alyafey22
Gold Member
MHB
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$$\int^{\infty}_{0} \frac{\log(t) \sin(t) }{t} \, dt$$
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 ?
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 ?