Welcome to our community

Be a part of something great, join today!

Prove or disprove convergence

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt\)

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
Jan 17, 2013
1,667
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
Jan 17, 2013
1,667
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-\(\displaystyle \big | \int^{\infty}_{\epsilon} \frac{\log(t)\cos(t) }{t^2}\,dt \big | \leq \int^{\infty}_{\epsilon} \frac{\sqrt{t}}{t^2}\,dt < \infty\)