Question

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

Answer 1

We want to show that,

$\int_{0}^{1} \frac{(\sin (1 / x))}{x^{n}} d x, x>0 \text { converges }$

Absolutely if $n<1$

Let, $f(x)=\frac{\sin (1 / x)}{x^{n}}, n>0$

0 is the only point of infinite discontinuity and f does net keep the same sign in the interval [0,1].

$\therefore \quad|f(x)| \&=\frac{\left|\sin \left(\frac{1}{x}\right)\right|}{x^{n}}<\frac{1}{x^{n}}$

Also, $\int_{0}^{1} \frac{1}{x^{n}} d x$ converges when $n <1$ .

Thus, $\int_{0}^{1}\left|\frac{\sin (1 / x)}{x^{n}}\right| d x$ converges if and only if$\quad n>0$

Thus, The Integration $\int_{0}^{1}\left|\frac{\sin (1 / x)}{x^{* n}}\right| d x$ converges if and only if $n<1$

Or $\int_{0}^{1} \frac{\sin (1 / x)}{x^{n}} d x$ converges absolutely if $n<1$.