infty}_{\epsilon} \frac{\cos(t) }{t^2}\,dt \leq \int^{\infty}_{\epsilon} \frac{1 }{t^2}\,dt < \infty$$

$$\int^{\infty}_{\epsilon} \frac{\log(t)\cos(t) }{t^2}\,dt \leq \int^{\infty}_{\epsilon} \frac{\sqrt{t}}{t^2}\,dt < \infty$$

so the integral converges .

What do you think guys ?

 

[alyafey22]

since $$\frac{\sin(t) }{t} \sim 1$$ near zero

1-$$\int^{\epsilon}_{0} \frac{\log(t) \sin(t) }{t} \, dt \sim \int^{\epsilon}_{0} \log(t) dt <\infty
$$

2-$$\big | \int^{\infty}_{\epsilon} \frac{\cos(t) }{t^2}\,dt \big | \leq \int^{\infty}_{\epsilon} \frac{1 }{t^2}\,dt < \infty$$

3-$$ \big | \int^{\infty}_{\epsilon} \frac{\log(t)\cos(t) }{t^2}\,dt \big | \leq \int^{\infty}_{\epsilon} \frac{\sqrt{t}}{t^2}\,dt < \infty$$