# Prove or disprove convergence

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt$$

Can we say the following :

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

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

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

If that is correct , how to check near infinity ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
If we use integration by parts we get the following

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

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

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

so the integral converges .

What do you think guys ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
since $$\displaystyle \frac{\sin(t) }{t} \sim 1$$ near zero

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

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

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