Question

Show that ∫¹₀ (sin(1/x))/xⁿ 𝑑𝑥 ; 𝑥 > 0 convergence absolutely, if 𝑛 < 1.

Answer 1

$u=1/x, du=-dx/x^2$

$∫^1_0 (sin(1/x))/x^n 𝑑𝑥=∫^{\infin}_0 (sin(u))u^{n-2} 𝑑u$

since n < 1:

$|(sin(u))u^{n-2}|\le 1/u^2$

since series $\sum 1/u^2$ converge, then $\sum (sin(u))u^{n-2}$ converge absolutely.

So,

$∫^1_0 (sin(1/x))/x^n 𝑑𝑥=∫^{\infin}_0 (sin(u))u^{n-2} 𝑑u$ convergences absolutely, if 𝑛 < 1.